diff --git a/HW4/README.md b/HW4/README.md
new file mode 100644
index 0000000..7f04a9a
--- /dev/null
+++ b/HW4/README.md
@@ -0,0 +1,37 @@
+# Homework #3
+## due 3/1/17 by 11:59pm
+
+
+1. Use your repository 'roots_and_optimization'. Document all the HW4 work under the
+heading `# Homework #4` in your `README.md` file
+
+ a. Create a function called 'collar_potential_energy' that computes the total
+ potential energy of a collar connected to a spring and sliding on a rod. As shown in
+ the figure given a position, xc, and angle, theta:
+
+ 
+
+ The spring is unstretched when x_C=0.5 m. The potential energy due to gravity is:
+
+ PE_g=m x_C\*g\*sin(theta)
+
+ where m=0.5 kg, and g is the acceleration due to gravity,
+
+ and the potential energy due to the spring is:
+
+ PE_s=1/2\*K \*(DL)^2
+
+ where DL = 0.5 - sqrt(0.5^2+(0.5-x_C)^2) and K=30 N/m.
+
+ b. Use the `goldmin.m` function to solve for the minimum potential energy at xc when
+ theta=0. *create an anonymous function with `@(x) collar_potential_energy(x,theta)` in
+ the input for goldmin. Be sure to include the script that solves for xc*
+
+ c. Create a for-loop that solves for the minimum potential energy position, xc, at a
+ given angle, theta, for theta = 0..90 degrees.
+
+ d. Include a plot of xc vs theta. `plot(theta,xc)` with
+
+ ``
+
+3. Commit your changes to your repository. Sync your local repository with github.
diff --git a/HW4/collar_mass.png b/HW4/collar_mass.png
new file mode 100644
index 0000000..f9b2836
Binary files /dev/null and b/HW4/collar_mass.png differ
diff --git a/HW4/collar_mass.svg b/HW4/collar_mass.svg
new file mode 100644
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diff --git a/README.md b/README.md
index 73ca14a..c7011f5 100644
--- a/README.md
+++ b/README.md
@@ -79,10 +79,10 @@ general, I will not post homework solutions.
| |2/16|8|Linear Algebra|
|6|2/21|9|Linear systems: Gauss elimination|
| |2/23|10|Linear Systems: LU factorization|
-|7|2/28||Midterm Review|
-| |3/2||Midterm|
-|8|3/7|11|Linear Systems: Error analysis|
-| |3/9|13|Eigenvalues|
+|7|2/28|11|Linear Systems: Error analysis|
+| |3/2|12|Eigenvalues|
+|8|3/7|1-10 |Midterm Review|
+| |3/9|1-10|Midterm|
|9|3/14| N/A |Spring Break!|
| |3/16| N/A |Spring Break!|
|10|3/21|12|Linear Systems: Iterative methods|
diff --git a/lecture_09/mass_springs.png b/lecture_09/mass_springs.png
new file mode 100644
index 0000000..d0dc65c
Binary files /dev/null and b/lecture_09/mass_springs.png differ
diff --git a/lecture_09/octave-workspace b/lecture_09/octave-workspace
new file mode 100644
index 0000000..f7e8b2b
Binary files /dev/null and b/lecture_09/octave-workspace differ
diff --git a/lecture_10/.ipynb_checkpoints/lecture_10-checkpoint.ipynb b/lecture_10/.ipynb_checkpoints/lecture_10-checkpoint.ipynb
new file mode 100644
index 0000000..2fd6442
--- /dev/null
+++ b/lecture_10/.ipynb_checkpoints/lecture_10-checkpoint.ipynb
@@ -0,0 +1,6 @@
+{
+ "cells": [],
+ "metadata": {},
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
diff --git a/lecture_10/GaussNaive.m b/lecture_10/GaussNaive.m
new file mode 100644
index 0000000..d8c60d3
--- /dev/null
+++ b/lecture_10/GaussNaive.m
@@ -0,0 +1,25 @@
+function [x,Aug] = GaussNaive(A,y)
+% GaussNaive: naive Gauss elimination
+% x = GaussNaive(A,b): Gauss elimination without pivoting.
+% input:
+% A = coefficient matrix
+% y = right hand side vector
+% output:
+% x = solution vector
+[m,n] = size(A);
+if m~=n, error('Matrix A must be square'); end
+nb = n+1;
+Aug = [A y];
+% forward elimination
+for k = 1:n-1
+ for i = k+1:n
+ factor = Aug(i,k)/Aug(k,k);
+ Aug(i,k:nb) = Aug(i,k:nb)-factor*Aug(k,k:nb);
+ end
+end
+% back substitution
+x = zeros(n,1);
+x(n) = Aug(n,nb)/Aug(n,n);
+for i = n-1:-1:1
+ x(i) = (Aug(i,nb)-Aug(i,i+1:n)*x(i+1:n))/Aug(i,i);
+end
diff --git a/lecture_10/GaussPivot.m b/lecture_10/GaussPivot.m
new file mode 100644
index 0000000..20df9e8
--- /dev/null
+++ b/lecture_10/GaussPivot.m
@@ -0,0 +1,33 @@
+function [x,Aug,npivots] = GaussPivot(A,b)
+% GaussPivot: Gauss elimination pivoting
+% x = GaussPivot(A,b): Gauss elimination with pivoting.
+% input:
+% A = coefficient matrix
+% b = right hand side vector
+% output:
+% x = solution vector
+[m,n]=size(A);
+if m~=n, error('Matrix A must be square'); end
+nb=n+1;
+Aug=[A b];
+npivots=0; % initially no pivots used
+% forward elimination
+for k = 1:n-1
+ % partial pivoting
+ [big,i]=max(abs(Aug(k:n,k)));
+ ipr=i+k-1;
+ if ipr~=k
+ npivots=npivots+1; % if the max is not the current index ipr, pivot count
+ Aug([k,ipr],:)=Aug([ipr,k],:);
+ end
+ for i = k+1:n
+ factor=Aug(i,k)/Aug(k,k);
+ Aug(i,k:nb)=Aug(i,k:nb)-factor*Aug(k,k:nb);
+ end
+end
+% back substitution
+x=zeros(n,1);
+x(n)=Aug(n,nb)/Aug(n,n);
+for i = n-1:-1:1
+ x(i)=(Aug(i,nb)-Aug(i,i+1:n)*x(i+1:n))/Aug(i,i);
+end
diff --git a/lecture_10/Tridiag.m b/lecture_10/Tridiag.m
new file mode 100644
index 0000000..ac4ec9b
--- /dev/null
+++ b/lecture_10/Tridiag.m
@@ -0,0 +1,22 @@
+function x = Tridiag(e,f,g,r)
+% Tridiag: Tridiagonal equation solver banded system
+% x = Tridiag(e,f,g,r): Tridiagonal system solver.
+% input:
+% e = subdiagonal vector
+% f = diagonal vector
+% g = superdiagonal vector
+% r = right hand side vector
+% output:
+% x = solution vector
+n=length(f);
+% forward elimination
+for k = 2:n
+ factor = e(k)/f(k-1);
+ f(k) = f(k) - factor*g(k-1);
+ r(k) = r(k) - factor*r(k-1);
+end
+% back substitution
+x(n) = r(n)/f(n);
+for k = n-1:-1:1
+ x(k) = (r(k)-g(k)*x(k+1))/f(k);
+end
diff --git a/lecture_10/lecture_10.aux b/lecture_10/lecture_10.aux
new file mode 100644
index 0000000..d8d4c62
--- /dev/null
+++ b/lecture_10/lecture_10.aux
@@ -0,0 +1,40 @@
+\relax
+\providecommand\hyper@newdestlabel[2]{}
+\providecommand\HyperFirstAtBeginDocument{\AtBeginDocument}
+\HyperFirstAtBeginDocument{\ifx\hyper@anchor\@undefined
+\global\let\oldcontentsline\contentsline
+\gdef\contentsline#1#2#3#4{\oldcontentsline{#1}{#2}{#3}}
+\global\let\oldnewlabel\newlabel
+\gdef\newlabel#1#2{\newlabelxx{#1}#2}
+\gdef\newlabelxx#1#2#3#4#5#6{\oldnewlabel{#1}{{#2}{#3}}}
+\AtEndDocument{\ifx\hyper@anchor\@undefined
+\let\contentsline\oldcontentsline
+\let\newlabel\oldnewlabel
+\fi}
+\fi}
+\global\let\hyper@last\relax
+\gdef\HyperFirstAtBeginDocument#1{#1}
+\providecommand\HyField@AuxAddToFields[1]{}
+\providecommand\HyField@AuxAddToCoFields[2]{}
+\providecommand \oddpage@label [2]{}
+\@writefile{toc}{\contentsline {section}{\numberline {1}Gauss Elimination}{1}{section.1}}
+\newlabel{gauss-elimination}{{1}{1}{Gauss Elimination}{section.1}{}}
+\@writefile{toc}{\contentsline {subsubsection}{\numberline {1.0.1}Solving sets of equations with matrix operations}{1}{subsubsection.1.0.1}}
+\newlabel{solving-sets-of-equations-with-matrix-operations}{{1.0.1}{1}{Solving sets of equations with matrix operations}{subsubsection.1.0.1}{}}
+\@writefile{toc}{\contentsline {subsection}{\numberline {1.1}Gauss elimination}{4}{subsection.1.1}}
+\newlabel{gauss-elimination}{{1.1}{4}{Gauss elimination}{subsection.1.1}{}}
+\@writefile{toc}{\contentsline {subsubsection}{\numberline {1.1.1}Solving sets of equations systematically}{4}{subsubsection.1.1.1}}
+\newlabel{solving-sets-of-equations-systematically}{{1.1.1}{4}{Solving sets of equations systematically}{subsubsection.1.1.1}{}}
+\@writefile{lof}{\contentsline {figure}{\numberline {1}{\ignorespaces Springs-masses\relax }}{4}{figure.caption.1}}
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diff --git a/lecture_10/lecture_10.ipynb b/lecture_10/lecture_10.ipynb
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+ {
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+ "execution_count": 1,
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+ "outputs": [],
+ "source": [
+ "%plot --format svg"
+ ]
+ },
+ {
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+ "metadata": {
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+ },
+ "outputs": [],
+ "source": [
+ "setdefaults"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Gauss Elimination\n",
+ "### Solving sets of equations with matrix operations\n",
+ "\n",
+ "The number of dimensions of a matrix indicate the degrees of freedom of the system you are solving. \n",
+ "\n",
+ "If you have a set of known output, $y_{1},~y_{2},~...y_{N}$ and a set of equations that\n",
+ "relate unknown inputs, $x_{1},~x_{2},~...x_{N}$, then these can be written in a vector\n",
+ "matrix format as:\n",
+ "\n",
+ "$y=Ax$\n",
+ "\n",
+ "Consider a problem with 2 DOF:\n",
+ "\n",
+ "$x_{1}+3x_{2}=1$\n",
+ "\n",
+ "$2x_{1}+x_{2}=1$\n",
+ "\n",
+ "$\\left[ \\begin{array}{cc}\n",
+ "1 & 3 \\\\\n",
+ "2 & 1 \\end{array} \\right]\n",
+ "\\left[\\begin{array}{c} \n",
+ "x_{1} \\\\ \n",
+ "x_{2} \\end{array}\\right]=\n",
+ "\\left[\\begin{array}{c} \n",
+ "1 \\\\\n",
+ "1\\end{array}\\right]$\n",
+ "\n",
+ "The solution for $x_{1}$ and $x_{2}$ is the intersection of two lines:"
+ ]
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+ "source": [
+ "x21=[-2:2];\n",
+ "x11=1-3*x21;\n",
+ "x21=[-2:2];\n",
+ "x22=1-2*x21;\n",
+ "plot(x11,x21,x21,x22)"
+ ]
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+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
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+ "ans =\n",
+ "\n",
+ " 0.40000\n",
+ " 0.20000\n",
+ "\n"
+ ]
+ }
+ ],
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+ "A=[1,3;2,1]; y=[1;1];\n",
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+ ]
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+ "metadata": {},
+ "source": [
+ "For a $3\\times3$ matrix, the solution is the intersection of the 3 planes.\n",
+ "\n",
+ "$10x_{1}+2x_{2}+x_{3}=1$\n",
+ "\n",
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+ "x_{2} \\\\\n",
+ "x_{3} \\end{array}\\right]=\n",
+ "\\left[\\begin{array}{c} \n",
+ "1 \\\\\n",
+ "1 \\\\\n",
+ "1\\end{array}\\right]$"
+ ]
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+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "\t<path d=\"M397.9,352.6 L397.9,281.6 \" stroke=\"black\"/>\t<g fill=\"rgb(0,0,0)\" font-family=\"{}\" font-size=\"16.00\" stroke=\"none\" text-anchor=\"middle\" transform=\"translate(239.0,389.8)\">\n",
+ "\t\t<text><tspan font-family=\"{}\">x1</tspan></text>\n",
+ "\t</g>\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "\t<g fill=\"rgb(0,0,0)\" font-family=\"{}\" font-size=\"16.00\" stroke=\"none\" text-anchor=\"middle\" transform=\"translate(507.1,296.6)\">\n",
+ "\t\t<text><tspan font-family=\"{}\">x2</tspan></text>\n",
+ "\t</g>\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "\t<g fill=\"rgb(0,0,0)\" font-family=\"{}\" font-size=\"16.00\" stroke=\"none\" text-anchor=\"middle\" transform=\"translate(59.6,257.9) rotate(-90)\">\n",
+ "\t\t<text><tspan font-family=\"{}\">x3</tspan></text>\n",
+ "\t</g>\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "</g>\n",
+ "</g>\n",
+ "</svg>"
+ ],
+ "text/plain": [
+ "<IPython.core.display.SVG object>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "N=25;\n",
+ "x11=linspace(-2,2,N);\n",
+ "x12=linspace(-2,2,N);\n",
+ "[X11,X12]=meshgrid(x11,x12);\n",
+ "X13=1-10*X11-2*X12;\n",
+ "\n",
+ "x21=linspace(-2,2,N);\n",
+ "x22=linspace(-2,2,N);\n",
+ "[X21,X22]=meshgrid(x21,x22);\n",
+ "X23=1-2*X11-X22;\n",
+ "\n",
+ "x31=linspace(-2,2,N);\n",
+ "x32=linspace(-2,2,N);\n",
+ "[X31,X32]=meshgrid(x31,x32);\n",
+ "X33=1/10*(1-X31-2*X32);\n",
+ "\n",
+ "mesh(X11,X12,X13);\n",
+ "hold on;\n",
+ "mesh(X21,X22,X23)\n",
+ "mesh(X31,X32,X33)\n",
+ "x=[10,2, 1;2,1, 1; 1, 2, 10]\\[1;1;1];\n",
+ "plot3(x(1),x(2),x(3),'o')\n",
+ "xlabel('x1')\n",
+ "ylabel('x2')\n",
+ "zlabel('x3')\n",
+ "view(10,45)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "After 3 DOF problems, the solutions are described as *hyperplane* intersections. Which are even harder to visualize"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Gauss elimination\n",
+ "### Solving sets of equations systematically\n",
+ "\n",
+ "$\\left[ \\begin{array}{ccc|c}\n",
+ " & A & & y \\\\\n",
+ "10 & 2 & 1 & 1\\\\\n",
+ "2 & 1 & 1 & 1 \\\\\n",
+ "1 & 2 & 10 & 1\\end{array} \n",
+ "\\right] $\n",
+ "\n",
+ "Ay(2,:)-Ay(1,:)/5 = ([2 1 1 1]-1/5[10 2 1 1])\n",
+ "\n",
+ "$\\left[ \\begin{array}{ccc|c}\n",
+ " & A & & y \\\\\n",
+ "10 & 2 & 1 & 1\\\\\n",
+ "0 & 3/5 & 4/5 & 4/5 \\\\\n",
+ "1 & 2 & 10 & 1\\end{array} \n",
+ "\\right] $\n",
+ "\n",
+ "Ay(3,:)-Ay(1,:)/10 = ([1 2 10 1]-1/10[10 2 1 1])\n",
+ "\n",
+ "$\\left[ \\begin{array}{ccc|c}\n",
+ " & A & & y \\\\\n",
+ "10 & 2 & 1 & 1\\\\\n",
+ "0 & 3/5 & 4/5 & 4/5 \\\\\n",
+ "0 & 1.8 & 9.9 & 0.9\\end{array} \n",
+ "\\right] $\n",
+ "\n",
+ "Ay(3,:)-1.8\\*5/3\\*Ay(2,:) = ([0 1.8 9.9 0.9]-3\\*[0 3/5 4/5 4/5])\n",
+ "\n",
+ "$\\left[ \\begin{array}{ccc|c}\n",
+ " & A & & y \\\\\n",
+ "10 & 2 & 1 & 1\\\\\n",
+ "0 & 3/5 & 4/5 & 4/5 \\\\\n",
+ "0 & 0 & 7.5 & -1.5\\end{array} \n",
+ "\\right] $\n",
+ "\n",
+ "now, $7.5x_{3}=-1.5$ so $x_{3}=-\\frac{1}{5}$\n",
+ "\n",
+ "then, $3/5x_{2}+4/5(-1/5)=1$ so $x_{2}=\\frac{8}{5}$\n",
+ "\n",
+ "finally, $10x_{1}+2(8/5)+1(-\\frac{1}{5})=1$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "Consider the problem again from the intro to Linear Algebra, 4 masses are connected in series to 4 springs with K=10 N/m. What are the final positions of the masses? \n",
+ "\n",
+ "\n",
+ "\n",
+ "The masses haves the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass:\n",
+ "\n",
+ "$m_{1}g+k(x_{2}-x_{1})-kx_{1}=0$\n",
+ "\n",
+ "$m_{2}g+k(x_{3}-x_{2})-k(x_{2}-x_{1})=0$\n",
+ "\n",
+ "$m_{3}g+k(x_{4}-x_{3})-k(x_{3}-x_{2})=0$\n",
+ "\n",
+ "$m_{4}g-k(x_{4}-x_{3})=0$\n",
+ "\n",
+ "in matrix form:\n",
+ "\n",
+ "$\\left[ \\begin{array}{cccc}\n",
+ "2k & -k & 0 & 0 \\\\\n",
+ "-k & 2k & -k & 0 \\\\\n",
+ "0 & -k & 2k & -k \\\\\n",
+ "0 & 0 & -k & k \\end{array} \\right]\n",
+ "\\left[ \\begin{array}{c}\n",
+ "x_{1} \\\\\n",
+ "x_{2} \\\\\n",
+ "x_{3} \\\\\n",
+ "x_{4} \\end{array} \\right]=\n",
+ "\\left[ \\begin{array}{c}\n",
+ "m_{1}g \\\\\n",
+ "m_{2}g \\\\\n",
+ "m_{3}g \\\\\n",
+ "m_{4}g \\end{array} \\right]$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "K =\n",
+ "\n",
+ " 20 -10 0 0\n",
+ " -10 20 -10 0\n",
+ " 0 -10 20 -10\n",
+ " 0 0 -10 10\n",
+ "\n",
+ "y =\n",
+ "\n",
+ " 9.8100\n",
+ " 19.6200\n",
+ " 29.4300\n",
+ " 39.2400\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "k=10; % N/m\n",
+ "m1=1; % kg\n",
+ "m2=2;\n",
+ "m3=3;\n",
+ "m4=4;\n",
+ "g=9.81; % m/s^2\n",
+ "K=[2*k -k 0 0; -k 2*k -k 0; 0 -k 2*k -k; 0 0 -k k]\n",
+ "y=[m1*g;m2*g;m3*g;m4*g]"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "K1 =\n",
+ "\n",
+ " 20.00000 -10.00000 0.00000 0.00000 9.81000\n",
+ " 0.00000 15.00000 -10.00000 0.00000 24.52500\n",
+ " 0.00000 -10.00000 20.00000 -10.00000 29.43000\n",
+ " 0.00000 0.00000 -10.00000 10.00000 39.24000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "K1=[K y];\n",
+ "K1(2,:)=K1(1,:)/2+K1(2,:)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "K2 =\n",
+ "\n",
+ " 20.00000 -10.00000 0.00000 0.00000 9.81000\n",
+ " 0.00000 15.00000 -10.00000 0.00000 24.52500\n",
+ " 0.00000 0.00000 13.33333 -10.00000 45.78000\n",
+ " 0.00000 0.00000 -10.00000 10.00000 39.24000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "K2=K1;\n",
+ "K2(3,:)=K1(2,:)*2/3+K1(3,:)\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "K2 =\n",
+ "\n",
+ " 20.00000 -10.00000 0.00000 0.00000 9.81000\n",
+ " 0.00000 15.00000 -10.00000 0.00000 24.52500\n",
+ " 0.00000 0.00000 13.33333 -10.00000 45.78000\n",
+ " 0.00000 0.00000 0.00000 2.50000 73.57500\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "K2(4,:)=-K2(3,:)*K2(4,3)/K2(3,3)+K2(4,:)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "x4 = 29.430\n",
+ "x3 = 25.506\n",
+ "x2 = 18.639\n",
+ "x1 = 9.8100\n"
+ ]
+ }
+ ],
+ "source": [
+ "yp=K2(:,5);\n",
+ "x4=yp(4)/K2(4,4)\n",
+ "x3=(yp(3)+10*x4)/K2(3,3)\n",
+ "x2=(yp(2)+10*x3)/K2(2,2)\n",
+ "x1=(yp(1)+10*x2)/K2(1,1)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans =\n",
+ "\n",
+ " 9.8100\n",
+ " 18.6390\n",
+ " 25.5060\n",
+ " 29.4300\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "K\\y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Automate Gauss Elimination\n",
+ "\n",
+ "We can automate Gauss elimination with a function whose input is A and y:\n",
+ "\n",
+ "`x=GaussNaive(A,y)`"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "x =\n",
+ "\n",
+ " 9.8100\n",
+ " 18.6390\n",
+ " 25.5060\n",
+ " 29.4300\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "x=GaussNaive(K,y)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Problem (Diagonal element is zero)\n",
+ "\n",
+ "If a diagonal element is 0 or very small either:\n",
+ "\n",
+ "1. no solution found\n",
+ "2. errors are introduced \n",
+ "\n",
+ "Therefore, we would want to pivot before applying Gauss elimination\n",
+ "\n",
+ "Consider:\n",
+ "\n",
+ "(a) $\\left[ \\begin{array}{cccc}\n",
+ "0 & 2 & 3 \\\\\n",
+ "4 & 6 & 7 \\\\\n",
+ "2 & -3 & 6 \\end{array} \\right]\n",
+ "\\left[ \\begin{array}{c}\n",
+ "x_{1} \\\\\n",
+ "x_{2} \\\\\n",
+ "x_{3} \\end{array} \\right]=\n",
+ "\\left[ \\begin{array}{c}\n",
+ "8 \\\\\n",
+ "-3 \\\\\n",
+ "5\\end{array} \\right]$\n",
+ "\n",
+ "(b) $\\left[ \\begin{array}{cccc}\n",
+ "0.0003 & 3.0000 \\\\\n",
+ "1.0000 & 1.0000 \\end{array} \\right]\n",
+ "\\left[ \\begin{array}{c}\n",
+ "x_{1} \\\\\n",
+ "x_{2} \\end{array} \\right]=\n",
+ "\\left[ \\begin{array}{c}\n",
+ "2.0001 \\\\\n",
+ "1.0000 \\end{array} \\right]$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "warning: division by zero\n",
+ "warning: called from\n",
+ " GaussNaive at line 16 column 12\n",
+ "warning: division by zero\n",
+ "warning: division by zero\n",
+ "ans =\n",
+ "\n",
+ " NaN\n",
+ " NaN\n",
+ " NaN\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " -5.423913\n",
+ " 0.021739\n",
+ " 2.652174\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "format short\n",
+ "Aa=[0,2,3;4,6,7;2,-3,6]; ya=[8;-3;5];\n",
+ "GaussNaive(Aa,ya)\n",
+ "Aa\\ya"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 32,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "x =\n",
+ "\n",
+ " -5.423913\n",
+ " 0.021739\n",
+ " 2.652174\n",
+ "\n",
+ "Aug =\n",
+ "\n",
+ " 4.00000 6.00000 7.00000 -3.00000\n",
+ " 0.00000 -6.00000 2.50000 6.50000\n",
+ " 0.00000 0.00000 3.83333 10.16667\n",
+ "\n",
+ "npivots = 2\n"
+ ]
+ }
+ ],
+ "source": [
+ "[x,Aug,npivots]=GaussPivot(Aa,ya)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 33,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans =\n",
+ "\n",
+ " 0.325665420556713\n",
+ " 0.666666666666667\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " 0.333333333333333\n",
+ " 0.666666666666667\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "format long\n",
+ "Ab=[0.3E-13,3.0000;1.0000,1.0000];yb=[2+0.1e-13;1.0000];\n",
+ "GaussNaive(Ab,yb)\n",
+ "Ab\\yb"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 34,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "x =\n",
+ "\n",
+ " 0.333333333333333\n",
+ " 0.666666666666667\n",
+ "\n",
+ "Aug =\n",
+ "\n",
+ " 1.000000000000000 1.000000000000000 1.000000000000000\n",
+ " 0.000000000000000 2.999999999999970 1.999999999999980\n",
+ "\n",
+ "npivots = 1\n",
+ "ans =\n",
+ "\n",
+ " 0.333333333333333\n",
+ " 0.666666666666667\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "[x,Aug,npivots]=GaussPivot(Ab,yb)\n",
+ "Ab\\yb\n",
+ "format short"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 36,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans = -3.0000\n",
+ "ans = 3.0000\n"
+ ]
+ }
+ ],
+ "source": [
+ "% determinant is (-1)^(number_of_pivots)*diagonal_elements\n",
+ "det(Ab)\n",
+ "Aug(1,1)*Aug(2,2)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "### Spring-Mass System again\n",
+ "Now, 4 masses are connected in series to 4 springs with $K_{1}$=10 N/m, $K_{2}$=5 N/m, \n",
+ "$K_{3}$=2 N/m \n",
+ "and $K_{4}$=1 N/m. What are the final positions of the masses? \n",
+ "\n",
+ "\n",
+ "\n",
+ "The masses have the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass:\n",
+ "\n",
+ "$m_{1}g+k_{2}(x_{2}-x_{1})-k_{1}x_{1}=0$\n",
+ "\n",
+ "$m_{2}g+k_{3}(x_{3}-x_{2})-k_{2}(x_{2}-x_{1})=0$\n",
+ "\n",
+ "$m_{3}g+k_{4}(x_{4}-x_{3})-k_{3}(x_{3}-x_{2})=0$\n",
+ "\n",
+ "$m_{4}g-k_{4}(x_{4}-x_{3})=0$\n",
+ "\n",
+ "in matrix form:\n",
+ "\n",
+ "$\\left[ \\begin{array}{cccc}\n",
+ "k_1+k_2 & -k_2 & 0 & 0 \\\\\n",
+ "-k_2 & k_2+k_3 & -k_3 & 0 \\\\\n",
+ "0 & -k_3 & k_3+k_4 & -k_4 \\\\\n",
+ "0 & 0 & -k_4 & k_4 \\end{array} \\right]\n",
+ "\\left[ \\begin{array}{c}\n",
+ "x_{1} \\\\\n",
+ "x_{2} \\\\\n",
+ "x_{3} \\\\\n",
+ "x_{4} \\end{array} \\right]=\n",
+ "\\left[ \\begin{array}{c}\n",
+ "m_{1}g \\\\\n",
+ "m_{2}g \\\\\n",
+ "m_{3}g \\\\\n",
+ "m_{4}g \\end{array} \\right]$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 24,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "K =\n",
+ "\n",
+ " 15 -5 0 0\n",
+ " -5 7 -2 0\n",
+ " 0 -2 3 -1\n",
+ " 0 0 -1 1\n",
+ "\n",
+ "y =\n",
+ "\n",
+ " 9.8100\n",
+ " 19.6200\n",
+ " 29.4300\n",
+ " 39.2400\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "k1=10; k2=5;k3=2;k4=1; % N/m\n",
+ "m1=1; % kg\n",
+ "m2=2;\n",
+ "m3=3;\n",
+ "m4=4;\n",
+ "g=9.81; % m/s^2\n",
+ "K=[k1+k2 -k2 0 0; -k2, k2+k3, -k3 0; 0 -k3, k3+k4, -k4; 0 0 -k4 k4]\n",
+ "y=[m1*g;m2*g;m3*g;m4*g]"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Tridiagonal matrix\n",
+ "\n",
+ "This matrix, K, could be rewritten as 3 vectors e, f and g\n",
+ "\n",
+ "$e=\\left[ \\begin{array}{c}\n",
+ "0 \\\\\n",
+ "-5 \\\\\n",
+ "-2 \\\\\n",
+ "-1 \\end{array} \\right]$\n",
+ "\n",
+ "$f=\\left[ \\begin{array}{c}\n",
+ "15 \\\\\n",
+ "7 \\\\\n",
+ "3 \\\\\n",
+ "1 \\end{array} \\right]$\n",
+ "\n",
+ "$g=\\left[ \\begin{array}{c}\n",
+ "-5 \\\\\n",
+ "-2 \\\\\n",
+ "-1 \\\\\n",
+ "0 \\end{array} \\right]$\n",
+ "\n",
+ "Where all other components are 0 and the length of the vectors are n and the first component of e and the last component of g are zero\n",
+ "\n",
+ "`e(1)=0` \n",
+ "\n",
+ "`g(end)=0`\n",
+ "\n",
+ "No need to pivot and number of calculations reduced enormously.\n",
+ "\n",
+ "|method |Number of Floating point operations for n$\\times$n-matrix|\n",
+ "|----------------|---------|\n",
+ "| Gauss | n-cubed |\n",
+ "| Tridiagonal | n |"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans =\n",
+ "\n",
+ " 9.8100 27.4680 61.8030 101.0430\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " 9.8100\n",
+ " 27.4680\n",
+ " 61.8030\n",
+ " 101.0430\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "e=[0;-5;-2;-1];\n",
+ "g=[-5;-2;-1;0];\n",
+ "f=[15;7;3;1];\n",
+ "Tridiag(e,f,g,y)\n",
+ "K\\y\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "% tic ... t=toc \n",
+ "% is Matlab timer used for debugging programs\n",
+ "t_GE = zeros(1,100);\n",
+ "t_GE_tridiag = zeros(1,100);\n",
+ "t_TD = zeros(1,100);\n",
+ "%for n = 1:200\n",
+ "for n=1:100\n",
+ " A = rand(n,n);\n",
+ " e = rand(n,1); e(1)=0;\n",
+ " f = rand(n,1);\n",
+ " g = rand(n,1); g(end)=0;\n",
+ " Atd=diag(f, 0) - diag(e(2:n), -1) - diag(g(1:n-1), 1);\n",
+ " b = rand(n,1);\n",
+ " tic;\n",
+ " x = GaussPivot(A,b);\n",
+ " t_GE(n) = toc;\n",
+ " tic;\n",
+ " x = A\\b;\n",
+ " t_GE_tridiag(n) = toc;\n",
+ " tic;\n",
+ " x = Tridiag(e,f,g,b);\n",
+ " t_TD(n) = toc;\n",
+ "end"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 28,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
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+ "\t</g>\n",
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+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"black\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "</g>\n",
+ "</g>\n",
+ "</svg>"
+ ],
+ "text/plain": [
+ "<IPython.core.display.SVG object>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "n=1:100;\n",
+ "loglog(n,t_GE,n,t_TD,n,t_GE_tridiag)\n",
+ "legend('Gauss elim','Matlab \\','TriDiag')\n",
+ "xlabel('number of elements')\n",
+ "ylabel('time (s)')"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "x =\n",
+ "\n",
+ " 9.8100\n",
+ " 27.4680\n",
+ " 61.8030\n",
+ " 101.0430\n",
+ "\n",
+ "Aug =\n",
+ "\n",
+ " 15.00000 -5.00000 0.00000 0.00000 9.81000\n",
+ " 0.00000 5.33333 -2.00000 0.00000 22.89000\n",
+ " 0.00000 0.00000 2.25000 -1.00000 38.01375\n",
+ " 0.00000 0.00000 0.00000 0.55556 56.13500\n",
+ "\n",
+ "npivots = 0\n"
+ ]
+ }
+ ],
+ "source": [
+ "[x,Aug,npivots]=GaussPivot(K,y)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 30,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A =\n",
+ "\n",
+ " 15.00000 -5.00000 0.00000 0.00000\n",
+ " 0.00000 5.33333 -2.00000 0.00000\n",
+ " 0.00000 0.00000 2.25000 -1.00000\n",
+ " 0.00000 0.00000 0.00000 0.55556\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "A=Aug(1:4,1:4)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 31,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans = 100.00\n",
+ "detA = 100.00\n"
+ ]
+ }
+ ],
+ "source": [
+ "det(A)\n",
+ "detA=A(1,1)*A(2,2)*A(3,3)*A(4,4)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Octave",
+ "language": "octave",
+ "name": "octave"
+ },
+ "language_info": {
+ "file_extension": ".m",
+ "help_links": [
+ {
+ "text": "MetaKernel Magics",
+ "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+ }
+ ],
+ "mimetype": "text/x-octave",
+ "name": "octave",
+ "version": "0.19.14"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
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+
+! LaTeX Error: Unknown graphics extension: .svg.
+
+See the LaTeX manual or LaTeX Companion for explanation.
+Type H <return> for immediate help.
+ ...
+
+l.735 ...egraphics{../lecture_09/mass_springs.svg}
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diff --git a/lecture_10/lecture_10.md b/lecture_10/lecture_10.md
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+
+
+```octave
+%plot --format svg
+```
+
+
+```octave
+setdefaults
+```
+
+# Gauss Elimination
+### Solving sets of equations with matrix operations
+
+The number of dimensions of a matrix indicate the degrees of freedom of the system you are solving.
+
+If you have a set of known output, $y_{1},~y_{2},~...y_{N}$ and a set of equations that
+relate unknown inputs, $x_{1},~x_{2},~...x_{N}$, then these can be written in a vector
+matrix format as:
+
+$y=Ax$
+
+Consider a problem with 2 DOF:
+
+$x_{1}+3x_{2}=1$
+
+$2x_{1}+x_{2}=1$
+
+$\left[ \begin{array}{cc}
+1 & 3 \\
+2 & 1 \end{array} \right]
+\left[\begin{array}{c}
+x_{1} \\
+x_{2} \end{array}\right]=
+\left[\begin{array}{c}
+1 \\
+1\end{array}\right]$
+
+The solution for $x_{1}$ and $x_{2}$ is the intersection of two lines:
+
+
+```octave
+x21=[-2:2];
+x11=1-3*x21;
+x21=[-2:2];
+x22=1-2*x21;
+plot(x11,x21,x21,x22)
+```
+
+
+
+
+
+For a 3$\times$3 matrix, the solution is the intersection of the 3 planes.
+
+$10x_{1}+2x_{2}+x_{3}=1$
+
+$2x_{1}+x_{2}+x_{3}=1$
+
+$x_{1}+2x_{2}+10x_{3}=1$
+
+$\left[ \begin{array}{cc}
+10 & 2 & 1\\
+2 & 1 & 1 \\
+1 & 2 & 10\end{array} \right]
+\left[\begin{array}{c}
+x_{1} \\
+x_{2} \\
+x_{3} \end{array}\right]=
+\left[\begin{array}{c}
+1 \\
+1 \\
+1\end{array}\right]$
+
+
+```octave
+x11=linspace(-2,2,5);
+x12=linspace(-2,2,5);
+[X11,X12]=meshgrid(x11,x12);
+X13=1-10*X11-2*X22;
+
+x21=linspace(-2,2,5);
+x22=linspace(-2,2,5);
+[X21,X22]=meshgrid(x21,x22);
+X23=1-2*X11-X22;
+
+x31=linspace(-2,2,5);
+x32=linspace(-2,2,5);
+[X31,X32]=meshgrid(x31,x32);
+X33=1/10*(1-X31-2*X32);
+
+mesh(X11,X12,X13);
+hold on;
+mesh(X21,X22,X23)
+mesh(X31,X32,X33)
+x=[10,2, 1;2,1, 1; 1, 2, 10]\[1;1;1];
+plot3(x(1),x(2),x(3),'o')
+view(45,45)
+```
+
+
+
+
+
+After 3 DOF problems, the solutions are described as *hyperplane* intersections. Which are even harder to visualize
+
+## Gauss elimination
+### Solving sets of equations systematically
+
+$\left[ \begin{array}{ccc|c}
+ & A & & y \\
+10 & 2 & 1 & 1\\
+2 & 1 & 1 & 1 \\
+1 & 2 & 10 & 1\end{array}
+\right] $
+
+Ay(2,:)-Ay(1,:)/5 = ([2 1 1 1]-1/5[10 2 1 1])
+
+$\left[ \begin{array}{ccc|c}
+ & A & & y \\
+10 & 2 & 1 & 1\\
+0 & 3/5 & 4/5 & 4/5 \\
+1 & 2 & 10 & 1\end{array}
+\right] $
+
+Ay(3,:)-Ay(1,:)/10 = ([1 2 10 1]-1/10[10 2 1 1])
+
+$\left[ \begin{array}{ccc|c}
+ & A & & y \\
+10 & 2 & 1 & 1\\
+0 & 3/5 & 4/5 & 4/5 \\
+0 & 1.8 & 9.9 & 0.9\end{array}
+\right] $
+
+Ay(3,:)-1.8\*5/3\*Ay(2,:) = ([0 1.8 9.9 0.9]-3\*[0 3/5 4/5 4/5])
+
+$\left[ \begin{array}{ccc|c}
+ & A & & y \\
+10 & 2 & 1 & 1\\
+0 & 3/5 & 4/5 & 4/5 \\
+0 & 0 & 7.5 & -1.5\end{array}
+\right] $
+
+now, $7.5x_{3}=-1.5$ so $x_{3}=-\frac{1}{5}$
+
+then, $3/5x_{2}+4/5(-1/5)=1$ so $x_{2}=\frac{8}{5}$
+
+finally, $10x_{1}+2(8/5)
+
+Consider the problem again from the intro to Linear Algebra, 4 masses are connected in series to 4 springs with K=10 N/m. What are the final positions of the masses?
+
+
+
+The masses haves the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass:
+
+$m_{1}g+k(x_{2}-x_{1})-kx_{1}=0$
+
+$m_{2}g+k(x_{3}-x_{2})-k(x_{2}-x_{1})=0$
+
+$m_{3}g+k(x_{4}-x_{3})-k(x_{3}-x_{2})=0$
+
+$m_{4}g-k(x_{4}-x_{3})=0$
+
+in matrix form:
+
+$\left[ \begin{array}{cccc}
+2k & -k & 0 & 0 \\
+-k & 2k & -k & 0 \\
+0 & -k & 2k & -k \\
+0 & 0 & -k & k \end{array} \right]
+\left[ \begin{array}{c}
+x_{1} \\
+x_{2} \\
+x_{3} \\
+x_{4} \end{array} \right]=
+\left[ \begin{array}{c}
+m_{1}g \\
+m_{2}g \\
+m_{3}g \\
+m_{4}g \end{array} \right]$
+
+
+```octave
+k=10; % N/m
+m1=1; % kg
+m2=2;
+m3=3;
+m4=4;
+g=9.81; % m/s^2
+K=[2*k -k 0 0; -k 2*k -k 0; 0 -k 2*k -k; 0 0 -k k]
+y=[m1*g;m2*g;m3*g;m4*g]
+```
+
+ K =
+
+ 20 -10 0 0
+ -10 20 -10 0
+ 0 -10 20 -10
+ 0 0 -10 10
+
+ y =
+
+ 9.8100
+ 19.6200
+ 29.4300
+ 39.2400
+
+
+
+
+```octave
+K1=[K y];
+K1(2,:)=K1(1,:)/2+K1(2,:)
+```
+
+ K1 =
+
+ 20.00000 -10.00000 0.00000 0.00000 9.81000
+ 0.00000 15.00000 -10.00000 0.00000 24.52500
+ 0.00000 -10.00000 20.00000 -10.00000 29.43000
+ 0.00000 0.00000 -10.00000 10.00000 39.24000
+
+
+
+
+```octave
+K2=K1;
+K2(3,:)=K1(2,:)*2/3+K1(3,:)
+
+```
+
+ K2 =
+
+ 20.00000 -10.00000 0.00000 0.00000 9.81000
+ 0.00000 15.00000 -10.00000 0.00000 24.52500
+ 0.00000 0.00000 13.33333 -10.00000 45.78000
+ 0.00000 0.00000 -10.00000 10.00000 39.24000
+
+
+
+
+```octave
+K2(4,:)=-K2(3,:)*K2(4,3)/K2(3,3)+K2(4,:)
+```
+
+ K2 =
+
+ 20.00000 -10.00000 0.00000 0.00000 9.81000
+ 0.00000 15.00000 -10.00000 0.00000 24.52500
+ 0.00000 0.00000 13.33333 -10.00000 45.78000
+ 0.00000 0.00000 0.00000 2.50000 73.57500
+
+
+
+
+```octave
+yp=K2(:,5);
+x4=yp(4)/K2(4,4)
+x3=(yp(3)+10*x4)/K2(3,3)
+x2=(yp(2)+10*x3)/K2(2,2)
+x1=(yp(1)+10*x2)/K2(1,1)
+```
+
+ x4 = 29.430
+ x3 = 25.506
+ x2 = 18.639
+ x1 = 9.8100
+
+
+
+```octave
+K\y
+```
+
+ ans =
+
+ 9.8100
+ 18.6390
+ 25.5060
+ 29.4300
+
+
+
+## Automate Gauss Elimination
+
+We can automate Gauss elimination with a function whose input is A and y:
+
+`x=GaussNaive(A,y)`
+
+
+```octave
+x=GaussNaive(K,y)
+```
+
+ x =
+
+ 9.8100
+ 18.6390
+ 25.5060
+ 29.4300
+
+
+
+## Problem (Diagonal element is zero)
+
+If a diagonal element is 0 or very small either:
+
+1. no solution found
+2. errors are introduced
+
+Therefore, we would want to pivot before applying Gauss elimination
+
+Consider:
+
+(a) $\left[ \begin{array}{cccc}
+0 & 2 & 3 \\
+4 & 6 & 7 \\
+2 & -3 & 6 \end{array} \right]
+\left[ \begin{array}{c}
+x_{1} \\
+x_{2} \\
+x_{3} \end{array} \right]=
+\left[ \begin{array}{c}
+8 \\
+-3 \\
+5\end{array} \right]$
+
+(b) $\left[ \begin{array}{cccc}
+0.0003 & 3.0000 \\
+1.0000 & 1.0000 \end{array} \right]
+\left[ \begin{array}{c}
+x_{1} \\
+x_{2} \end{array} \right]=
+\left[ \begin{array}{c}
+2.0001 \\
+1.0000 \end{array} \right]$
+
+
+```octave
+format short
+Aa=[0,2,3;4,6,7;2,-3,6]; ya=[8;-3;5];
+GaussNaive(Aa,ya)
+Aa\ya
+```
+
+ warning: division by zero
+ warning: called from
+ GaussNaive at line 16 column 12
+ warning: division by zero
+ warning: division by zero
+ ans =
+
+ NaN
+ NaN
+ NaN
+
+ ans =
+
+ -5.423913
+ 0.021739
+ 2.652174
+
+
+
+
+```octave
+[x,Aug,npivots]=GaussPivot(Aa,ya)
+```
+
+ x =
+
+ -5.423913
+ 0.021739
+ 2.652174
+
+ Aug =
+
+ 4.00000 6.00000 7.00000 -3.00000
+ 0.00000 -6.00000 2.50000 6.50000
+ 0.00000 0.00000 3.83333 10.16667
+
+ npivots = 2
+
+
+
+```octave
+format long
+Ab=[0.3E-13,3.0000;1.0000,1.0000];yb=[2+0.1e-13;1.0000];
+GaussNaive(Ab,yb)
+Ab\yb
+```
+
+ ans =
+
+ 0.325665420556713
+ 0.666666666666667
+
+ ans =
+
+ 0.333333333333333
+ 0.666666666666667
+
+
+
+
+```octave
+[x,Aug,npivots]=GaussPivot(Ab,yb)
+Ab\yb
+format short
+```
+
+ x =
+
+ 0.333333333333333
+ 0.666666666666667
+
+ Aug =
+
+ 1.000000000000000 1.000000000000000 1.000000000000000
+ 0.000000000000000 2.999999999999970 1.999999999999980
+
+ npivots = 1
+ ans =
+
+ 0.333333333333333
+ 0.666666666666667
+
+
+
+### Spring-Mass System again
+Now, 4 masses are connected in series to 4 springs with $K_{1}$=10 N/m, $K_{2}$=5 N/m,
+$K_{3}$=2 N/m
+and $K_{4}$=1 N/m. What are the final positions of the masses?
+
+
+
+The masses have the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass:
+
+$m_{1}g+k_{2}(x_{2}-x_{1})-k_{1}x_{1}=0$
+
+$m_{2}g+k_{3}(x_{3}-x_{2})-k_{2}(x_{2}-x_{1})=0$
+
+$m_{3}g+k_{4}(x_{4}-x_{3})-k_{3}(x_{3}-x_{2})=0$
+
+$m_{4}g-k_{4}(x_{4}-x_{3})=0$
+
+in matrix form:
+
+$\left[ \begin{array}{cccc}
+k_1+k_2 & -k_2 & 0 & 0 \\
+-k_2 & k_2+k_3 & -k_3 & 0 \\
+0 & -k_3 & k_3+k_4 & -k_4 \\
+0 & 0 & -k_4 & k_4 \end{array} \right]
+\left[ \begin{array}{c}
+x_{1} \\
+x_{2} \\
+x_{3} \\
+x_{4} \end{array} \right]=
+\left[ \begin{array}{c}
+m_{1}g \\
+m_{2}g \\
+m_{3}g \\
+m_{4}g \end{array} \right]$
+
+
+```octave
+k1=10; k2=5;k3=2;k4=1; % N/m
+m1=1; % kg
+m2=2;
+m3=3;
+m4=4;
+g=9.81; % m/s^2
+K=[k1+k2 -k2 0 0; -k2, k2+k3, -k3 0; 0 -k3, k3+k4, -k4; 0 0 -k4 k4]
+y=[m1*g;m2*g;m3*g;m4*g]
+```
+
+ K =
+
+ 15 -5 0 0
+ -5 7 -2 0
+ 0 -2 3 -1
+ 0 0 -1 1
+
+ y =
+
+ 9.8100
+ 19.6200
+ 29.4300
+ 39.2400
+
+
+
+## Tridiagonal matrix
+
+This matrix, K, could be rewritten as 3 vectors e, f and g
+
+$e=\left[ \begin{array}{c}
+0 \\
+-5 \\
+-2 \\
+-1 \end{array} \right]$
+
+$f=\left[ \begin{array}{c}
+15 \\
+7 \\
+3 \\
+1 \end{array} \right]$
+
+$g=\left[ \begin{array}{c}
+-5 \\
+-2 \\
+-1 \\
+0 \end{array} \right]$
+
+Where all other components are 0 and the length of the vectors are n and the first component of e and the last component of g are zero
+
+`e(1)=0`
+
+`g(end)=0`
+
+No need to pivot and number of calculations reduced enormously.
+
+|method |Number of Floating point operations for n$\times$n-matrix|
+|----------------|---------|
+| Naive Gauss | n-cubed |
+| Tridiagonal | n |
+
+
+```octave
+e=[0;-5;-2;-1];
+g=[-5;-2;-1;0];
+f=[15;7;3;1];
+Tridiag(e,f,g,y)
+
+```
+
+ ans =
+
+ 9.8100 27.4680 61.8030 101.0430
+
+
+
+
+```octave
+% tic ... t=toc
+% is Matlab timer used for debugging programs
+t_GE = zeros(1,100);
+t_GE_tridiag = zeros(1,100);
+t_TD = zeros(1,100);
+for n = 1:200
+ A = rand(n,n);
+ e = rand(n,1); e(1)=0;
+ f = rand(n,1);
+ g = rand(n,1); g(end)=0;
+ Atd=diag(f, 0) - diag(e(2:n), -1) - diag(g(1:n-1), 1);
+ b = rand(n,1);
+ tic;
+ x = GaussPivot(A,b);
+ t_GE(n) = toc;
+ tic;
+ x = GaussPivot(Atd,b);
+ t_GE_tridiag(n) = toc;
+ tic;
+ x = Tridiag(e,f,g,b);
+ t_TD(n) = toc;
+end
+```
+
+
+```octave
+n=1:200;
+loglog(n,t_GE,n,t_TD,n,t_GE_tridiag)
+xlabel('number of elements')
+ylabel('time (s)')
+```
+
+
+
+
+
+
+```octave
+plot(t_TD)
+```
+
+
+
+
+
+
+```octave
+
+```
diff --git a/lecture_10/lecture_10.out b/lecture_10/lecture_10.out
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+\BOOKMARK [1][-]{section.1}{Gauss Elimination}{}% 1
+\BOOKMARK [2][-]{subsubsection.1.0.1}{Solving sets of equations with matrix operations}{section.1}% 2
+\BOOKMARK [2][-]{subsection.1.1}{Gauss elimination}{section.1}% 3
+\BOOKMARK [3][-]{subsubsection.1.1.1}{Solving sets of equations systematically}{subsection.1.1}% 4
+\BOOKMARK [2][-]{subsection.1.2}{Automate Gauss Elimination}{section.1}% 5
+\BOOKMARK [2][-]{subsection.1.3}{Problem \(Diagonal element is zero\)}{section.1}% 6
+\BOOKMARK [3][-]{subsubsection.1.3.1}{Spring-Mass System again}{subsection.1.3}% 7
+\BOOKMARK [2][-]{subsection.1.4}{Tridiagonal matrix}{section.1}% 8
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+
+% Default to the notebook output style
+
+
+
+
+% Inherit from the specified cell style.
+
+
+
+
+
+\documentclass[11pt]{article}
+
+
+
+ \usepackage[T1]{fontenc}
+ % Nicer default font (+ math font) than Computer Modern for most use cases
+ \usepackage{mathpazo}
+
+ % Basic figure setup, for now with no caption control since it's done
+ % automatically by Pandoc (which extracts  syntax from Markdown).
+ \usepackage{graphicx}
+ % We will generate all images so they have a width \maxwidth. This means
+ % that they will get their normal width if they fit onto the page, but
+ % are scaled down if they would overflow the margins.
+ \makeatletter
+ \def\maxwidth{\ifdim\Gin@nat@width>\linewidth\linewidth
+ \else\Gin@nat@width\fi}
+ \makeatother
+ \let\Oldincludegraphics\includegraphics
+ % Set max figure width to be 80% of text width, for now hardcoded.
+ \renewcommand{\includegraphics}[1]{\Oldincludegraphics[width=.8\maxwidth]{#1}}
+ % Ensure that by default, figures have no caption (until we provide a
+ % proper Figure object with a Caption API and a way to capture that
+ % in the conversion process - todo).
+ \usepackage{caption}
+ \DeclareCaptionLabelFormat{nolabel}{}
+ \captionsetup{labelformat=nolabel}
+
+ \usepackage{adjustbox} % Used to constrain images to a maximum size
+ \usepackage{xcolor} % Allow colors to be defined
+ \usepackage{enumerate} % Needed for markdown enumerations to work
+ \usepackage{geometry} % Used to adjust the document margins
+ \usepackage{amsmath} % Equations
+ \usepackage{amssymb} % Equations
+ \usepackage{textcomp} % defines textquotesingle
+ % Hack from http://tex.stackexchange.com/a/47451/13684:
+ \AtBeginDocument{%
+ \def\PYZsq{\textquotesingle}% Upright quotes in Pygmentized code
+ }
+ \usepackage{upquote} % Upright quotes for verbatim code
+ \usepackage{eurosym} % defines \euro
+ \usepackage[mathletters]{ucs} % Extended unicode (utf-8) support
+ \usepackage[utf8x]{inputenc} % Allow utf-8 characters in the tex document
+ \usepackage{fancyvrb} % verbatim replacement that allows latex
+ \usepackage{grffile} % extends the file name processing of package graphics
+ % to support a larger range
+ % The hyperref package gives us a pdf with properly built
+ % internal navigation ('pdf bookmarks' for the table of contents,
+ % internal cross-reference links, web links for URLs, etc.)
+ \usepackage{hyperref}
+ \usepackage{longtable} % longtable support required by pandoc >1.10
+ \usepackage{booktabs} % table support for pandoc > 1.12.2
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+
+
+ % Define a nice break command that doesn't care if a line doesn't already
+ % exist.
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+ \def\lt{<}
+ % Document parameters
+ \title{lecture\_10}
+
+
+
+
+ % Pygments definitions
+
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+
+\def\PYZbs{\char`\\}
+\def\PYZus{\char`\_}
+\def\PYZob{\char`\{}
+\def\PYZcb{\char`\}}
+\def\PYZca{\char`\^}
+\def\PYZam{\char`\&}
+\def\PYZlt{\char`\<}
+\def\PYZgt{\char`\>}
+\def\PYZsh{\char`\#}
+\def\PYZpc{\char`\%}
+\def\PYZdl{\char`\$}
+\def\PYZhy{\char`\-}
+\def\PYZsq{\char`\'}
+\def\PYZdq{\char`\"}
+\def\PYZti{\char`\~}
+% for compatibility with earlier versions
+\def\PYZat{@}
+\def\PYZlb{[}
+\def\PYZrb{]}
+\makeatother
+
+
+ % Exact colors from NB
+ \definecolor{incolor}{rgb}{0.0, 0.0, 0.5}
+ \definecolor{outcolor}{rgb}{0.545, 0.0, 0.0}
+
+
+
+
+ % Prevent overflowing lines due to hard-to-break entities
+ \sloppy
+ % Setup hyperref package
+ \hypersetup{
+ breaklinks=true, % so long urls are correctly broken across lines
+ colorlinks=true,
+ urlcolor=urlcolor,
+ linkcolor=linkcolor,
+ citecolor=citecolor,
+ }
+ % Slightly bigger margins than the latex defaults
+
+ \geometry{verbose,tmargin=1in,bmargin=1in,lmargin=1in,rmargin=1in}
+
+
+
+ \begin{document}
+
+
+ \maketitle
+
+
+
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}1}]:} \PY{c}{\PYZpc{}plot \PYZhy{}\PYZhy{}format svg}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}2}]:} \PY{n}{setdefaults}
+\end{Verbatim}
+
+ \section{Gauss Elimination}\label{gauss-elimination}
+
+\subsubsection{Solving sets of equations with matrix
+operations}\label{solving-sets-of-equations-with-matrix-operations}
+
+The number of dimensions of a matrix indicate the degrees of freedom of
+the system you are solving.
+
+If you have a set of known output, \(y_{1},~y_{2},~...y_{N}\) and a set
+of equations that relate unknown inputs, \(x_{1},~x_{2},~...x_{N}\),
+then these can be written in a vector matrix format as:
+
+\(y=Ax\)
+
+Consider a problem with 2 DOF:
+
+\(x_{1}+3x_{2}=1\)
+
+\(2x_{1}+x_{2}=1\)
+
+\(\left[ \begin{array}{cc} 1 & 3 \\ 2 & 1 \end{array} \right] \left[\begin{array}{c} x_{1} \\ x_{2} \end{array}\right]= \left[\begin{array}{c} 1 \\ 1\end{array}\right]\)
+
+The solution for \(x_{1}\) and \(x_{2}\) is the intersection of two
+lines:
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}3}]:} \PY{n}{x21}\PY{p}{=}\PY{p}{[}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{p}{:}\PY{l+m+mi}{2}\PY{p}{]}\PY{p}{;}
+ \PY{n}{x11}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{\PYZhy{}}\PY{l+m+mi}{3}\PY{o}{*}\PY{n}{x21}\PY{p}{;}
+ \PY{n}{x21}\PY{p}{=}\PY{p}{[}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{p}{:}\PY{l+m+mi}{2}\PY{p}{]}\PY{p}{;}
+ \PY{n}{x22}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{x21}\PY{p}{;}
+ \PY{n+nb}{plot}\PY{p}{(}\PY{n}{x11}\PY{p}{,}\PY{n}{x21}\PY{p}{,}\PY{n}{x21}\PY{p}{,}\PY{n}{x22}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{center}
+ \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_10_files/lecture_10_3_0.pdf}
+ \end{center}
+ { \hspace*{\fill} \\}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}4}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{;} \PY{n}{y}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{;}
+ \PY{n}{A}\PY{o}{\PYZbs{}}\PY{n}{y} \PY{c}{\PYZpc{} matlab\PYZsq{}s Ax=y solution for x}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+ans =
+
+ 0.40000
+ 0.20000
+
+
+ \end{Verbatim}
+
+ For a \(3\times3\) matrix, the solution is the intersection of the 3
+planes.
+
+\(10x_{1}+2x_{2}+x_{3}=1\)
+
+\(2x_{1}+x_{2}+x_{3}=1\)
+
+\(x_{1}+2x_{2}+10x_{3}=1\)
+
+$\left[ \begin{array}{ccc} 10 & 2 & 1\\ 2 & 1 & 1 \\ 1 & 2 & 10\end{array} \right]
+ \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right]=
+ \left[\begin{array}{c} 1 \\ 1 \\ 1\end{array}\right]$
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}9}]:} \PY{n}{N}\PY{p}{=}\PY{l+m+mi}{25}\PY{p}{;}
+ \PY{n}{x11}\PY{p}{=}\PY{n+nb}{linspace}\PY{p}{(}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{n}{N}\PY{p}{)}\PY{p}{;}
+ \PY{n}{x12}\PY{p}{=}\PY{n+nb}{linspace}\PY{p}{(}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{n}{N}\PY{p}{)}\PY{p}{;}
+ \PY{p}{[}\PY{n}{X11}\PY{p}{,}\PY{n}{X12}\PY{p}{]}\PY{p}{=}\PY{n+nb}{meshgrid}\PY{p}{(}\PY{n}{x11}\PY{p}{,}\PY{n}{x12}\PY{p}{)}\PY{p}{;}
+ \PY{n}{X13}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{\PYZhy{}}\PY{l+m+mi}{10}\PY{o}{*}\PY{n}{X11}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{X12}\PY{p}{;}
+
+ \PY{n}{x21}\PY{p}{=}\PY{n+nb}{linspace}\PY{p}{(}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{n}{N}\PY{p}{)}\PY{p}{;}
+ \PY{n}{x22}\PY{p}{=}\PY{n+nb}{linspace}\PY{p}{(}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{n}{N}\PY{p}{)}\PY{p}{;}
+ \PY{p}{[}\PY{n}{X21}\PY{p}{,}\PY{n}{X22}\PY{p}{]}\PY{p}{=}\PY{n+nb}{meshgrid}\PY{p}{(}\PY{n}{x21}\PY{p}{,}\PY{n}{x22}\PY{p}{)}\PY{p}{;}
+ \PY{n}{X23}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{X11}\PY{o}{\PYZhy{}}\PY{n}{X22}\PY{p}{;}
+
+ \PY{n}{x31}\PY{p}{=}\PY{n+nb}{linspace}\PY{p}{(}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{n}{N}\PY{p}{)}\PY{p}{;}
+ \PY{n}{x32}\PY{p}{=}\PY{n+nb}{linspace}\PY{p}{(}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{n}{N}\PY{p}{)}\PY{p}{;}
+ \PY{p}{[}\PY{n}{X31}\PY{p}{,}\PY{n}{X32}\PY{p}{]}\PY{p}{=}\PY{n+nb}{meshgrid}\PY{p}{(}\PY{n}{x31}\PY{p}{,}\PY{n}{x32}\PY{p}{)}\PY{p}{;}
+ \PY{n}{X33}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{10}\PY{o}{*}\PY{p}{(}\PY{l+m+mi}{1}\PY{o}{\PYZhy{}}\PY{n}{X31}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{X32}\PY{p}{)}\PY{p}{;}
+
+ \PY{n+nb}{mesh}\PY{p}{(}\PY{n}{X11}\PY{p}{,}\PY{n}{X12}\PY{p}{,}\PY{n}{X13}\PY{p}{)}\PY{p}{;}
+ \PY{n+nb}{hold} \PY{n}{on}\PY{p}{;}
+ \PY{n+nb}{mesh}\PY{p}{(}\PY{n}{X21}\PY{p}{,}\PY{n}{X22}\PY{p}{,}\PY{n}{X23}\PY{p}{)}
+ \PY{n+nb}{mesh}\PY{p}{(}\PY{n}{X31}\PY{p}{,}\PY{n}{X32}\PY{p}{,}\PY{n}{X33}\PY{p}{)}
+ \PY{n}{x}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{10}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,} \PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{,} \PY{l+m+mi}{1}\PY{p}{;} \PY{l+m+mi}{1}\PY{p}{,} \PY{l+m+mi}{2}\PY{p}{,} \PY{l+m+mi}{10}\PY{p}{]}\PY{o}{\PYZbs{}}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{;}
+ \PY{n}{plot3}\PY{p}{(}\PY{n}{x}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{,}\PY{n}{x}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{,}\PY{n}{x}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{o\PYZsq{}}\PY{p}{)}
+ \PY{n+nb}{xlabel}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{x1\PYZsq{}}\PY{p}{)}
+ \PY{n+nb}{ylabel}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{x2\PYZsq{}}\PY{p}{)}
+ \PY{n+nb}{zlabel}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{x3\PYZsq{}}\PY{p}{)}
+ \PY{n+nb}{view}\PY{p}{(}\PY{l+m+mi}{10}\PY{p}{,}\PY{l+m+mi}{45}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{center}
+ \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_10_files/lecture_10_6_0.pdf}
+ \end{center}
+ { \hspace*{\fill} \\}
+
+ After 3 DOF problems, the solutions are described as \emph{hyperplane}
+intersections. Which are even harder to visualize
+
+ \subsection{Gauss elimination}\label{gauss-elimination}
+
+\subsubsection{Solving sets of equations
+systematically}\label{solving-sets-of-equations-systematically}
+
+$\left[
+\begin{array}{ccc|c}
+ & A & & y \\
+10 & 2 & 1 & 1\\
+2 & 1 & 1 & 1 \\
+1 & 2 & 10 & 1\end{array}
+\right] $
+
+Ay(2,:)-Ay(1,:)/5 = ({[}2 1 1 1{]}-1/5{[}10 2 1 1{]})
+
+$\left[
+\begin{array}{ccc|c}
+ & A & & y \\
+10 & 2 & 1 & 1\\
+0 & 3/5 & 4/5 & 4/5 \\
+1 & 2 & 10 & 1\end{array}
+\right] $
+
+Ay(3,:)-Ay(1,:)/10 = ({[}1 2 10 1{]}-1/10{[}10 2 1 1{]})
+
+$\left[
+\begin{array}{ccc|c}
+ & A & & y \\
+10 & 2 & 1 & 1\\
+0 & 3/5 & 4/5 & 4/5 \\
+0 & 1.8 & 9.9 & 0.9\end{array}
+\right] $
+
+Ay(3,:)-1.8*5/3*Ay(2,:) = ({[}0 1.8 9.9 0.9{]}-3*{[}0 3/5 4/5 4/5{]})
+
+$\left[
+\begin{array}{ccc|c}
+ & A & & y \\
+10 & 2 & 1 & 1\\
+0 & 3/5 & 4/5 & 4/5 \\
+0 & 0 & 7.5 & -1.5\end{array}
+\right]$
+
+now, \(7.5x_{3}=-1.5\) so \(x_{3}=-\frac{1}{5}\)
+
+then, \(3/5x_{2}+4/5(-1/5)=1\) so \(x_{2}=\frac{8}{5}\)
+
+finally, \(10x_{1}+2(8/5)+1(-\frac{1}{5})=1\)
+
+ Consider the problem again from the intro to Linear Algebra, 4 masses
+are connected in series to 4 springs with K=10 N/m. What are the final
+positions of the masses?
+
+\begin{figure}[htbp]
+\centering
+\includegraphics{../lecture_09/mass_springs.svg}
+\caption{Springs-masses}
+\end{figure}
+
+The masses haves the following amounts, 1, 2, 3, and 4 kg for masses
+1-4. Using a FBD for each mass:
+
+\(m_{1}g+k(x_{2}-x_{1})-kx_{1}=0\)
+
+\(m_{2}g+k(x_{3}-x_{2})-k(x_{2}-x_{1})=0\)
+
+\(m_{3}g+k(x_{4}-x_{3})-k(x_{3}-x_{2})=0\)
+
+\(m_{4}g-k(x_{4}-x_{3})=0\)
+
+in matrix form:
+
+\(\left[ \begin{array}{cccc} 2k & -k & 0 & 0 \\ -k & 2k & -k & 0 \\ 0 & -k & 2k & -k \\ 0 & 0 & -k & k \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right]= \left[ \begin{array}{c} m_{1}g \\ m_{2}g \\ m_{3}g \\ m_{4}g \end{array} \right]\)
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}10}]:} \PY{n}{k}\PY{p}{=}\PY{l+m+mi}{10}\PY{p}{;} \PY{c}{\PYZpc{} N/m}
+ \PY{n}{m1}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} \PY{c}{\PYZpc{} kg}
+ \PY{n}{m2}\PY{p}{=}\PY{l+m+mi}{2}\PY{p}{;}
+ \PY{n}{m3}\PY{p}{=}\PY{l+m+mi}{3}\PY{p}{;}
+ \PY{n}{m4}\PY{p}{=}\PY{l+m+mi}{4}\PY{p}{;}
+ \PY{n}{g}\PY{p}{=}\PY{l+m+mf}{9.81}\PY{p}{;} \PY{c}{\PYZpc{} m/s\PYZca{}2}
+ \PY{n}{K}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{k} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{l+m+mi}{0} \PY{l+m+mi}{0}\PY{p}{;} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{l+m+mi}{2}\PY{o}{*}\PY{n}{k} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{l+m+mi}{0}\PY{p}{;} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{l+m+mi}{2}\PY{o}{*}\PY{n}{k} \PY{o}{\PYZhy{}}\PY{n}{k}\PY{p}{;} \PY{l+m+mi}{0} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{n}{k}\PY{p}{]}
+ \PY{n}{y}\PY{p}{=}\PY{p}{[}\PY{n}{m1}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m2}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m3}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m4}\PY{o}{*}\PY{n}{g}\PY{p}{]}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+K =
+
+ 20 -10 0 0
+ -10 20 -10 0
+ 0 -10 20 -10
+ 0 0 -10 10
+
+y =
+
+ 9.8100
+ 19.6200
+ 29.4300
+ 39.2400
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}11}]:} \PY{n}{K1}\PY{p}{=}\PY{p}{[}\PY{n}{K} \PY{n}{y}\PY{p}{]}\PY{p}{;}
+ \PY{n}{K1}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{K1}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{/}\PY{l+m+mi}{2}\PY{o}{+}\PY{n}{K1}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+K1 =
+
+ 20.00000 -10.00000 0.00000 0.00000 9.81000
+ 0.00000 15.00000 -10.00000 0.00000 24.52500
+ 0.00000 -10.00000 20.00000 -10.00000 29.43000
+ 0.00000 0.00000 -10.00000 10.00000 39.24000
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}12}]:} \PY{n}{K2}\PY{p}{=}\PY{n}{K1}\PY{p}{;}
+ \PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{K1}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{*}\PY{l+m+mi}{2}\PY{o}{/}\PY{l+m+mi}{3}\PY{o}{+}\PY{n}{K1}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+K2 =
+
+ 20.00000 -10.00000 0.00000 0.00000 9.81000
+ 0.00000 15.00000 -10.00000 0.00000 24.52500
+ 0.00000 0.00000 13.33333 -10.00000 45.78000
+ 0.00000 0.00000 -10.00000 10.00000 39.24000
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}13}]:} \PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{o}{\PYZhy{}}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{*}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{/}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{+}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{p}{:}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+K2 =
+
+ 20.00000 -10.00000 0.00000 0.00000 9.81000
+ 0.00000 15.00000 -10.00000 0.00000 24.52500
+ 0.00000 0.00000 13.33333 -10.00000 45.78000
+ 0.00000 0.00000 0.00000 2.50000 73.57500
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}14}]:} \PY{n}{yp}\PY{p}{=}\PY{n}{K2}\PY{p}{(}\PY{p}{:}\PY{p}{,}\PY{l+m+mi}{5}\PY{p}{)}\PY{p}{;}
+ \PY{n}{x4}\PY{p}{=}\PY{n}{yp}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{)}\PY{o}{/}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{)}
+ \PY{n}{x3}\PY{p}{=}\PY{p}{(}\PY{n}{yp}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{+}\PY{l+m+mi}{10}\PY{o}{*}\PY{n}{x4}\PY{p}{)}\PY{o}{/}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}
+ \PY{n}{x2}\PY{p}{=}\PY{p}{(}\PY{n}{yp}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{+}\PY{l+m+mi}{10}\PY{o}{*}\PY{n}{x3}\PY{p}{)}\PY{o}{/}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}
+ \PY{n}{x1}\PY{p}{=}\PY{p}{(}\PY{n}{yp}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{+}\PY{l+m+mi}{10}\PY{o}{*}\PY{n}{x2}\PY{p}{)}\PY{o}{/}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+x4 = 29.430
+x3 = 25.506
+x2 = 18.639
+x1 = 9.8100
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}15}]:} \PY{n}{K}\PY{o}{\PYZbs{}}\PY{n}{y}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+ans =
+
+ 9.8100
+ 18.6390
+ 25.5060
+ 29.4300
+
+
+ \end{Verbatim}
+
+ \subsection{Automate Gauss
+Elimination}\label{automate-gauss-elimination}
+
+We can automate Gauss elimination with a function whose input is A and
+y:
+
+\texttt{x=GaussNaive(A,y)}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}16}]:} \PY{n}{x}\PY{p}{=}\PY{n}{GaussNaive}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{n}{y}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+x =
+
+ 9.8100
+ 18.6390
+ 25.5060
+ 29.4300
+
+
+ \end{Verbatim}
+
+ \subsection{Problem (Diagonal element is
+zero)}\label{problem-diagonal-element-is-zero}
+
+If a diagonal element is 0 or very small either:
+
+\begin{enumerate}
+\def\labelenumi{\arabic{enumi}.}
+\tightlist
+\item
+ no solution found
+\item
+ errors are introduced
+\end{enumerate}
+
+Therefore, we would want to pivot before applying Gauss elimination
+
+Consider:
+
+\begin{enumerate}
+\def\labelenumi{(\alph{enumi})}
+\item
+ \(\left[ \begin{array}{cccc} 0 & 2 & 3 \\ 4 & 6 & 7 \\ 2 & -3 & 6 \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array} \right]= \left[ \begin{array}{c} 8 \\ -3 \\ 5\end{array} \right]\)
+\item
+ \(\left[ \begin{array}{cccc} 0.0003 & 3.0000 \\ 1.0000 & 1.0000 \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \end{array} \right]= \left[ \begin{array}{c} 2.0001 \\ 1.0000 \end{array} \right]\)
+\end{enumerate}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}17}]:} \PY{n}{format} \PY{n}{short}
+ \PY{n}{Aa}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{6}\PY{p}{,}\PY{l+m+mi}{7}\PY{p}{;}\PY{l+m+mi}{2}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{6}\PY{p}{]}\PY{p}{;} \PY{n}{ya}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{8}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{5}\PY{p}{]}\PY{p}{;}
+ \PY{n}{GaussNaive}\PY{p}{(}\PY{n}{Aa}\PY{p}{,}\PY{n}{ya}\PY{p}{)}
+ \PY{n}{Aa}\PY{o}{\PYZbs{}}\PY{n}{ya}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+warning: division by zero
+warning: called from
+ GaussNaive at line 16 column 12
+warning: division by zero
+warning: division by zero
+ans =
+
+ NaN
+ NaN
+ NaN
+
+ans =
+
+ -5.423913
+ 0.021739
+ 2.652174
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}32}]:} \PY{p}{[}\PY{n}{x}\PY{p}{,}\PY{n}{Aug}\PY{p}{,}\PY{n}{npivots}\PY{p}{]}\PY{p}{=}\PY{n}{GaussPivot}\PY{p}{(}\PY{n}{Aa}\PY{p}{,}\PY{n}{ya}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+x =
+
+ -5.423913
+ 0.021739
+ 2.652174
+
+Aug =
+
+ 4.00000 6.00000 7.00000 -3.00000
+ 0.00000 -6.00000 2.50000 6.50000
+ 0.00000 0.00000 3.83333 10.16667
+
+npivots = 2
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}33}]:} \PY{n}{format} \PY{n}{long}
+ \PY{n}{Ab}\PY{p}{=}\PY{p}{[}\PY{l+m+mf}{0.3E\PYZhy{}13}\PY{p}{,}\PY{l+m+mf}{3.0000}\PY{p}{;}\PY{l+m+mf}{1.0000}\PY{p}{,}\PY{l+m+mf}{1.0000}\PY{p}{]}\PY{p}{;}\PY{n}{yb}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{2}\PY{o}{+}\PY{l+m+mf}{0.1e\PYZhy{}13}\PY{p}{;}\PY{l+m+mf}{1.0000}\PY{p}{]}\PY{p}{;}
+ \PY{n}{GaussNaive}\PY{p}{(}\PY{n}{Ab}\PY{p}{,}\PY{n}{yb}\PY{p}{)}
+ \PY{n}{Ab}\PY{o}{\PYZbs{}}\PY{n}{yb}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+ans =
+
+ 0.325665420556713
+ 0.666666666666667
+
+ans =
+
+ 0.333333333333333
+ 0.666666666666667
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}34}]:} \PY{p}{[}\PY{n}{x}\PY{p}{,}\PY{n}{Aug}\PY{p}{,}\PY{n}{npivots}\PY{p}{]}\PY{p}{=}\PY{n}{GaussPivot}\PY{p}{(}\PY{n}{Ab}\PY{p}{,}\PY{n}{yb}\PY{p}{)}
+ \PY{n}{Ab}\PY{o}{\PYZbs{}}\PY{n}{yb}
+ \PY{n}{format} \PY{n}{short}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+x =
+
+ 0.333333333333333
+ 0.666666666666667
+
+Aug =
+
+ 1.000000000000000 1.000000000000000 1.000000000000000
+ 0.000000000000000 2.999999999999970 1.999999999999980
+
+npivots = 1
+ans =
+
+ 0.333333333333333
+ 0.666666666666667
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}36}]:} \PY{c}{\PYZpc{} determinant is (\PYZhy{}1)\PYZca{}(number\PYZus{}of\PYZus{}pivots)*diagonal\PYZus{}elements}
+ \PY{n+nb}{det}\PY{p}{(}\PY{n}{Ab}\PY{p}{)}
+ \PY{n}{Aug}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{Aug}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+ans = -3.0000
+ans = 3.0000
+
+ \end{Verbatim}
+
+ \subsubsection{Spring-Mass System again}\label{spring-mass-system-again}
+
+Now, 4 masses are connected in series to 4 springs with \(K_{1}\)=10
+N/m, \(K_{2}\)=5 N/m, \(K_{3}\)=2 N/m and \(K_{4}\)=1 N/m. What are the
+final positions of the masses?
+
+\begin{figure}[htbp]
+\centering
+\includegraphics{../lecture_09/mass_springs.svg}
+\caption{Springs-masses}
+\end{figure}
+
+The masses have the following amounts, 1, 2, 3, and 4 kg for masses 1-4.
+Using a FBD for each mass:
+
+\(m_{1}g+k_{2}(x_{2}-x_{1})-k_{1}x_{1}=0\)
+
+\(m_{2}g+k_{3}(x_{3}-x_{2})-k_{2}(x_{2}-x_{1})=0\)
+
+\(m_{3}g+k_{4}(x_{4}-x_{3})-k_{3}(x_{3}-x_{2})=0\)
+
+\(m_{4}g-k_{4}(x_{4}-x_{3})=0\)
+
+in matrix form:
+
+\(\left[ \begin{array}{cccc} k_1+k_2 & -k_2 & 0 & 0 \\ -k_2 & k_2+k_3 & -k_3 & 0 \\ 0 & -k_3 & k_3+k_4 & -k_4 \\ 0 & 0 & -k_4 & k_4 \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right]= \left[ \begin{array}{c} m_{1}g \\ m_{2}g \\ m_{3}g \\ m_{4}g \end{array} \right]\)
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}24}]:} \PY{n}{k1}\PY{p}{=}\PY{l+m+mi}{10}\PY{p}{;} \PY{n}{k2}\PY{p}{=}\PY{l+m+mi}{5}\PY{p}{;}\PY{n}{k3}\PY{p}{=}\PY{l+m+mi}{2}\PY{p}{;}\PY{n}{k4}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} \PY{c}{\PYZpc{} N/m}
+ \PY{n}{m1}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} \PY{c}{\PYZpc{} kg}
+ \PY{n}{m2}\PY{p}{=}\PY{l+m+mi}{2}\PY{p}{;}
+ \PY{n}{m3}\PY{p}{=}\PY{l+m+mi}{3}\PY{p}{;}
+ \PY{n}{m4}\PY{p}{=}\PY{l+m+mi}{4}\PY{p}{;}
+ \PY{n}{g}\PY{p}{=}\PY{l+m+mf}{9.81}\PY{p}{;} \PY{c}{\PYZpc{} m/s\PYZca{}2}
+ \PY{n}{K}\PY{p}{=}\PY{p}{[}\PY{n}{k1}\PY{o}{+}\PY{n}{k2} \PY{o}{\PYZhy{}}\PY{n}{k2} \PY{l+m+mi}{0} \PY{l+m+mi}{0}\PY{p}{;} \PY{o}{\PYZhy{}}\PY{n}{k2}\PY{p}{,} \PY{n}{k2}\PY{o}{+}\PY{n}{k3}\PY{p}{,} \PY{o}{\PYZhy{}}\PY{n}{k3} \PY{l+m+mi}{0}\PY{p}{;} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k3}\PY{p}{,} \PY{n}{k3}\PY{o}{+}\PY{n}{k4}\PY{p}{,} \PY{o}{\PYZhy{}}\PY{n}{k4}\PY{p}{;} \PY{l+m+mi}{0} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k4} \PY{n}{k4}\PY{p}{]}
+ \PY{n}{y}\PY{p}{=}\PY{p}{[}\PY{n}{m1}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m2}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m3}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m4}\PY{o}{*}\PY{n}{g}\PY{p}{]}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+K =
+
+ 15 -5 0 0
+ -5 7 -2 0
+ 0 -2 3 -1
+ 0 0 -1 1
+
+y =
+
+ 9.8100
+ 19.6200
+ 29.4300
+ 39.2400
+
+
+ \end{Verbatim}
+
+ \subsection{Tridiagonal matrix}\label{tridiagonal-matrix}
+
+This matrix, K, could be rewritten as 3 vectors e, f and g
+
+\(e=\left[ \begin{array}{c} 0 \\ -5 \\ -2 \\ -1 \end{array} \right]\)
+
+\(f=\left[ \begin{array}{c} 15 \\ 7 \\ 3 \\ 1 \end{array} \right]\)
+
+\(g=\left[ \begin{array}{c} -5 \\ -2 \\ -1 \\ 0 \end{array} \right]\)
+
+Where all other components are 0 and the length of the vectors are n and
+the first component of e and the last component of g are zero
+
+\texttt{e(1)=0}
+
+\texttt{g(end)=0}
+
+No need to pivot and number of calculations reduced enormously.
+
+\begin{longtable}[c]{@{}ll@{}}
+\toprule
+method & Number of Floating point operations for
+n\(\times\)n-matrix\tabularnewline
+\midrule
+\endhead
+Gauss & n-cubed\tabularnewline
+Tridiagonal & n\tabularnewline
+\bottomrule
+\end{longtable}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}25}]:} \PY{n+nb}{e}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{5}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{;}
+ \PY{n}{g}\PY{p}{=}\PY{p}{[}\PY{o}{\PYZhy{}}\PY{l+m+mi}{5}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{;}
+ \PY{n}{f}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{15}\PY{p}{;}\PY{l+m+mi}{7}\PY{p}{;}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{;}
+ \PY{n}{Tridiag}\PY{p}{(}\PY{n+nb}{e}\PY{p}{,}\PY{n}{f}\PY{p}{,}\PY{n}{g}\PY{p}{,}\PY{n}{y}\PY{p}{)}
+ \PY{n}{K}\PY{o}{\PYZbs{}}\PY{n}{y}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+ans =
+
+ 9.8100 27.4680 61.8030 101.0430
+
+ans =
+
+ 9.8100
+ 27.4680
+ 61.8030
+ 101.0430
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}26}]:} \PY{c}{\PYZpc{} tic ... t=toc }
+ \PY{c}{\PYZpc{} is Matlab timer used for debugging programs}
+ \PY{n}{t\PYZus{}GE} \PY{p}{=} \PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{100}\PY{p}{)}\PY{p}{;}
+ \PY{n}{t\PYZus{}GE\PYZus{}tridiag} \PY{p}{=} \PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{100}\PY{p}{)}\PY{p}{;}
+ \PY{n}{t\PYZus{}TD} \PY{p}{=} \PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{100}\PY{p}{)}\PY{p}{;}
+ \PY{c}{\PYZpc{}for n = 1:200}
+ \PY{k}{for} \PY{n}{n}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{:}\PY{l+m+mi}{100}
+ \PY{n}{A} \PY{p}{=} \PY{n+nb}{rand}\PY{p}{(}\PY{n}{n}\PY{p}{,}\PY{n}{n}\PY{p}{)}\PY{p}{;}
+ \PY{n+nb}{e} \PY{p}{=} \PY{n+nb}{rand}\PY{p}{(}\PY{n}{n}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} \PY{n+nb}{e}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{p}{;}
+ \PY{n}{f} \PY{p}{=} \PY{n+nb}{rand}\PY{p}{(}\PY{n}{n}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{g} \PY{p}{=} \PY{n+nb}{rand}\PY{p}{(}\PY{n}{n}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} \PY{n}{g}\PY{p}{(}\PY{k}{end}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{p}{;}
+ \PY{n}{Atd}\PY{p}{=}\PY{n+nb}{diag}\PY{p}{(}\PY{n}{f}\PY{p}{,} \PY{l+m+mi}{0}\PY{p}{)} \PY{o}{\PYZhy{}} \PY{n+nb}{diag}\PY{p}{(}\PY{n+nb}{e}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{:}\PY{n}{n}\PY{p}{)}\PY{p}{,} \PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{)} \PY{o}{\PYZhy{}} \PY{n+nb}{diag}\PY{p}{(}\PY{n}{g}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{:}\PY{n}{n}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{,} \PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{b} \PY{p}{=} \PY{n+nb}{rand}\PY{p}{(}\PY{n}{n}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n+nb}{tic}\PY{p}{;}
+ \PY{n}{x} \PY{p}{=} \PY{n}{GaussPivot}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{n}{b}\PY{p}{)}\PY{p}{;}
+ \PY{n}{t\PYZus{}GE}\PY{p}{(}\PY{n}{n}\PY{p}{)} \PY{p}{=} \PY{n+nb}{toc}\PY{p}{;}
+ \PY{n+nb}{tic}\PY{p}{;}
+ \PY{n}{x} \PY{p}{=} \PY{n}{A}\PY{o}{\PYZbs{}}\PY{n}{b}\PY{p}{;}
+ \PY{n}{t\PYZus{}GE\PYZus{}tridiag}\PY{p}{(}\PY{n}{n}\PY{p}{)} \PY{p}{=} \PY{n+nb}{toc}\PY{p}{;}
+ \PY{n+nb}{tic}\PY{p}{;}
+ \PY{n}{x} \PY{p}{=} \PY{n}{Tridiag}\PY{p}{(}\PY{n+nb}{e}\PY{p}{,}\PY{n}{f}\PY{p}{,}\PY{n}{g}\PY{p}{,}\PY{n}{b}\PY{p}{)}\PY{p}{;}
+ \PY{n}{t\PYZus{}TD}\PY{p}{(}\PY{n}{n}\PY{p}{)} \PY{p}{=} \PY{n+nb}{toc}\PY{p}{;}
+ \PY{k}{end}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}28}]:} \PY{n}{n}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{:}\PY{l+m+mi}{100}\PY{p}{;}
+ \PY{n+nb}{loglog}\PY{p}{(}\PY{n}{n}\PY{p}{,}\PY{n}{t\PYZus{}GE}\PY{p}{,}\PY{n}{n}\PY{p}{,}\PY{n}{t\PYZus{}TD}\PY{p}{,}\PY{n}{n}\PY{p}{,}\PY{n}{t\PYZus{}GE\PYZus{}tridiag}\PY{p}{)}
+ \PY{n+nb}{legend}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{Gauss elim\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{Matlab \PYZbs{}\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{TriDiag\PYZsq{}}\PY{p}{)}
+ \PY{n+nb}{xlabel}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{number of elements\PYZsq{}}\PY{p}{)}
+ \PY{n+nb}{ylabel}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{time (s)\PYZsq{}}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{center}
+ \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_10_files/lecture_10_29_0.pdf}
+ \end{center}
+ { \hspace*{\fill} \\}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}29}]:} \PY{p}{[}\PY{n}{x}\PY{p}{,}\PY{n}{Aug}\PY{p}{,}\PY{n}{npivots}\PY{p}{]}\PY{p}{=}\PY{n}{GaussPivot}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{n}{y}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+x =
+
+ 9.8100
+ 27.4680
+ 61.8030
+ 101.0430
+
+Aug =
+
+ 15.00000 -5.00000 0.00000 0.00000 9.81000
+ 0.00000 5.33333 -2.00000 0.00000 22.89000
+ 0.00000 0.00000 2.25000 -1.00000 38.01375
+ 0.00000 0.00000 0.00000 0.55556 56.13500
+
+npivots = 0
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}30}]:} \PY{n}{A}\PY{p}{=}\PY{n}{Aug}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{:}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{:}\PY{l+m+mi}{4}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+A =
+
+ 15.00000 -5.00000 0.00000 0.00000
+ 0.00000 5.33333 -2.00000 0.00000
+ 0.00000 0.00000 2.25000 -1.00000
+ 0.00000 0.00000 0.00000 0.55556
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}31}]:} \PY{n+nb}{det}\PY{p}{(}\PY{n}{A}\PY{p}{)}
+ \PY{n}{detA}\PY{p}{=}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+ans = 100.00
+detA = 100.00
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor} }]:}
+\end{Verbatim}
+
+
+ % Add a bibliography block to the postdoc
+
+
+
+ \end{document}
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+ </g>
+ <g id="gnuplot_plot_2a"><title>gnuplot_plot_2a</title>
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+ </g>
+ <g id="gnuplot_plot_3a"><title>gnuplot_plot_3a</title>
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+ <g id="gnuplot_plot_4a"><title>gnuplot_plot_4a</title>
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diff --git a/lecture_10/nohup.out b/lecture_10/nohup.out
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diff --git a/lecture_10/octave-workspace b/lecture_10/octave-workspace
new file mode 100644
index 0000000..8a9aba2
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diff --git a/lecture_11/.ipynb_checkpoints/lecture_11-checkpoint.ipynb b/lecture_11/.ipynb_checkpoints/lecture_11-checkpoint.ipynb
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+{
+ "cells": [
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "%plot --format svg"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "setdefaults"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "What is the determinant of A?\n",
+ "\n",
+ "$A=\\left[ \\begin{array}{ccc}\n",
+ "10 & 2 & 1 \\\\\n",
+ "0 & 1 & 1 \\\\\n",
+ "0 & 0 & 10\\end{array} \\right]$\n",
+ "\n",
+ ""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# LU Decomposition\n",
+ "### efficient storage of matrices for solutions\n",
+ "\n",
+ "Considering the same solution set:\n",
+ "\n",
+ "$y=Ax$\n",
+ "\n",
+ "Assume that we can perform Gauss elimination and achieve this formula:\n",
+ "\n",
+ "$Ux=d$ \n",
+ "\n",
+ "Where, $U$ is an upper triangular matrix that we derived from Gauss elimination and $d$ is the set of dependent variables after Gauss elimination. \n",
+ "\n",
+ "Assume there is a lower triangular matrix, $L$, with ones on the diagonal and same dimensions of $U$ and the following is true:\n",
+ "\n",
+ "$L(Ux-d)=Ax-y=0$\n",
+ "\n",
+ "Now, $Ax=LUx$, so $A=LU$, and $y=Ld$.\n",
+ "\n",
+ "$2x_{1}+x_{2}=1$\n",
+ "\n",
+ "$x_{1}+3x_{2}=1$\n",
+ "\n",
+ "\n",
+ "$\\left[ \\begin{array}{cc}\n",
+ "2 & 1 \\\\\n",
+ "1 & 3 \\end{array} \\right]\n",
+ "\\left[\\begin{array}{c} \n",
+ "x_{1} \\\\ \n",
+ "x_{2} \\end{array}\\right]=\n",
+ "\\left[\\begin{array}{c} \n",
+ "1 \\\\\n",
+ "1\\end{array}\\right]$\n",
+ "\n",
+ "f21=0.5\n",
+ "\n",
+ "A(2,1)=1-1 = 0 \n",
+ "\n",
+ "A(2,2)=3-0.5=2.5\n",
+ "\n",
+ "y(2)=1-0.5=0.5\n",
+ "\n",
+ "$L(Ux-d)=\n",
+ "\\left[ \\begin{array}{cc}\n",
+ "1 & 0 \\\\\n",
+ "0.5 & 1 \\end{array} \\right]\n",
+ "\\left(\\left[ \\begin{array}{cc}\n",
+ "2 & 1 \\\\\n",
+ "0 & 2.5 \\end{array} \\right]\n",
+ "\\left[\\begin{array}{c} \n",
+ "x_{1} \\\\ \n",
+ "x_{2} \\end{array}\\right]-\n",
+ "\\left[\\begin{array}{c} \n",
+ "1 \\\\\n",
+ "0.5\\end{array}\\right]\\right)=0$\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A =\n",
+ "\n",
+ " 2 1\n",
+ " 1 3\n",
+ "\n",
+ "L =\n",
+ "\n",
+ " 1.00000 0.00000\n",
+ " 0.50000 1.00000\n",
+ "\n",
+ "U =\n",
+ "\n",
+ " 2.00000 1.00000\n",
+ " 0.00000 2.50000\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " 2 1\n",
+ " 1 3\n",
+ "\n",
+ "d =\n",
+ "\n",
+ " 1.00000\n",
+ " 0.50000\n",
+ "\n",
+ "y =\n",
+ "\n",
+ " 1\n",
+ " 1\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "A=[2,1;1,3]\n",
+ "L=[1,0;0.5,1]\n",
+ "U=[2,1;0,2.5]\n",
+ "L*U\n",
+ "\n",
+ "d=[1;0.5]\n",
+ "y=L*d"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Pivoting for LU factorization\n",
+ "\n",
+ "LU factorization uses the same method as Gauss elimination so it is also necessary to perform partial pivoting when creating the lower and upper triangular matrices. \n",
+ "\n",
+ "Matlab and Octave use pivoting in the command \n",
+ "\n",
+ "`[L,U,P]=lu(A)`\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "'lu' is a built-in function from the file libinterp/corefcn/lu.cc\n",
+ "\n",
+ " -- Built-in Function: [L, U] = lu (A)\n",
+ " -- Built-in Function: [L, U, P] = lu (A)\n",
+ " -- Built-in Function: [L, U, P, Q] = lu (S)\n",
+ " -- Built-in Function: [L, U, P, Q, R] = lu (S)\n",
+ " -- Built-in Function: [...] = lu (S, THRES)\n",
+ " -- Built-in Function: Y = lu (...)\n",
+ " -- Built-in Function: [...] = lu (..., \"vector\")\n",
+ " Compute the LU decomposition of A.\n",
+ "\n",
+ " If A is full subroutines from LAPACK are used and if A is sparse\n",
+ " then UMFPACK is used.\n",
+ "\n",
+ " The result is returned in a permuted form, according to the\n",
+ " optional return value P. For example, given the matrix 'a = [1, 2;\n",
+ " 3, 4]',\n",
+ "\n",
+ " [l, u, p] = lu (A)\n",
+ "\n",
+ " returns\n",
+ "\n",
+ " l =\n",
+ "\n",
+ " 1.00000 0.00000\n",
+ " 0.33333 1.00000\n",
+ "\n",
+ " u =\n",
+ "\n",
+ " 3.00000 4.00000\n",
+ " 0.00000 0.66667\n",
+ "\n",
+ " p =\n",
+ "\n",
+ " 0 1\n",
+ " 1 0\n",
+ "\n",
+ " The matrix is not required to be square.\n",
+ "\n",
+ " When called with two or three output arguments and a spare input\n",
+ " matrix, 'lu' does not attempt to perform sparsity preserving column\n",
+ " permutations. Called with a fourth output argument, the sparsity\n",
+ " preserving column transformation Q is returned, such that 'P * A *\n",
+ " Q = L * U'.\n",
+ "\n",
+ " Called with a fifth output argument and a sparse input matrix, 'lu'\n",
+ " attempts to use a scaling factor R on the input matrix such that 'P\n",
+ " * (R \\ A) * Q = L * U'. This typically leads to a sparser and more\n",
+ " stable factorization.\n",
+ "\n",
+ " An additional input argument THRES, that defines the pivoting\n",
+ " threshold can be given. THRES can be a scalar, in which case it\n",
+ " defines the UMFPACK pivoting tolerance for both symmetric and\n",
+ " unsymmetric cases. If THRES is a 2-element vector, then the first\n",
+ " element defines the pivoting tolerance for the unsymmetric UMFPACK\n",
+ " pivoting strategy and the second for the symmetric strategy. By\n",
+ " default, the values defined by 'spparms' are used ([0.1, 0.001]).\n",
+ "\n",
+ " Given the string argument \"vector\", 'lu' returns the values of P\n",
+ " and Q as vector values, such that for full matrix, 'A (P,:) = L *\n",
+ " U', and 'R(P,:) * A (:, Q) = L * U'.\n",
+ "\n",
+ " With two output arguments, returns the permuted forms of the upper\n",
+ " and lower triangular matrices, such that 'A = L * U'. With one\n",
+ " output argument Y, then the matrix returned by the LAPACK routines\n",
+ " is returned. If the input matrix is sparse then the matrix L is\n",
+ " embedded into U to give a return value similar to the full case.\n",
+ " For both full and sparse matrices, 'lu' loses the permutation\n",
+ " information.\n",
+ "\n",
+ " See also: luupdate, ilu, chol, hess, qr, qz, schur, svd.\n",
+ "\n",
+ "Additional help for built-in functions and operators is\n",
+ "available in the online version of the manual. Use the command\n",
+ "'doc <topic>' to search the manual index.\n",
+ "\n",
+ "Help and information about Octave is also available on the WWW\n",
+ "at http://www.octave.org and via the help@octave.org\n",
+ "mailing list.\n"
+ ]
+ }
+ ],
+ "source": [
+ "help lu"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 22,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
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+ "\t<path d=\"M78.8,16.7 L78.8,384.0 L535.0,384.0 L535.0,16.7 L78.8,16.7 Z \" stroke=\"black\"/></g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"black\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"1.00\">\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"1.00\">\n",
+ "\t<path d=\"M349.7,121.7 L349.7,25.7 L526.7,25.7 L526.7,121.7 L349.7,121.7 Z \" stroke=\"black\"/></g>\n",
+ "\t<g id=\"gnuplot_plot_1a\"><title>LU decomp</title>\n",
+ "<g color=\"white\" fill=\"none\" stroke=\"black\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"3.00\">\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"3.00\">\n",
+ "\t<g fill=\"rgb(0,0,0)\" font-family=\"{}\" font-size=\"16.00\" stroke=\"none\" text-anchor=\"end\" transform=\"translate(526.7,55.7)\">\n",
+ "\t\t<text><tspan font-family=\"{}\">LU decomp</tspan></text>\n",
+ "\t</g>\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"3.00\">\n",
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+ "\t</g>\n",
+ "\t<g id=\"gnuplot_plot_2a\"><title>Octave \\\\</title>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"3.00\">\n",
+ "\t<g fill=\"rgb(0,0,0)\" font-family=\"{}\" font-size=\"16.00\" stroke=\"none\" text-anchor=\"end\" transform=\"translate(526.7,103.7)\">\n",
+ "\t\t<text><tspan font-family=\"{}\">Octave \\</tspan></text>\n",
+ "\t</g>\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"3.00\">\n",
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+ "\t</g>\n",
+ "<g color=\"white\" fill=\"none\" stroke=\"rgb( 0, 128, 0)\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"2.00\">\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"2.00\">\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"black\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "</g>\n",
+ "</g>\n",
+ "</svg>"
+ ],
+ "text/plain": [
+ "<IPython.core.display.SVG object>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "% time LU solution vs backslash\n",
+ "t_lu=zeros(100,1);\n",
+ "t_bs=zeros(100,1);\n",
+ "for N=1:100\n",
+ " A=rand(N,N);\n",
+ " y=rand(N,1);\n",
+ " [L,U,P]=lu(A);\n",
+ "\n",
+ " tic; d=L\\y; x=U\\d; t_lu(N)=toc;\n",
+ "\n",
+ " tic; x=A\\y; t_bs(N)=toc;\n",
+ "end\n",
+ "plot([1:100],t_lu,[1:100],t_bs) \n",
+ "legend('LU decomp','Octave \\\\')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "Consider the problem again from the intro to Linear Algebra, 4 masses are connected in series to 4 springs with K=10 N/m. What are the final positions of the masses? \n",
+ "\n",
+ "\n",
+ "\n",
+ "The masses haves the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass:\n",
+ "\n",
+ "$m_{1}g+k(x_{2}-x_{1})-kx_{1}=0$\n",
+ "\n",
+ "$m_{2}g+k(x_{3}-x_{2})-k(x_{2}-x_{1})=0$\n",
+ "\n",
+ "$m_{3}g+k(x_{4}-x_{3})-k(x_{3}-x_{2})=0$\n",
+ "\n",
+ "$m_{4}g-k(x_{4}-x_{3})=0$\n",
+ "\n",
+ "in matrix form:\n",
+ "\n",
+ "$\\left[ \\begin{array}{cccc}\n",
+ "2k & -k & 0 & 0 \\\\\n",
+ "-k & 2k & -k & 0 \\\\\n",
+ "0 & -k & 2k & -k \\\\\n",
+ "0 & 0 & -k & k \\end{array} \\right]\n",
+ "\\left[ \\begin{array}{c}\n",
+ "x_{1} \\\\\n",
+ "x_{2} \\\\\n",
+ "x_{3} \\\\\n",
+ "x_{4} \\end{array} \\right]=\n",
+ "\\left[ \\begin{array}{c}\n",
+ "m_{1}g \\\\\n",
+ "m_{2}g \\\\\n",
+ "m_{3}g \\\\\n",
+ "m_{4}g \\end{array} \\right]$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 24,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "K =\n",
+ "\n",
+ " 20 -10 0 0\n",
+ " -10 20 -10 0\n",
+ " 0 -10 20 -10\n",
+ " 0 0 -10 10\n",
+ "\n",
+ "y =\n",
+ "\n",
+ " 9.8100\n",
+ " 19.6200\n",
+ " 29.4300\n",
+ " 39.2400\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "k=10; % N/m\n",
+ "m1=1; % kg\n",
+ "m2=2;\n",
+ "m3=3;\n",
+ "m4=4;\n",
+ "g=9.81; % m/s^2\n",
+ "K=[2*k -k 0 0; -k 2*k -k 0; 0 -k 2*k -k; 0 0 -k k]\n",
+ "y=[m1*g;m2*g;m3*g;m4*g]"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This matrix, K, is symmetric. \n",
+ "\n",
+ "`K(i,j)==K(j,i)`\n",
+ "\n",
+ "Now we can use,\n",
+ "\n",
+ "## Cholesky Factorization\n",
+ "\n",
+ "We can decompose the matrix, K into two matrices, $U$ and $U^{T}$, where\n",
+ "\n",
+ "$K=U^{T}U$\n",
+ "\n",
+ "each of the components of U can be calculated with the following equations:\n",
+ "\n",
+ "$u_{ii}=\\sqrt{a_{ii}-\\sum_{k=1}^{i-1}u_{ki}^{2}}$\n",
+ "\n",
+ "$u_{ij}=\\frac{a_{ij}-\\sum_{k=1}^{i-1}u_{ki}u_{kj}}{u_{ii}}$\n",
+ "\n",
+ "so for K"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "K =\n",
+ "\n",
+ " 20 -10 0 0\n",
+ " -10 20 -10 0\n",
+ " 0 -10 20 -10\n",
+ " 0 0 -10 10\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "K"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "u11 = 4.4721\n",
+ "u12 = -2.2361\n",
+ "u13 = 0\n",
+ "u14 = 0\n",
+ "u22 = 3.8730\n",
+ "u23 = -2.5820\n",
+ "u24 = 0\n",
+ "u33 = 3.6515\n",
+ "u34 = -2.7386\n",
+ "u44 = 1.5811\n",
+ "U =\n",
+ "\n",
+ " 4.47214 -2.23607 0.00000 0.00000\n",
+ " 0.00000 3.87298 -2.58199 0.00000\n",
+ " 0.00000 0.00000 3.65148 -2.73861\n",
+ " 0.00000 0.00000 0.00000 1.58114\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "u11=sqrt(K(1,1))\n",
+ "u12=(K(1,2))/u11\n",
+ "u13=(K(1,3))/u11\n",
+ "u14=(K(1,4))/u11\n",
+ "u22=sqrt(K(2,2)-u12^2)\n",
+ "u23=(K(2,3)-u12*u13)/u22\n",
+ "u24=(K(2,4)-u12*u14)/u22\n",
+ "u33=sqrt(K(3,3)-u13^2-u23^2)\n",
+ "u34=(K(3,4)-u13*u14-u23*u24)/u33\n",
+ "u44=sqrt(K(4,4)-u14^2-u24^2-u34^2)\n",
+ "U=[u11,u12,u13,u14;0,u22,u23,u24;0,0,u33,u34;0,0,0,u44]"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 27,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans =\n",
+ "\n",
+ " 20.00000 -10.00000 0.00000 0.00000\n",
+ " -10.00000 20.00000 -10.00000 0.00000\n",
+ " 0.00000 -10.00000 20.00000 -10.00000\n",
+ " 0.00000 0.00000 -10.00000 10.00000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "U'*U"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 37,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "average time spent for Cholesky factored solution = 1.623154e-05+/-1.166726e-05\n",
+ "average time spent for backslash solution = 1.675844e-05+/-1.187234e-05\n"
+ ]
+ }
+ ],
+ "source": [
+ "% time solution for Cholesky vs backslash\n",
+ "t_chol=zeros(1000,1);\n",
+ "t_bs=zeros(1000,1);\n",
+ "for i=1:1000\n",
+ " tic; d=U'*y; x=U\\d; t_chol(i)=toc;\n",
+ " tic; x=K\\y; t_bs(i)=toc;\n",
+ "end\n",
+ "fprintf('average time spent for Cholesky factored solution = %e+/-%e',mean(t_chol),std(t_chol))\n",
+ "\n",
+ "fprintf('average time spent for backslash solution = %e+/-%e',mean(t_bs),std(t_bs))"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Octave",
+ "language": "octave",
+ "name": "octave"
+ },
+ "language_info": {
+ "file_extension": ".m",
+ "help_links": [
+ {
+ "text": "MetaKernel Magics",
+ "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+ }
+ ],
+ "mimetype": "text/x-octave",
+ "name": "octave",
+ "version": "0.19.14"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
diff --git a/lecture_11/LU_naive.m b/lecture_11/LU_naive.m
new file mode 100644
index 0000000..92efde6
--- /dev/null
+++ b/lecture_11/LU_naive.m
@@ -0,0 +1,27 @@
+function [L, U] = LU_naive(A)
+% GaussNaive: naive Gauss elimination
+% x = GaussNaive(A,b): Gauss elimination without pivoting.
+% input:
+% A = coefficient matrix
+% y = right hand side vector
+% output:
+% x = solution vector
+[m,n] = size(A);
+if m~=n, error('Matrix A must be square'); end
+nb = n;
+L=diag(ones(n,1));
+U=A;
+% forward elimination
+for k = 1:n-1
+ for i = k+1:n
+ fik = U(i,k)/U(k,k);
+ L(i,k)=fik;
+ U(i,k:nb) = U(i,k:nb)-fik*U(k,k:nb);
+ end
+end
+%% back substitution
+%x = zeros(n,1);
+%x(n) = Aug(n,nb)/Aug(n,n);
+%for i = n-1:-1:1
+% x(i) = (Aug(i,nb)-Aug(i,i+1:n)*x(i+1:n))/Aug(i,i);
+%end
diff --git a/lecture_11/LU_pivot.m b/lecture_11/LU_pivot.m
new file mode 100644
index 0000000..37abb26
--- /dev/null
+++ b/lecture_11/LU_pivot.m
@@ -0,0 +1,36 @@
+function [L,U,P] = LU_pivot(A)
+% GaussPivot: Gauss elimination pivoting
+% x = GaussPivot(A,b): Gauss elimination with pivoting.
+% input:
+% A = coefficient matrix
+% b = right hand side vector
+% output:
+% x = solution vector
+[m,n]=size(A);
+if m~=n, error('Matrix A must be square'); end
+nb = n;
+L=diag(ones(n,1));
+U=A;
+P=diag(ones(n,1));
+% forward elimination
+for k = 1:n-1
+ % partial pivoting
+ [big,i]=max(abs(U(k:n,k)));
+ ipr=i+k-1;
+ if ipr~=k
+ P([k,ipr],:)=P([ipr,k],:); % if the max is not the current index ipr, pivot count
+ %L([k,ipr],:)=L([ipr,k],:);
+ %U([k,ipr],:)=U([ipr,k],:);
+ end
+ for i = k+1:n
+ fik=U(i,k)/U(k,k);
+ L(i,k)=fik;
+ U(i,k:nb)=U(i,k:nb)-fik*U(k,k:nb);
+ end
+end
+% back substitution
+%x=zeros(n,1);
+%x(n)=Aug(n,nb)/Aug(n,n);
+%for i = n-1:-1:1
+% x(i)=(Aug(i,nb)-Aug(i,i+1:n)*x(i+1:n))/Aug(i,i);
+%end
diff --git a/lecture_11/det_A.png b/lecture_11/det_A.png
new file mode 100644
index 0000000..50f6ac1
Binary files /dev/null and b/lecture_11/det_A.png differ
diff --git a/lecture_11/lecture_11.aux b/lecture_11/lecture_11.aux
new file mode 100644
index 0000000..a9c4771
--- /dev/null
+++ b/lecture_11/lecture_11.aux
@@ -0,0 +1,38 @@
+\relax
+\providecommand\hyper@newdestlabel[2]{}
+\providecommand\HyperFirstAtBeginDocument{\AtBeginDocument}
+\HyperFirstAtBeginDocument{\ifx\hyper@anchor\@undefined
+\global\let\oldcontentsline\contentsline
+\gdef\contentsline#1#2#3#4{\oldcontentsline{#1}{#2}{#3}}
+\global\let\oldnewlabel\newlabel
+\gdef\newlabel#1#2{\newlabelxx{#1}#2}
+\gdef\newlabelxx#1#2#3#4#5#6{\oldnewlabel{#1}{{#2}{#3}}}
+\AtEndDocument{\ifx\hyper@anchor\@undefined
+\let\contentsline\oldcontentsline
+\let\newlabel\oldnewlabel
+\fi}
+\fi}
+\global\let\hyper@last\relax
+\gdef\HyperFirstAtBeginDocument#1{#1}
+\providecommand\HyField@AuxAddToFields[1]{}
+\providecommand\HyField@AuxAddToCoFields[2]{}
+\providecommand \oddpage@label [2]{}
+\@writefile{toc}{\contentsline {subsection}{\numberline {0.1}My question from last class}{1}{subsection.0.1}}
+\newlabel{my-question-from-last-class}{{0.1}{1}{My question from last class}{subsection.0.1}{}}
+\@writefile{lof}{\contentsline {figure}{\numberline {1}{\ignorespaces responses to determinant of A\relax }}{1}{figure.caption.1}}
+\@writefile{toc}{\contentsline {subsection}{\numberline {0.2}Your questions from last class}{1}{subsection.0.2}}
+\newlabel{your-questions-from-last-class}{{0.2}{1}{Your questions from last class}{subsection.0.2}{}}
+\@writefile{toc}{\contentsline {subsection}{\numberline {0.3}Midterm preference}{2}{subsection.0.3}}
+\newlabel{midterm-preference}{{0.3}{2}{Midterm preference}{subsection.0.3}{}}
+\@writefile{lof}{\contentsline {figure}{\numberline {2}{\ignorespaces responses to midterm date\relax }}{2}{figure.caption.2}}
+\@writefile{toc}{\contentsline {subsubsection}{\numberline {0.3.1}Midterm Questions}{2}{subsubsection.0.3.1}}
+\newlabel{midterm-questions}{{0.3.1}{2}{Midterm Questions}{subsubsection.0.3.1}{}}
+\@writefile{toc}{\contentsline {section}{\numberline {1}LU Decomposition}{2}{section.1}}
+\newlabel{lu-decomposition}{{1}{2}{LU Decomposition}{section.1}{}}
+\@writefile{toc}{\contentsline {subsubsection}{\numberline {1.0.1}efficient storage of matrices for solutions}{2}{subsubsection.1.0.1}}
+\newlabel{efficient-storage-of-matrices-for-solutions}{{1.0.1}{2}{efficient storage of matrices for solutions}{subsubsection.1.0.1}{}}
+\@writefile{toc}{\contentsline {subsection}{\numberline {1.1}Pivoting for LU factorization}{4}{subsection.1.1}}
+\newlabel{pivoting-for-lu-factorization}{{1.1}{4}{Pivoting for LU factorization}{subsection.1.1}{}}
+\@writefile{lof}{\contentsline {figure}{\numberline {3}{\ignorespaces Springs-masses\relax }}{7}{figure.caption.3}}
+\@writefile{toc}{\contentsline {subsection}{\numberline {1.2}Cholesky Factorization}{8}{subsection.1.2}}
+\newlabel{cholesky-factorization}{{1.2}{8}{Cholesky Factorization}{subsection.1.2}{}}
diff --git a/lecture_11/lecture_11.bbl b/lecture_11/lecture_11.bbl
new file mode 100644
index 0000000..e69de29
diff --git a/lecture_11/lecture_11.blg b/lecture_11/lecture_11.blg
new file mode 100644
index 0000000..f6400af
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diff --git a/lecture_11/lecture_11.ipynb b/lecture_11/lecture_11.ipynb
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+{
+ "cells": [
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "%plot --format svg"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "setdefaults"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## My question from last class \n",
+ "\n",
+ "$A=\\left[ \\begin{array}{ccc}\n",
+ "10 & 2 & 1 \\\\\n",
+ "0 & 1 & 1 \\\\\n",
+ "0 & 0 & 10\\end{array} \\right]$\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Your questions from last class\n",
+ "\n",
+ "1. Need more linear algebra review\n",
+ " \n",
+ " -We will keep doing Linear Algebra, try the practice problems in 'linear_algebra'\n",
+ "\n",
+ "2. How do I do HW3? \n",
+ " \n",
+ " -demo today\n",
+ "\n",
+ "3. For hw4 is the spring constant (K) suppose to be given? \n",
+ " \n",
+ " -yes, its 30 N/m\n",
+ " \n",
+ "4. Deapool or Joker?\n",
+ "\n",
+ "\n",
+ "## Midterm preference\n",
+ "\n",
+ "\n",
+ "\n",
+ "### Midterm Questions\n",
+ "\n",
+ "1. Notes allowed\n",
+ " \n",
+ " -no\n",
+ "\n",
+ "2. Will there be a review/study sheet\n",
+ "\n",
+ " -yes"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# LU Decomposition\n",
+ "### efficient storage of matrices for solutions\n",
+ "\n",
+ "Considering the same solution set:\n",
+ "\n",
+ "$y=Ax$\n",
+ "\n",
+ "Assume that we can perform Gauss elimination and achieve this formula:\n",
+ "\n",
+ "$Ux=d$ \n",
+ "\n",
+ "Where, $U$ is an upper triangular matrix that we derived from Gauss elimination and $d$ is the set of dependent variables after Gauss elimination. \n",
+ "\n",
+ "Assume there is a lower triangular matrix, $L$, with ones on the diagonal and same dimensions of $U$ and the following is true:\n",
+ "\n",
+ "$L(Ux-d)=Ax-y=0$\n",
+ "\n",
+ "Now, $Ax=LUx$, so $A=LU$, and $y=Ld$.\n",
+ "\n",
+ "$2x_{1}+x_{2}=1$\n",
+ "\n",
+ "$x_{1}+3x_{2}=1$\n",
+ "\n",
+ "\n",
+ "$\\left[ \\begin{array}{cc}\n",
+ "2 & 1 \\\\\n",
+ "1 & 3 \\end{array} \\right]\n",
+ "\\left[\\begin{array}{c} \n",
+ "x_{1} \\\\ \n",
+ "x_{2} \\end{array}\\right]=\n",
+ "\\left[\\begin{array}{c} \n",
+ "1 \\\\\n",
+ "1\\end{array}\\right]$\n",
+ "\n",
+ "f21=0.5\n",
+ "\n",
+ "A(2,1)=1-1 = 0 \n",
+ "\n",
+ "A(2,2)=3-0.5=2.5\n",
+ "\n",
+ "y(2)=1-0.5=0.5\n",
+ "\n",
+ "$L(Ux-d)=\n",
+ "\\left[ \\begin{array}{cc}\n",
+ "1 & 0 \\\\\n",
+ "0.5 & 1 \\end{array} \\right]\n",
+ "\\left(\\left[ \\begin{array}{cc}\n",
+ "2 & 1 \\\\\n",
+ "0 & 2.5 \\end{array} \\right]\n",
+ "\\left[\\begin{array}{c} \n",
+ "x_{1} \\\\ \n",
+ "x_{2} \\end{array}\\right]-\n",
+ "\\left[\\begin{array}{c} \n",
+ "1 \\\\\n",
+ "0.5\\end{array}\\right]\\right)=0$\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A =\n",
+ "\n",
+ " 2 1\n",
+ " 1 3\n",
+ "\n",
+ "L =\n",
+ "\n",
+ " 1.00000 0.00000\n",
+ " 0.50000 1.00000\n",
+ "\n",
+ "U =\n",
+ "\n",
+ " 2.00000 1.00000\n",
+ " 0.00000 2.50000\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " 2 1\n",
+ " 1 3\n",
+ "\n",
+ "d =\n",
+ "\n",
+ " 1.00000\n",
+ " 0.50000\n",
+ "\n",
+ "y =\n",
+ "\n",
+ " 1\n",
+ " 1\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "A=[2,1;1,3]\n",
+ "L=[1,0;0.5,1]\n",
+ "U=[2,1;0,2.5]\n",
+ "L*U\n",
+ "\n",
+ "d=[1;0.5]\n",
+ "y=L*d"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans = 5.0000\n",
+ "ans = 1\n",
+ "ans = 5\n",
+ "ans = 5\n",
+ "ans = 5.0000\n"
+ ]
+ }
+ ],
+ "source": [
+ "% what is the determinant of L, U and A?\n",
+ "\n",
+ "det(A)\n",
+ "det(L)\n",
+ "det(U)\n",
+ "det(L)*det(U)\n",
+ "det(L*U)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Pivoting for LU factorization\n",
+ "\n",
+ "LU factorization uses the same method as Gauss elimination so it is also necessary to perform partial pivoting when creating the lower and upper triangular matrices. \n",
+ "\n",
+ "Matlab and Octave use pivoting in the command \n",
+ "\n",
+ "`[L,U,P]=lu(A)`\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "'lu' is a built-in function from the file libinterp/corefcn/lu.cc\n",
+ "\n",
+ " -- Built-in Function: [L, U] = lu (A)\n",
+ " -- Built-in Function: [L, U, P] = lu (A)\n",
+ " -- Built-in Function: [L, U, P, Q] = lu (S)\n",
+ " -- Built-in Function: [L, U, P, Q, R] = lu (S)\n",
+ " -- Built-in Function: [...] = lu (S, THRES)\n",
+ " -- Built-in Function: Y = lu (...)\n",
+ " -- Built-in Function: [...] = lu (..., \"vector\")\n",
+ " Compute the LU decomposition of A.\n",
+ "\n",
+ " If A is full subroutines from LAPACK are used and if A is sparse\n",
+ " then UMFPACK is used.\n",
+ "\n",
+ " The result is returned in a permuted form, according to the\n",
+ " optional return value P. For example, given the matrix 'a = [1, 2;\n",
+ " 3, 4]',\n",
+ "\n",
+ " [l, u, p] = lu (A)\n",
+ "\n",
+ " returns\n",
+ "\n",
+ " l =\n",
+ "\n",
+ " 1.00000 0.00000\n",
+ " 0.33333 1.00000\n",
+ "\n",
+ " u =\n",
+ "\n",
+ " 3.00000 4.00000\n",
+ " 0.00000 0.66667\n",
+ "\n",
+ " p =\n",
+ "\n",
+ " 0 1\n",
+ " 1 0\n",
+ "\n",
+ " The matrix is not required to be square.\n",
+ "\n",
+ " When called with two or three output arguments and a spare input\n",
+ " matrix, 'lu' does not attempt to perform sparsity preserving column\n",
+ " permutations. Called with a fourth output argument, the sparsity\n",
+ " preserving column transformation Q is returned, such that 'P * A *\n",
+ " Q = L * U'.\n",
+ "\n",
+ " Called with a fifth output argument and a sparse input matrix, 'lu'\n",
+ " attempts to use a scaling factor R on the input matrix such that 'P\n",
+ " * (R \\ A) * Q = L * U'. This typically leads to a sparser and more\n",
+ " stable factorization.\n",
+ "\n",
+ " An additional input argument THRES, that defines the pivoting\n",
+ " threshold can be given. THRES can be a scalar, in which case it\n",
+ " defines the UMFPACK pivoting tolerance for both symmetric and\n",
+ " unsymmetric cases. If THRES is a 2-element vector, then the first\n",
+ " element defines the pivoting tolerance for the unsymmetric UMFPACK\n",
+ " pivoting strategy and the second for the symmetric strategy. By\n",
+ " default, the values defined by 'spparms' are used ([0.1, 0.001]).\n",
+ "\n",
+ " Given the string argument \"vector\", 'lu' returns the values of P\n",
+ " and Q as vector values, such that for full matrix, 'A (P,:) = L *\n",
+ " U', and 'R(P,:) * A (:, Q) = L * U'.\n",
+ "\n",
+ " With two output arguments, returns the permuted forms of the upper\n",
+ " and lower triangular matrices, such that 'A = L * U'. With one\n",
+ " output argument Y, then the matrix returned by the LAPACK routines\n",
+ " is returned. If the input matrix is sparse then the matrix L is\n",
+ " embedded into U to give a return value similar to the full case.\n",
+ " For both full and sparse matrices, 'lu' loses the permutation\n",
+ " information.\n",
+ "\n",
+ " See also: luupdate, ilu, chol, hess, qr, qz, schur, svd.\n",
+ "\n",
+ "Additional help for built-in functions and operators is\n",
+ "available in the online version of the manual. Use the command\n",
+ "'doc <topic>' to search the manual index.\n",
+ "\n",
+ "Help and information about Octave is also available on the WWW\n",
+ "at http://www.octave.org and via the help@octave.org\n",
+ "mailing list.\n"
+ ]
+ }
+ ],
+ "source": [
+ "help lu"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
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+ "</g>\n",
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+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "</g>\n",
+ "</g>\n",
+ "</svg>"
+ ],
+ "text/plain": [
+ "<IPython.core.display.SVG object>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "% time LU solution vs backslash\n",
+ "t_lu=zeros(100,1);\n",
+ "t_bs=zeros(100,1);\n",
+ "for N=1:100\n",
+ " A=rand(N,N);\n",
+ " y=rand(N,1);\n",
+ " [L,U,P]=lu(A);\n",
+ "\n",
+ " tic; d=inv(L)*y; x=inv(U)*d; t_lu(N)=toc;\n",
+ "\n",
+ " tic; x=inv(A)*y; t_bs(N)=toc;\n",
+ "end\n",
+ "plot([1:100],t_lu,[1:100],t_bs) \n",
+ "legend('LU decomp','Octave \\\\')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "Consider the problem again from the intro to Linear Algebra, 4 masses are connected in series to 4 springs with K=10 N/m. What are the final positions of the masses? \n",
+ "\n",
+ "\n",
+ "\n",
+ "The masses haves the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass:\n",
+ "\n",
+ "$m_{1}g+k(x_{2}-x_{1})-kx_{1}=0$\n",
+ "\n",
+ "$m_{2}g+k(x_{3}-x_{2})-k(x_{2}-x_{1})=0$\n",
+ "\n",
+ "$m_{3}g+k(x_{4}-x_{3})-k(x_{3}-x_{2})=0$\n",
+ "\n",
+ "$m_{4}g-k(x_{4}-x_{3})=0$\n",
+ "\n",
+ "in matrix form:\n",
+ "\n",
+ "$\\left[ \\begin{array}{cccc}\n",
+ "2k & -k & 0 & 0 \\\\\n",
+ "-k & 2k & -k & 0 \\\\\n",
+ "0 & -k & 2k & -k \\\\\n",
+ "0 & 0 & -k & k \\end{array} \\right]\n",
+ "\\left[ \\begin{array}{c}\n",
+ "x_{1} \\\\\n",
+ "x_{2} \\\\\n",
+ "x_{3} \\\\\n",
+ "x_{4} \\end{array} \\right]=\n",
+ "\\left[ \\begin{array}{c}\n",
+ "m_{1}g \\\\\n",
+ "m_{2}g \\\\\n",
+ "m_{3}g \\\\\n",
+ "m_{4}g \\end{array} \\right]$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "K =\n",
+ "\n",
+ " 20 -10 0 0\n",
+ " -10 20 -10 0\n",
+ " 0 -10 20 -10\n",
+ " 0 0 -10 10\n",
+ "\n",
+ "y =\n",
+ "\n",
+ " 9.8100\n",
+ " 19.6200\n",
+ " 29.4300\n",
+ " 39.2400\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "k=10; % N/m\n",
+ "m1=1; % kg\n",
+ "m2=2;\n",
+ "m3=3;\n",
+ "m4=4;\n",
+ "g=9.81; % m/s^2\n",
+ "K=[2*k -k 0 0; -k 2*k -k 0; 0 -k 2*k -k; 0 0 -k k]\n",
+ "y=[m1*g;m2*g;m3*g;m4*g]"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This matrix, K, is symmetric. \n",
+ "\n",
+ "`K(i,j)==K(j,i)`\n",
+ "\n",
+ "Now we can use,\n",
+ "\n",
+ "## Cholesky Factorization\n",
+ "\n",
+ "We can decompose the matrix, K into two matrices, $U$ and $U^{T}$, where\n",
+ "\n",
+ "$K=U^{T}U$\n",
+ "\n",
+ "each of the components of U can be calculated with the following equations:\n",
+ "\n",
+ "$u_{ii}=\\sqrt{a_{ii}-\\sum_{k=1}^{i-1}u_{ki}^{2}}$\n",
+ "\n",
+ "$u_{ij}=\\frac{a_{ij}-\\sum_{k=1}^{i-1}u_{ki}u_{kj}}{u_{ii}}$\n",
+ "\n",
+ "so for K"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "K =\n",
+ "\n",
+ " 20 -10 0 0\n",
+ " -10 20 -10 0\n",
+ " 0 -10 20 -10\n",
+ " 0 0 -10 10\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "K"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "u11 = 4.4721\n",
+ "u12 = -2.2361\n",
+ "u13 = 0\n",
+ "u14 = 0\n",
+ "u22 = 3.8730\n",
+ "u23 = -2.5820\n",
+ "u24 = 0\n",
+ "u33 = 3.6515\n",
+ "u34 = -2.7386\n",
+ "u44 = 1.5811\n",
+ "U =\n",
+ "\n",
+ " 4.47214 -2.23607 0.00000 0.00000\n",
+ " 0.00000 3.87298 -2.58199 0.00000\n",
+ " 0.00000 0.00000 3.65148 -2.73861\n",
+ " 0.00000 0.00000 0.00000 1.58114\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "u11=sqrt(K(1,1))\n",
+ "u12=(K(1,2))/u11\n",
+ "u13=(K(1,3))/u11\n",
+ "u14=(K(1,4))/u11\n",
+ "u22=sqrt(K(2,2)-u12^2)\n",
+ "u23=(K(2,3)-u12*u13)/u22\n",
+ "u24=(K(2,4)-u12*u14)/u22\n",
+ "u33=sqrt(K(3,3)-u13^2-u23^2)\n",
+ "u34=(K(3,4)-u13*u14-u23*u24)/u33\n",
+ "u44=sqrt(K(4,4)-u14^2-u24^2-u34^2)\n",
+ "U=[u11,u12,u13,u14;0,u22,u23,u24;0,0,u33,u34;0,0,0,u44]"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans =\n",
+ "\n",
+ " 1 1 1 1\n",
+ " 1 1 1 1\n",
+ " 1 1 1 1\n",
+ " 1 1 1 1\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "(U'*U)'==U'*U"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "average time spent for Cholesky factored solution = 1.465964e-05+/-9.806001e-06\n",
+ "average time spent for backslash solution = 1.555967e-05+/-1.048561e-05\n"
+ ]
+ }
+ ],
+ "source": [
+ "% time solution for Cholesky vs backslash\n",
+ "t_chol=zeros(1000,1);\n",
+ "t_bs=zeros(1000,1);\n",
+ "for i=1:1000\n",
+ " tic; d=U'*y; x=U\\d; t_chol(i)=toc;\n",
+ " tic; x=K\\y; t_bs(i)=toc;\n",
+ "end\n",
+ "fprintf('average time spent for Cholesky factored solution = %e+/-%e',mean(t_chol),std(t_chol))\n",
+ "\n",
+ "fprintf('average time spent for backslash solution = %e+/-%e',mean(t_bs),std(t_bs))"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Octave",
+ "language": "octave",
+ "name": "octave"
+ },
+ "language_info": {
+ "file_extension": ".m",
+ "help_links": [
+ {
+ "text": "MetaKernel Magics",
+ "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+ }
+ ],
+ "mimetype": "text/x-octave",
+ "name": "octave",
+ "version": "0.19.14"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
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diff --git a/lecture_11/lecture_11.out b/lecture_11/lecture_11.out
new file mode 100644
index 0000000..481e738
--- /dev/null
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+\BOOKMARK [2][-]{subsection.0.1}{My question from last class}{}% 1
+\BOOKMARK [2][-]{subsection.0.2}{Your questions from last class}{}% 2
+\BOOKMARK [2][-]{subsection.0.3}{Midterm preference}{}% 3
+\BOOKMARK [3][-]{subsubsection.0.3.1}{Midterm Questions}{subsection.0.3}% 4
+\BOOKMARK [1][-]{section.1}{LU Decomposition}{}% 5
+\BOOKMARK [2][-]{subsubsection.1.0.1}{efficient storage of matrices for solutions}{section.1}% 6
+\BOOKMARK [2][-]{subsection.1.1}{Pivoting for LU factorization}{section.1}% 7
+\BOOKMARK [2][-]{subsection.1.2}{Cholesky Factorization}{section.1}% 8
diff --git a/lecture_11/lecture_11.pdf b/lecture_11/lecture_11.pdf
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diff --git a/lecture_11/lecture_11.tex b/lecture_11/lecture_11.tex
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+\expandafter\def\csname PY@tok@m\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
+\expandafter\def\csname PY@tok@gp\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.00,0.50}{##1}}}
+\expandafter\def\csname PY@tok@sh\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
+\expandafter\def\csname PY@tok@ow\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.67,0.13,1.00}{##1}}}
+\expandafter\def\csname PY@tok@sx\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@bp\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@c1\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
+\expandafter\def\csname PY@tok@o\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
+\expandafter\def\csname PY@tok@kc\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@c\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
+\expandafter\def\csname PY@tok@mf\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
+\expandafter\def\csname PY@tok@err\endcsname{\def\PY@bc##1{\setlength{\fboxsep}{0pt}\fcolorbox[rgb]{1.00,0.00,0.00}{1,1,1}{\strut ##1}}}
+\expandafter\def\csname PY@tok@mb\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
+\expandafter\def\csname PY@tok@ss\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
+\expandafter\def\csname PY@tok@sr\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.40,0.53}{##1}}}
+\expandafter\def\csname PY@tok@mo\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
+\expandafter\def\csname PY@tok@kd\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@mi\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
+\expandafter\def\csname PY@tok@kn\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@cpf\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
+\expandafter\def\csname PY@tok@kr\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@s\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
+\expandafter\def\csname PY@tok@kp\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@w\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.73,0.73}{##1}}}
+\expandafter\def\csname PY@tok@kt\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.69,0.00,0.25}{##1}}}
+\expandafter\def\csname PY@tok@sc\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
+\expandafter\def\csname PY@tok@sb\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
+\expandafter\def\csname PY@tok@k\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@se\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.73,0.40,0.13}{##1}}}
+\expandafter\def\csname PY@tok@sd\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
+
+\def\PYZbs{\char`\\}
+\def\PYZus{\char`\_}
+\def\PYZob{\char`\{}
+\def\PYZcb{\char`\}}
+\def\PYZca{\char`\^}
+\def\PYZam{\char`\&}
+\def\PYZlt{\char`\<}
+\def\PYZgt{\char`\>}
+\def\PYZsh{\char`\#}
+\def\PYZpc{\char`\%}
+\def\PYZdl{\char`\$}
+\def\PYZhy{\char`\-}
+\def\PYZsq{\char`\'}
+\def\PYZdq{\char`\"}
+\def\PYZti{\char`\~}
+% for compatibility with earlier versions
+\def\PYZat{@}
+\def\PYZlb{[}
+\def\PYZrb{]}
+\makeatother
+
+
+ % Exact colors from NB
+ \definecolor{incolor}{rgb}{0.0, 0.0, 0.5}
+ \definecolor{outcolor}{rgb}{0.545, 0.0, 0.0}
+
+
+
+
+ % Prevent overflowing lines due to hard-to-break entities
+ \sloppy
+ % Setup hyperref package
+ \hypersetup{
+ breaklinks=true, % so long urls are correctly broken across lines
+ colorlinks=true,
+ urlcolor=urlcolor,
+ linkcolor=linkcolor,
+ citecolor=citecolor,
+ }
+ % Slightly bigger margins than the latex defaults
+
+ \geometry{verbose,tmargin=1in,bmargin=1in,lmargin=1in,rmargin=1in}
+
+
+
+ \begin{document}
+
+
+ \maketitle
+
+
+
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}1}]:} \PY{c}{\PYZpc{}plot \PYZhy{}\PYZhy{}format svg}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}2}]:} \PY{n}{setdefaults}
+\end{Verbatim}
+
+ \subsection{My question from last
+class}\label{my-question-from-last-class}
+
+\(A=\left[ \begin{array}{ccc} 10 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 10\end{array} \right]\)
+
+\begin{figure}[htbp]
+\centering
+\includegraphics{det_A.png}
+\caption{responses to determinant of A}
+\end{figure}
+
+\subsection{Your questions from last
+class}\label{your-questions-from-last-class}
+
+\begin{enumerate}
+\def\labelenumi{\arabic{enumi}.}
+\item
+ Need more linear algebra review
+
+ -We will keep doing Linear Algebra, try the practice problems in
+ 'linear\_algebra'
+\item
+ How do I do HW3?
+
+ -demo today
+\item
+ For hw4 is the spring constant (K) suppose to be given?
+
+ -yes, its 30 N/m
+\item
+ Deapool or Joker?
+\end{enumerate}
+
+\subsection{Midterm preference}\label{midterm-preference}
+
+\begin{figure}[htbp]
+\centering
+\includegraphics{midterm_date.png}
+\caption{responses to midterm date}
+\end{figure}
+
+\subsubsection{Midterm Questions}\label{midterm-questions}
+
+\begin{enumerate}
+\def\labelenumi{\arabic{enumi}.}
+\item
+ Notes allowed
+
+ -no
+\item
+ Will there be a review/study sheet
+
+ -yes
+\end{enumerate}
+
+ \section{LU Decomposition}\label{lu-decomposition}
+
+\subsubsection{efficient storage of matrices for
+solutions}\label{efficient-storage-of-matrices-for-solutions}
+
+Considering the same solution set:
+
+\(y=Ax\)
+
+Assume that we can perform Gauss elimination and achieve this formula:
+
+\(Ux=d\)
+
+Where, \(U\) is an upper triangular matrix that we derived from Gauss
+elimination and \(d\) is the set of dependent variables after Gauss
+elimination.
+
+Assume there is a lower triangular matrix, \(L\), with ones on the
+diagonal and same dimensions of \(U\) and the following is true:
+
+\(L(Ux-d)=Ax-y=0\)
+
+Now, \(Ax=LUx\), so \(A=LU\), and \(y=Ld\).
+
+\(2x_{1}+x_{2}=1\)
+
+\(x_{1}+3x_{2}=1\)
+
+\(\left[ \begin{array}{cc} 2 & 1 \\ 1 & 3 \end{array} \right] \left[\begin{array}{c} x_{1} \\ x_{2} \end{array}\right]= \left[\begin{array}{c} 1 \\ 1\end{array}\right]\)
+
+f21=0.5
+
+A(2,1)=1-1 = 0
+
+A(2,2)=3-0.5=2.5
+
+y(2)=1-0.5=0.5
+
+\(L(Ux-d)= \left[ \begin{array}{cc} 1 & 0 \\ 0.5 & 1 \end{array} \right] \left(\left[ \begin{array}{cc} 2 & 1 \\ 0 & 2.5 \end{array} \right] \left[\begin{array}{c} x_{1} \\ x_{2} \end{array}\right]- \left[\begin{array}{c} 1 \\ 0.5\end{array}\right]\right)=0\)
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}3}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{]}
+ \PY{n}{L}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mf}{0.5}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{]}
+ \PY{n}{U}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mf}{2.5}\PY{p}{]}
+ \PY{n}{L}\PY{o}{*}\PY{n}{U}
+
+ \PY{n}{d}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mf}{0.5}\PY{p}{]}
+ \PY{n}{y}\PY{p}{=}\PY{n}{L}\PY{o}{*}\PY{n}{d}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+A =
+
+ 2 1
+ 1 3
+
+L =
+
+ 1.00000 0.00000
+ 0.50000 1.00000
+
+U =
+
+ 2.00000 1.00000
+ 0.00000 2.50000
+
+ans =
+
+ 2 1
+ 1 3
+
+d =
+
+ 1.00000
+ 0.50000
+
+y =
+
+ 1
+ 1
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}5}]:} \PY{c}{\PYZpc{} what is the determinant of L, U and A?}
+
+ \PY{n+nb}{det}\PY{p}{(}\PY{n}{A}\PY{p}{)}
+ \PY{n+nb}{det}\PY{p}{(}\PY{n}{L}\PY{p}{)}
+ \PY{n+nb}{det}\PY{p}{(}\PY{n}{U}\PY{p}{)}
+ \PY{n+nb}{det}\PY{p}{(}\PY{n}{L}\PY{p}{)}\PY{o}{*}\PY{n+nb}{det}\PY{p}{(}\PY{n}{U}\PY{p}{)}
+ \PY{n+nb}{det}\PY{p}{(}\PY{n}{L}\PY{o}{*}\PY{n}{U}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+ans = 5.0000
+ans = 1
+ans = 5
+ans = 5
+ans = 5.0000
+
+ \end{Verbatim}
+
+ \subsection{Pivoting for LU
+factorization}\label{pivoting-for-lu-factorization}
+
+LU factorization uses the same method as Gauss elimination so it is also
+necessary to perform partial pivoting when creating the lower and upper
+triangular matrices.
+
+Matlab and Octave use pivoting in the command
+
+\texttt{{[}L,U,P{]}=lu(A)}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}4}]:} \PY{n}{help} \PY{n+nb}{lu}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+'lu' is a built-in function from the file libinterp/corefcn/lu.cc
+
+ -- Built-in Function: [L, U] = lu (A)
+ -- Built-in Function: [L, U, P] = lu (A)
+ -- Built-in Function: [L, U, P, Q] = lu (S)
+ -- Built-in Function: [L, U, P, Q, R] = lu (S)
+ -- Built-in Function: [{\ldots}] = lu (S, THRES)
+ -- Built-in Function: Y = lu ({\ldots})
+ -- Built-in Function: [{\ldots}] = lu ({\ldots}, "vector")
+ Compute the LU decomposition of A.
+
+ If A is full subroutines from LAPACK are used and if A is sparse
+ then UMFPACK is used.
+
+ The result is returned in a permuted form, according to the
+ optional return value P. For example, given the matrix 'a = [1, 2;
+ 3, 4]',
+
+ [l, u, p] = lu (A)
+
+ returns
+
+ l =
+
+ 1.00000 0.00000
+ 0.33333 1.00000
+
+ u =
+
+ 3.00000 4.00000
+ 0.00000 0.66667
+
+ p =
+
+ 0 1
+ 1 0
+
+ The matrix is not required to be square.
+
+ When called with two or three output arguments and a spare input
+ matrix, 'lu' does not attempt to perform sparsity preserving column
+ permutations. Called with a fourth output argument, the sparsity
+ preserving column transformation Q is returned, such that 'P * A *
+ Q = L * U'.
+
+ Called with a fifth output argument and a sparse input matrix, 'lu'
+ attempts to use a scaling factor R on the input matrix such that 'P
+ * (R \textbackslash{} A) * Q = L * U'. This typically leads to a sparser and more
+ stable factorization.
+
+ An additional input argument THRES, that defines the pivoting
+ threshold can be given. THRES can be a scalar, in which case it
+ defines the UMFPACK pivoting tolerance for both symmetric and
+ unsymmetric cases. If THRES is a 2-element vector, then the first
+ element defines the pivoting tolerance for the unsymmetric UMFPACK
+ pivoting strategy and the second for the symmetric strategy. By
+ default, the values defined by 'spparms' are used ([0.1, 0.001]).
+
+ Given the string argument "vector", 'lu' returns the values of P
+ and Q as vector values, such that for full matrix, 'A (P,:) = L *
+ U', and 'R(P,:) * A (:, Q) = L * U'.
+
+ With two output arguments, returns the permuted forms of the upper
+ and lower triangular matrices, such that 'A = L * U'. With one
+ output argument Y, then the matrix returned by the LAPACK routines
+ is returned. If the input matrix is sparse then the matrix L is
+ embedded into U to give a return value similar to the full case.
+ For both full and sparse matrices, 'lu' loses the permutation
+ information.
+
+ See also: luupdate, ilu, chol, hess, qr, qz, schur, svd.
+
+Additional help for built-in functions and operators is
+available in the online version of the manual. Use the command
+'doc <topic>' to search the manual index.
+
+Help and information about Octave is also available on the WWW
+at http://www.octave.org and via the help@octave.org
+mailing list.
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}9}]:} \PY{c}{\PYZpc{} time LU solution vs backslash}
+ \PY{n}{t\PYZus{}lu}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{100}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{t\PYZus{}bs}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{100}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{k}{for} \PY{n}{N}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{:}\PY{l+m+mi}{100}
+ \PY{n}{A}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{n}{N}\PY{p}{,}\PY{n}{N}\PY{p}{)}\PY{p}{;}
+ \PY{n}{y}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{n}{N}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{p}{[}\PY{n}{L}\PY{p}{,}\PY{n}{U}\PY{p}{,}\PY{n}{P}\PY{p}{]}\PY{p}{=}\PY{n+nb}{lu}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{;}
+
+ \PY{n+nb}{tic}\PY{p}{;} \PY{n}{d}\PY{p}{=}\PY{n+nb}{inv}\PY{p}{(}\PY{n}{L}\PY{p}{)}\PY{o}{*}\PY{n}{y}\PY{p}{;} \PY{n}{x}\PY{p}{=}\PY{n+nb}{inv}\PY{p}{(}\PY{n}{U}\PY{p}{)}\PY{o}{*}\PY{n}{d}\PY{p}{;} \PY{n}{t\PYZus{}lu}\PY{p}{(}\PY{n}{N}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;}
+
+ \PY{n+nb}{tic}\PY{p}{;} \PY{n}{x}\PY{p}{=}\PY{n+nb}{inv}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{o}{*}\PY{n}{y}\PY{p}{;} \PY{n}{t\PYZus{}bs}\PY{p}{(}\PY{n}{N}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;}
+ \PY{k}{end}
+ \PY{n+nb}{plot}\PY{p}{(}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{:}\PY{l+m+mi}{100}\PY{p}{]}\PY{p}{,}\PY{n}{t\PYZus{}lu}\PY{p}{,}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{:}\PY{l+m+mi}{100}\PY{p}{]}\PY{p}{,}\PY{n}{t\PYZus{}bs}\PY{p}{)}
+ \PY{n+nb}{legend}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{LU decomp\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{Octave \PYZbs{}\PYZbs{}\PYZsq{}}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{center}
+ \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_11_files/lecture_11_8_0.pdf}
+ \end{center}
+ { \hspace*{\fill} \\}
+
+ Consider the problem again from the intro to Linear Algebra, 4 masses
+are connected in series to 4 springs with K=10 N/m. What are the final
+positions of the masses?
+
+\begin{figure}[htbp]
+\centering
+\includegraphics{../lecture_09/mass_springs.png}
+\caption{Springs-masses}
+\end{figure}
+
+The masses haves the following amounts, 1, 2, 3, and 4 kg for masses
+1-4. Using a FBD for each mass:
+
+\(m_{1}g+k(x_{2}-x_{1})-kx_{1}=0\)
+
+\(m_{2}g+k(x_{3}-x_{2})-k(x_{2}-x_{1})=0\)
+
+\(m_{3}g+k(x_{4}-x_{3})-k(x_{3}-x_{2})=0\)
+
+\(m_{4}g-k(x_{4}-x_{3})=0\)
+
+in matrix form:
+
+\(\left[ \begin{array}{cccc} 2k & -k & 0 & 0 \\ -k & 2k & -k & 0 \\ 0 & -k & 2k & -k \\ 0 & 0 & -k & k \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right]= \left[ \begin{array}{c} m_{1}g \\ m_{2}g \\ m_{3}g \\ m_{4}g \end{array} \right]\)
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}10}]:} \PY{n}{k}\PY{p}{=}\PY{l+m+mi}{10}\PY{p}{;} \PY{c}{\PYZpc{} N/m}
+ \PY{n}{m1}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} \PY{c}{\PYZpc{} kg}
+ \PY{n}{m2}\PY{p}{=}\PY{l+m+mi}{2}\PY{p}{;}
+ \PY{n}{m3}\PY{p}{=}\PY{l+m+mi}{3}\PY{p}{;}
+ \PY{n}{m4}\PY{p}{=}\PY{l+m+mi}{4}\PY{p}{;}
+ \PY{n}{g}\PY{p}{=}\PY{l+m+mf}{9.81}\PY{p}{;} \PY{c}{\PYZpc{} m/s\PYZca{}2}
+ \PY{n}{K}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{k} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{l+m+mi}{0} \PY{l+m+mi}{0}\PY{p}{;} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{l+m+mi}{2}\PY{o}{*}\PY{n}{k} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{l+m+mi}{0}\PY{p}{;} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{l+m+mi}{2}\PY{o}{*}\PY{n}{k} \PY{o}{\PYZhy{}}\PY{n}{k}\PY{p}{;} \PY{l+m+mi}{0} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{n}{k}\PY{p}{]}
+ \PY{n}{y}\PY{p}{=}\PY{p}{[}\PY{n}{m1}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m2}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m3}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m4}\PY{o}{*}\PY{n}{g}\PY{p}{]}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+K =
+
+ 20 -10 0 0
+ -10 20 -10 0
+ 0 -10 20 -10
+ 0 0 -10 10
+
+y =
+
+ 9.8100
+ 19.6200
+ 29.4300
+ 39.2400
+
+
+ \end{Verbatim}
+
+ This matrix, K, is symmetric.
+
+\texttt{K(i,j)==K(j,i)}
+
+Now we can use,
+
+\subsection{Cholesky Factorization}\label{cholesky-factorization}
+
+We can decompose the matrix, K into two matrices, \(U\) and \(U^{T}\),
+where
+
+\(K=U^{T}U\)
+
+each of the components of U can be calculated with the following
+equations:
+
+\(u_{ii}=\sqrt{a_{ii}-\sum_{k=1}^{i-1}u_{ki}^{2}}\)
+
+\(u_{ij}=\frac{a_{ij}-\sum_{k=1}^{i-1}u_{ki}u_{kj}}{u_{ii}}\)
+
+so for K
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}11}]:} \PY{n}{K}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+K =
+
+ 20 -10 0 0
+ -10 20 -10 0
+ 0 -10 20 -10
+ 0 0 -10 10
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}12}]:} \PY{n}{u11}\PY{p}{=}\PY{n+nb}{sqrt}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{)}
+ \PY{n}{u12}\PY{p}{=}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{o}{/}\PY{n}{u11}
+ \PY{n}{u13}\PY{p}{=}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)}\PY{o}{/}\PY{n}{u11}
+ \PY{n}{u14}\PY{p}{=}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{)}\PY{p}{)}\PY{o}{/}\PY{n}{u11}
+ \PY{n}{u22}\PY{p}{=}\PY{n+nb}{sqrt}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{u12}\PYZca{}\PY{l+m+mi}{2}\PY{p}{)}
+ \PY{n}{u23}\PY{p}{=}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{u12}\PY{o}{*}\PY{n}{u13}\PY{p}{)}\PY{o}{/}\PY{n}{u22}
+ \PY{n}{u24}\PY{p}{=}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{u12}\PY{o}{*}\PY{n}{u14}\PY{p}{)}\PY{o}{/}\PY{n}{u22}
+ \PY{n}{u33}\PY{p}{=}\PY{n+nb}{sqrt}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{u13}\PYZca{}\PY{l+m+mi}{2}\PY{o}{\PYZhy{}}\PY{n}{u23}\PYZca{}\PY{l+m+mi}{2}\PY{p}{)}
+ \PY{n}{u34}\PY{p}{=}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{u13}\PY{o}{*}\PY{n}{u14}\PY{o}{\PYZhy{}}\PY{n}{u23}\PY{o}{*}\PY{n}{u24}\PY{p}{)}\PY{o}{/}\PY{n}{u33}
+ \PY{n}{u44}\PY{p}{=}\PY{n+nb}{sqrt}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{u14}\PYZca{}\PY{l+m+mi}{2}\PY{o}{\PYZhy{}}\PY{n}{u24}\PYZca{}\PY{l+m+mi}{2}\PY{o}{\PYZhy{}}\PY{n}{u34}\PYZca{}\PY{l+m+mi}{2}\PY{p}{)}
+ \PY{n}{U}\PY{p}{=}\PY{p}{[}\PY{n}{u11}\PY{p}{,}\PY{n}{u12}\PY{p}{,}\PY{n}{u13}\PY{p}{,}\PY{n}{u14}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{,}\PY{n}{u22}\PY{p}{,}\PY{n}{u23}\PY{p}{,}\PY{n}{u24}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{,}\PY{n}{u33}\PY{p}{,}\PY{n}{u34}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{,}\PY{n}{u44}\PY{p}{]}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+u11 = 4.4721
+u12 = -2.2361
+u13 = 0
+u14 = 0
+u22 = 3.8730
+u23 = -2.5820
+u24 = 0
+u33 = 3.6515
+u34 = -2.7386
+u44 = 1.5811
+U =
+
+ 4.47214 -2.23607 0.00000 0.00000
+ 0.00000 3.87298 -2.58199 0.00000
+ 0.00000 0.00000 3.65148 -2.73861
+ 0.00000 0.00000 0.00000 1.58114
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}17}]:} \PY{p}{(}\PY{n}{U}\PY{o}{\PYZsq{}}\PY{o}{*}\PY{n}{U}\PY{p}{)}\PY{o}{\PYZsq{}}\PY{o}{==}\PY{n}{U}\PY{o}{\PYZsq{}}\PY{o}{*}\PY{n}{U}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+ans =
+
+ 1 1 1 1
+ 1 1 1 1
+ 1 1 1 1
+ 1 1 1 1
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}18}]:} \PY{c}{\PYZpc{} time solution for Cholesky vs backslash}
+ \PY{n}{t\PYZus{}chol}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{1000}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{t\PYZus{}bs}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{1000}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{k}{for} \PY{n}{i}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{:}\PY{l+m+mi}{1000}
+ \PY{n+nb}{tic}\PY{p}{;} \PY{n}{d}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZsq{}}\PY{o}{*}\PY{n}{y}\PY{p}{;} \PY{n}{x}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZbs{}}\PY{n}{d}\PY{p}{;} \PY{n}{t\PYZus{}chol}\PY{p}{(}\PY{n}{i}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;}
+ \PY{n+nb}{tic}\PY{p}{;} \PY{n}{x}\PY{p}{=}\PY{n}{K}\PY{o}{\PYZbs{}}\PY{n}{y}\PY{p}{;} \PY{n}{t\PYZus{}bs}\PY{p}{(}\PY{n}{i}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;}
+ \PY{k}{end}
+ \PY{n+nb}{fprintf}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{average time spent for Cholesky factored solution = \PYZpc{}e+/\PYZhy{}\PYZpc{}e\PYZsq{}}\PY{p}{,}\PY{n+nb}{mean}\PY{p}{(}\PY{n}{t\PYZus{}chol}\PY{p}{)}\PY{p}{,}\PY{n+nb}{std}\PY{p}{(}\PY{n}{t\PYZus{}chol}\PY{p}{)}\PY{p}{)}
+
+ \PY{n+nb}{fprintf}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{average time spent for backslash solution = \PYZpc{}e+/\PYZhy{}\PYZpc{}e\PYZsq{}}\PY{p}{,}\PY{n+nb}{mean}\PY{p}{(}\PY{n}{t\PYZus{}bs}\PY{p}{)}\PY{p}{,}\PY{n+nb}{std}\PY{p}{(}\PY{n}{t\PYZus{}bs}\PY{p}{)}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+average time spent for Cholesky factored solution = 1.465964e-05+/-9.806001e-06
+average time spent for backslash solution = 1.555967e-05+/-1.048561e-05
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor} }]:}
+\end{Verbatim}
+
+
+ % Add a bibliography block to the postdoc
+
+
+
+ \end{document}
diff --git a/lecture_11/lecture_11_files/lecture_11_6_0.pdf b/lecture_11/lecture_11_files/lecture_11_6_0.pdf
new file mode 100644
index 0000000..40e3f28
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new file mode 100644
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diff --git a/lecture_11/midterm_date.png b/lecture_11/midterm_date.png
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+
+(evince:15386): Gtk-WARNING **: gtk_widget_size_allocate(): attempt to allocate widget with width -71 and height 20
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diff --git a/lecture_12/.ipynb_checkpoints/lecture_12-checkpoint.ipynb b/lecture_12/.ipynb_checkpoints/lecture_12-checkpoint.ipynb
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@@ -0,0 +1,1483 @@
+{
+ "cells": [
+ {
+ "cell_type": "code",
+ "execution_count": 47,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "%plot --format svg"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 48,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "setdefaults"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## My question from last class \n",
+ "\n",
+ "\n",
+ "## Your questions from last class\n",
+ "\n",
+ "1. Need more linear algebra review\n",
+ " \n",
+ " -We will keep doing Linear Algebra, try the practice problems in 'linear_algebra'\n",
+ "\n",
+ "2. How do I do HW3? \n",
+ " \n",
+ " -demo today\n",
+ "\n",
+ "3. For hw4 is the spring constant (K) suppose to be given? \n",
+ " \n",
+ " -yes, its 30 N/m\n",
+ " \n",
+ "4. Deapool or Joker?"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Matrix Inverse and Condition\n",
+ "\n",
+ "\n",
+ "Considering the same solution set:\n",
+ "\n",
+ "$y=Ax$\n",
+ "\n",
+ "If we know that $A^{-1}A=I$, then \n",
+ "\n",
+ "$A^{-1}y=A^{-1}Ax=x$\n",
+ "\n",
+ "so \n",
+ "\n",
+ "$x=A^{-1}y$\n",
+ "\n",
+ "Where, $A^{-1}$ is the inverse of matrix $A$.\n",
+ "\n",
+ "$2x_{1}+x_{2}=1$\n",
+ "\n",
+ "$x_{1}+3x_{2}=1$\n",
+ "\n",
+ "$Ax=y$\n",
+ "\n",
+ "$\\left[ \\begin{array}{cc}\n",
+ "2 & 1 \\\\\n",
+ "1 & 3 \\end{array} \\right]\n",
+ "\\left[\\begin{array}{c} \n",
+ "x_{1} \\\\ \n",
+ "x_{2} \\end{array}\\right]=\n",
+ "\\left[\\begin{array}{c} \n",
+ "1 \\\\\n",
+ "1\\end{array}\\right]$\n",
+ "\n",
+ "$A^{-1}=\\frac{1}{2*3-1*1}\\left[ \\begin{array}{cc}\n",
+ "3 & 1 \\\\\n",
+ "-1 & 2 \\end{array} \\right]=\n",
+ "\\left[ \\begin{array}{cc}\n",
+ "3/5 & -1/5 \\\\\n",
+ "-1/5 & 2/5 \\end{array} \\right]$\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A =\n",
+ "\n",
+ " 2 1\n",
+ " 1 3\n",
+ "\n",
+ "invA =\n",
+ "\n",
+ " 0.60000 -0.20000\n",
+ " -0.20000 0.40000\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " 1.00000 0.00000\n",
+ " 0.00000 1.00000\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " 1.00000 0.00000\n",
+ " 0.00000 1.00000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "A=[2,1;1,3]\n",
+ "invA=1/5*[3,-1;-1,2]\n",
+ "\n",
+ "A*invA\n",
+ "invA*A"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "How did we know the inverse of A? \n",
+ "\n",
+ "for 2$\\times$2 matrices, it is always:\n",
+ "\n",
+ "$A=\\left[ \\begin{array}{cc}\n",
+ "A_{11} & A_{12} \\\\\n",
+ "A_{21} & A_{22} \\end{array} \\right]$\n",
+ "\n",
+ "$A^{-1}=\\frac{1}{det(A)}\\left[ \\begin{array}{cc}\n",
+ "A_{22} & -A_{12} \\\\\n",
+ "-A_{21} & A_{11} \\end{array} \\right]$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$AA^{-1}=\\frac{1}{A_{11}A_{22}-A_{21}A_{12}}\\left[ \\begin{array}{cc}\n",
+ "A_{11}A_{22}-A_{21}A_{12} & -A_{11}A_{12}+A_{12}A_{11} \\\\\n",
+ "A_{21}A_{22}-A_{22}A_{21} & -A_{21}A_{12}+A_{22}A_{11} \\end{array} \\right]\n",
+ "=\\left[ \\begin{array}{cc}\n",
+ "1 & 0 \\\\\n",
+ "0 & 1 \\end{array} \\right]$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "What about bigger matrices?\n",
+ "\n",
+ "We can use the LU-decomposition\n",
+ "\n",
+ "$A=LU$\n",
+ "\n",
+ "$A^{-1}=(LU)^{-1}=U^{-1}L^{-1}$\n",
+ "\n",
+ "if we divide $A^{-1}$ into n-column vectors, $a_{n}$, then\n",
+ "\n",
+ "$Aa_{1}=\\left[\\begin{array}{c} \n",
+ "1 \\\\ \n",
+ "0 \\\\ \n",
+ "\\vdots \\\\\n",
+ "0 \\end{array} \\right]$\n",
+ "$Aa_{2}=\\left[\\begin{array}{c} \n",
+ "0 \\\\ \n",
+ "1 \\\\ \n",
+ "\\vdots \\\\\n",
+ "0 \\end{array} \\right]$\n",
+ "$Aa_{n}=\\left[\\begin{array}{c} \n",
+ "0 \\\\ \n",
+ "0 \\\\ \n",
+ "\\vdots \\\\\n",
+ "1 \\end{array} \\right]$\n",
+ "\n",
+ "\n",
+ "Which we can solve for each $a_{n}$ with LU-decomposition, knowing the lower and upper triangular decompositions, then \n",
+ "\n",
+ "$A^{-1}=\\left[ \\begin{array}{cccc}\n",
+ "| & | & & | \\\\\n",
+ "a_{1} & a_{2} & \\cdots & a_{3} \\\\\n",
+ "| & | & & | \\end{array} \\right]$\n",
+ "\n",
+ "\n",
+ "$Ld_{1}=\\left[\\begin{array}{c} \n",
+ "1 \\\\ \n",
+ "0 \\\\ \n",
+ "\\vdots \\\\\n",
+ "0 \\end{array} \\right]$\n",
+ "$;~Ua_{1}=d_{1}$\n",
+ "\n",
+ "$Ld_{2}=\\left[\\begin{array}{c} \n",
+ "0 \\\\ \n",
+ "1 \\\\ \n",
+ "\\vdots \\\\\n",
+ "0 \\end{array} \\right]$\n",
+ "$;~Ua_{2}=d_{2}$\n",
+ "\n",
+ "$Ld_{n}=\\left[\\begin{array}{c} \n",
+ "0 \\\\ \n",
+ "1 \\\\ \n",
+ "\\vdots \\\\\n",
+ "n \\end{array} \\right]$\n",
+ "$;~Ua_{n}=d_{n}$\n",
+ "\n",
+ "Consider the following matrix:\n",
+ "\n",
+ "$A=\\left[ \\begin{array}{ccc}\n",
+ "2 & -1 & 0\\\\\n",
+ "-1 & 2 & -1\\\\\n",
+ "0 & -1 & 1 \\end{array} \\right]$\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 21,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A =\n",
+ "\n",
+ " 2 -1 0\n",
+ " -1 2 -1\n",
+ " 0 -1 1\n",
+ "\n",
+ "U =\n",
+ "\n",
+ " 2.00000 -1.00000 0.00000\n",
+ " 0.00000 1.50000 -1.00000\n",
+ " 0.00000 -1.00000 1.00000\n",
+ "\n",
+ "L =\n",
+ "\n",
+ " 1.00000 0.00000 0.00000\n",
+ " -0.50000 1.00000 0.00000\n",
+ " 0.00000 0.00000 1.00000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "A=[2,-1,0;-1,2,-1;0,-1,1]\n",
+ "U=A;\n",
+ "L=eye(3,3);\n",
+ "U(2,:)=U(2,:)-U(2,1)/U(1,1)*U(1,:)\n",
+ "L(2,1)=A(2,1)/A(1,1)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 22,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "L =\n",
+ "\n",
+ " 1.00000 0.00000 0.00000\n",
+ " -0.50000 1.00000 0.00000\n",
+ " 0.00000 -0.66667 1.00000\n",
+ "\n",
+ "U =\n",
+ "\n",
+ " 2.00000 -1.00000 0.00000\n",
+ " 0.00000 1.50000 -1.00000\n",
+ " 0.00000 0.00000 0.33333\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "L(3,2)=U(3,2)/U(2,2)\n",
+ "U(3,:)=U(3,:)-U(3,2)/U(2,2)*U(2,:)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Now solve for $d_1$ then $a_1$, $d_2$ then $a_2$, and $d_3$ then $a_{3}$\n",
+ "\n",
+ "$Ld_{1}=\\left[\\begin{array}{c} \n",
+ "1 \\\\ \n",
+ "0 \\\\ \n",
+ "\\vdots \\\\\n",
+ "0 \\end{array} \\right]$\n",
+ "$;~Ua_{1}=d_{1}$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "d1 =\n",
+ "\n",
+ " 1.00000\n",
+ " 0.50000\n",
+ " 0.33333\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "d1=zeros(3,1);\n",
+ "d1(1)=1;\n",
+ "d1(2)=0-L(2,1)*d1(1);\n",
+ "d1(3)=0-L(3,1)*d1(1)-L(3,2)*d1(2)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 30,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "a1 =\n",
+ "\n",
+ " 1.00000\n",
+ " 1.00000\n",
+ " 1.00000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "a1=zeros(3,1);\n",
+ "a1(3)=d1(3)/U(3,3);\n",
+ "a1(2)=1/U(2,2)*(d1(2)-U(2,3)*a1(3));\n",
+ "a1(1)=1/U(1,1)*(d1(1)-U(1,2)*a1(2)-U(1,3)*a1(3))"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 28,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "d2 =\n",
+ "\n",
+ " 0.00000\n",
+ " 1.00000\n",
+ " 0.66667\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "d2=zeros(3,1);\n",
+ "d2(1)=0;\n",
+ "d2(2)=1-L(2,1)*d2(1);\n",
+ "d2(3)=0-L(3,1)*d2(1)-L(3,2)*d2(2)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 31,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "a2 =\n",
+ "\n",
+ " 1.0000\n",
+ " 2.0000\n",
+ " 2.0000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "a2=zeros(3,1);\n",
+ "a2(3)=d2(3)/U(3,3);\n",
+ "a2(2)=1/U(2,2)*(d2(2)-U(2,3)*a2(3));\n",
+ "a2(1)=1/U(1,1)*(d2(1)-U(1,2)*a2(2)-U(1,3)*a2(3))"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 37,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "d3 =\n",
+ "\n",
+ " 0\n",
+ " 0\n",
+ " 1\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "d3=zeros(3,1);\n",
+ "d3(1)=0;\n",
+ "d3(2)=0-L(2,1)*d3(1);\n",
+ "d3(3)=1-L(3,1)*d3(1)-L(3,2)*d3(2)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 38,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "a3 =\n",
+ "\n",
+ " 1.00000\n",
+ " 2.00000\n",
+ " 3.00000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "a3=zeros(3,1);\n",
+ "a3(3)=d3(3)/U(3,3);\n",
+ "a3(2)=1/U(2,2)*(d3(2)-U(2,3)*a3(3));\n",
+ "a3(1)=1/U(1,1)*(d3(1)-U(1,2)*a3(2)-U(1,3)*a3(3))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Final solution for $A^{-1}$ is $[a_{1}~a_{2}~a_{3}]$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 40,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "invA =\n",
+ "\n",
+ " 1.00000 1.00000 1.00000\n",
+ " 1.00000 2.00000 2.00000\n",
+ " 1.00000 2.00000 3.00000\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " 1.00000 0.00000 0.00000\n",
+ " 0.00000 1.00000 -0.00000\n",
+ " -0.00000 -0.00000 1.00000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "invA=[a1,a2,a3]\n",
+ "A*invA"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Now the solution of $x$ to $Ax=y$ is $x=A^{-1}y$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 44,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "y =\n",
+ "\n",
+ " 1\n",
+ " 2\n",
+ " 3\n",
+ "\n",
+ "x =\n",
+ "\n",
+ " 6.0000\n",
+ " 11.0000\n",
+ " 14.0000\n",
+ "\n",
+ "xbs =\n",
+ "\n",
+ " 6.0000\n",
+ " 11.0000\n",
+ " 14.0000\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " -3.5527e-15\n",
+ " -8.8818e-15\n",
+ " -1.0658e-14\n",
+ "\n",
+ "ans = 2.2204e-16\n"
+ ]
+ }
+ ],
+ "source": [
+ "y=[1;2;3]\n",
+ "x=invA*y\n",
+ "xbs=A\\y\n",
+ "x-xbs\n",
+ "eps"
+ ]
+ },
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+ "source": [
+ "N=100;\n",
+ "n=[1:N];\n",
+ "t_inv=zeros(N,1);\n",
+ "t_bs=zeros(N,1);\n",
+ "t_mult=zeros(N,1);\n",
+ "for i=1:N\n",
+ " A=rand(i,i);\n",
+ " tic\n",
+ " invA=inv(A);\n",
+ " t_inv(i)=toc;\n",
+ " b=rand(i,1);\n",
+ " tic;\n",
+ " x=A\\b;\n",
+ " t_bs(i)=toc;\n",
+ " tic;\n",
+ " x=invA*b;\n",
+ " t_mult(i)=toc;\n",
+ "end\n",
+ "plot(n,t_inv,n,t_bs,n,t_mult)\n",
+ "axis([0 100 0 0.002])\n",
+ "legend('inversion','backslash','multiplication','Location','NorthWest')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Condition of a matrix \n",
+ "### *just checked in to see what condition my condition was in*\n",
+ "### Matrix norms\n",
+ "\n",
+ "The Euclidean norm of a vector is measure of the magnitude (in 3D this would be: $|x|=\\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$) in general the equation is:\n",
+ "\n",
+ "$||x||_{e}=\\sqrt{\\sum_{i=1}^{n}x_{i}^{2}}$\n",
+ "\n",
+ "For a matrix, A, the same norm is called the Frobenius norm:\n",
+ "\n",
+ "$||A||_{f}=\\sqrt{\\sum_{i=1}^{n}\\sum_{i=1}^{m}A_{i,j}^{2}}$\n",
+ "\n",
+ "In general we can calculate any $p$-norm where\n",
+ "\n",
+ "$||A||_{p}=\\sqrt{\\sum_{i=1}^{n}\\sum_{i=1}^{m}A_{i,j}^{p}}$\n",
+ "\n",
+ "so the p=1, 1-norm is \n",
+ "\n",
+ "$||A||_{1}=\\sqrt{\\sum_{i=1}^{n}\\sum_{i=1}^{m}A_{i,j}^{1}}=\\sum_{i=1}^{n}\\sum_{i=1}^{m}|A_{i,j}|$\n",
+ "\n",
+ "$||A||_{\\infty}=\\sqrt{\\sum_{i=1}^{n}\\sum_{i=1}^{m}A_{i,j}^{\\infty}}=\\max_{1\\le i \\le n}\\sum_{j=1}^{m}|A_{i,j}|$\n",
+ "\n",
+ "### Condition of Matrix\n",
+ "\n",
+ "The matrix condition is the product of \n",
+ "\n",
+ "$Cond(A) = ||A||\\cdot||A^{-1}||$ \n",
+ "\n",
+ "So each norm will have a different condition number, but the limit is $Cond(A)\\ge 1$\n",
+ "\n",
+ "An estimate of the rounding error is based on the condition of A:\n",
+ "\n",
+ "$\\frac{||\\Delta x||}{x} \\le Cond(A) \\frac{||\\Delta A||}{||A||}$\n",
+ "\n",
+ "So if the coefficients of A have accuracy to $10^{-t}\n",
+ "\n",
+ "and the condition of A, $Cond(A)=10^{c}$\n",
+ "\n",
+ "then the solution for x can have rounding errors up to $10^{c-t}$\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A =\n",
+ "\n",
+ " 1.00000 0.50000 0.33333\n",
+ " 0.50000 0.33333 0.25000\n",
+ " 0.33333 0.25000 0.20000\n",
+ "\n",
+ "L =\n",
+ "\n",
+ " 1.00000 0.00000 0.00000\n",
+ " 0.50000 1.00000 0.00000\n",
+ " 0.33333 1.00000 1.00000\n",
+ "\n",
+ "U =\n",
+ "\n",
+ " 1.00000 0.50000 0.33333\n",
+ " 0.00000 0.08333 0.08333\n",
+ " 0.00000 -0.00000 0.00556\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "A=[1,1/2,1/3;1/2,1/3,1/4;1/3,1/4,1/5]\n",
+ "[L,U]=LU_naive(A)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "invA="
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A =\n",
+ "\n",
+ " 2 1\n",
+ " 1 3\n",
+ "\n",
+ "L =\n",
+ "\n",
+ " 1.00000 0.00000\n",
+ " 0.50000 1.00000\n",
+ "\n",
+ "U =\n",
+ "\n",
+ " 2.00000 1.00000\n",
+ " 0.00000 2.50000\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " 2 1\n",
+ " 1 3\n",
+ "\n",
+ "d =\n",
+ "\n",
+ " 1.00000\n",
+ " 0.50000\n",
+ "\n",
+ "y =\n",
+ "\n",
+ " 1\n",
+ " 1\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "A=[2,1;1,3]\n",
+ "L=[1,0;0.5,1]\n",
+ "U=[2,1;0,2.5]\n",
+ "L*U\n",
+ "\n",
+ "d=[1;0.5]\n",
+ "y=L*d"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans = 5.0000\n",
+ "ans = 1\n",
+ "ans = 5\n",
+ "ans = 5\n",
+ "ans = 5.0000\n"
+ ]
+ }
+ ],
+ "source": [
+ "% what is the determinant of L, U and A?\n",
+ "\n",
+ "det(A)\n",
+ "det(L)\n",
+ "det(U)\n",
+ "det(L)*det(U)\n",
+ "det(L*U)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Pivoting for LU factorization\n",
+ "\n",
+ "LU factorization uses the same method as Gauss elimination so it is also necessary to perform partial pivoting when creating the lower and upper triangular matrices. \n",
+ "\n",
+ "Matlab and Octave use pivoting in the command \n",
+ "\n",
+ "`[L,U,P]=lu(A)`\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "'lu' is a built-in function from the file libinterp/corefcn/lu.cc\n",
+ "\n",
+ " -- Built-in Function: [L, U] = lu (A)\n",
+ " -- Built-in Function: [L, U, P] = lu (A)\n",
+ " -- Built-in Function: [L, U, P, Q] = lu (S)\n",
+ " -- Built-in Function: [L, U, P, Q, R] = lu (S)\n",
+ " -- Built-in Function: [...] = lu (S, THRES)\n",
+ " -- Built-in Function: Y = lu (...)\n",
+ " -- Built-in Function: [...] = lu (..., \"vector\")\n",
+ " Compute the LU decomposition of A.\n",
+ "\n",
+ " If A is full subroutines from LAPACK are used and if A is sparse\n",
+ " then UMFPACK is used.\n",
+ "\n",
+ " The result is returned in a permuted form, according to the\n",
+ " optional return value P. For example, given the matrix 'a = [1, 2;\n",
+ " 3, 4]',\n",
+ "\n",
+ " [l, u, p] = lu (A)\n",
+ "\n",
+ " returns\n",
+ "\n",
+ " l =\n",
+ "\n",
+ " 1.00000 0.00000\n",
+ " 0.33333 1.00000\n",
+ "\n",
+ " u =\n",
+ "\n",
+ " 3.00000 4.00000\n",
+ " 0.00000 0.66667\n",
+ "\n",
+ " p =\n",
+ "\n",
+ " 0 1\n",
+ " 1 0\n",
+ "\n",
+ " The matrix is not required to be square.\n",
+ "\n",
+ " When called with two or three output arguments and a spare input\n",
+ " matrix, 'lu' does not attempt to perform sparsity preserving column\n",
+ " permutations. Called with a fourth output argument, the sparsity\n",
+ " preserving column transformation Q is returned, such that 'P * A *\n",
+ " Q = L * U'.\n",
+ "\n",
+ " Called with a fifth output argument and a sparse input matrix, 'lu'\n",
+ " attempts to use a scaling factor R on the input matrix such that 'P\n",
+ " * (R \\ A) * Q = L * U'. This typically leads to a sparser and more\n",
+ " stable factorization.\n",
+ "\n",
+ " An additional input argument THRES, that defines the pivoting\n",
+ " threshold can be given. THRES can be a scalar, in which case it\n",
+ " defines the UMFPACK pivoting tolerance for both symmetric and\n",
+ " unsymmetric cases. If THRES is a 2-element vector, then the first\n",
+ " element defines the pivoting tolerance for the unsymmetric UMFPACK\n",
+ " pivoting strategy and the second for the symmetric strategy. By\n",
+ " default, the values defined by 'spparms' are used ([0.1, 0.001]).\n",
+ "\n",
+ " Given the string argument \"vector\", 'lu' returns the values of P\n",
+ " and Q as vector values, such that for full matrix, 'A (P,:) = L *\n",
+ " U', and 'R(P,:) * A (:, Q) = L * U'.\n",
+ "\n",
+ " With two output arguments, returns the permuted forms of the upper\n",
+ " and lower triangular matrices, such that 'A = L * U'. With one\n",
+ " output argument Y, then the matrix returned by the LAPACK routines\n",
+ " is returned. If the input matrix is sparse then the matrix L is\n",
+ " embedded into U to give a return value similar to the full case.\n",
+ " For both full and sparse matrices, 'lu' loses the permutation\n",
+ " information.\n",
+ "\n",
+ " See also: luupdate, ilu, chol, hess, qr, qz, schur, svd.\n",
+ "\n",
+ "Additional help for built-in functions and operators is\n",
+ "available in the online version of the manual. Use the command\n",
+ "'doc <topic>' to search the manual index.\n",
+ "\n",
+ "Help and information about Octave is also available on the WWW\n",
+ "at http://www.octave.org and via the help@octave.org\n",
+ "mailing list.\n"
+ ]
+ }
+ ],
+ "source": [
+ "help lu"
+ ]
+ },
+ {
+ "cell_type": "code",
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+ "\t<path d=\"M442.1,384.0 L442.1,371.5 M442.1,16.7 L442.1,29.2 \" stroke=\"black\"/>\t<g fill=\"rgb(0,0,0)\" font-family=\"{}\" font-size=\"16.00\" stroke=\"none\" text-anchor=\"middle\" transform=\"translate(442.1,408.0)\">\n",
+ "\t\t<text><tspan font-family=\"{}\">80</tspan></text>\n",
+ "\t</g>\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "\t<path d=\"M535.0,384.0 L535.0,371.5 M535.0,16.7 L535.0,29.2 \" stroke=\"black\"/>\t<g fill=\"rgb(0,0,0)\" font-family=\"{}\" font-size=\"16.00\" stroke=\"none\" text-anchor=\"middle\" transform=\"translate(535.0,408.0)\">\n",
+ "\t\t<text><tspan font-family=\"{}\">100</tspan></text>\n",
+ "\t</g>\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "\t<path d=\"M70.5,16.7 L70.5,384.0 L535.0,384.0 L535.0,16.7 L70.5,16.7 Z \" stroke=\"black\"/></g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"black\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"1.00\">\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"1.00\">\n",
+ "\t<path d=\"M349.7,121.7 L349.7,25.7 L526.7,25.7 L526.7,121.7 L349.7,121.7 Z \" stroke=\"black\"/></g>\n",
+ "\t<g id=\"gnuplot_plot_1a\"><title>LU decomp</title>\n",
+ "<g color=\"white\" fill=\"none\" stroke=\"black\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"3.00\">\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"3.00\">\n",
+ "\t<g fill=\"rgb(0,0,0)\" font-family=\"{}\" font-size=\"16.00\" stroke=\"none\" text-anchor=\"end\" transform=\"translate(526.7,55.7)\">\n",
+ "\t\t<text><tspan font-family=\"{}\">LU decomp</tspan></text>\n",
+ "\t</g>\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"3.00\">\n",
+ "\t<path d=\"M360.9,49.7 L414.7,49.7 M75.1,357.4 L79.8,366.4 L84.4,370.3 L89.1,369.9 L93.7,363.8 L98.4,369.5 L103.0,369.5 L107.7,362.2 L112.3,368.4 L117.0,368.0 L121.6,360.8 L126.2,366.5 L130.9,366.6 L135.5,355.4 L140.2,355.5 L144.8,353.0 L149.5,346.7 L154.1,360.0 L158.8,362.3 L163.4,361.2 L168.0,362.3 L172.7,355.7 L177.3,360.3 L182.0,356.9 L186.6,355.1 L191.3,357.9 L195.9,351.3 L200.6,356.6 L205.2,355.0 L209.9,355.4 L214.5,362.2 L219.1,360.9 L223.8,360.8 L228.4,359.7 L233.1,359.0 L237.7,356.5 L242.4,358.3 L247.0,356.6 L251.7,360.7 L256.3,360.3 L260.9,360.0 L265.6,359.0 L270.2,359.0 L274.9,357.6 L279.5,357.4 L284.2,356.2 L288.8,354.2 L293.5,359.6 L298.1,358.5 L302.8,357.9 L307.4,356.5 L312.0,356.5 L316.7,355.9 L321.3,355.1 L326.0,353.9 L330.6,352.4 L335.3,357.0 L339.9,356.6 L344.6,355.9 L349.2,355.2 L353.8,354.1 L358.5,353.1 L363.1,352.6 L367.8,351.5 L372.4,335.7 L377.1,356.3 L381.7,356.6 L386.4,352.6 L391.0,355.5 L395.7,355.5 L400.3,354.6 L404.9,355.1 L409.6,224.4 L414.2,352.4 L418.9,338.8 L423.5,351.5 L428.2,350.8 L432.8,350.4 L437.5,349.7 L442.1,350.4 L446.7,348.4 L451.4,347.8 L456.0,346.2 L460.7,347.0 L465.3,340.7 L470.0,338.6 L474.6,346.2 L479.3,347.3 L483.9,344.2 L488.6,343.8 L493.2,342.5 L497.8,343.2 L502.5,341.2 L507.1,342.5 L511.8,338.8 L516.4,342.3 L521.1,337.9 L525.7,338.3 L530.4,337.8 L535.0,339.2 \" stroke=\"rgb( 0, 0, 255)\"/></g>\n",
+ "\t</g>\n",
+ "\t<g id=\"gnuplot_plot_2a\"><title>Octave \\\\</title>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"3.00\">\n",
+ "\t<g fill=\"rgb(0,0,0)\" font-family=\"{}\" font-size=\"16.00\" stroke=\"none\" text-anchor=\"end\" transform=\"translate(526.7,103.7)\">\n",
+ "\t\t<text><tspan font-family=\"{}\">Octave \\</tspan></text>\n",
+ "\t</g>\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"3.00\">\n",
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+ "\t</g>\n",
+ "<g color=\"white\" fill=\"none\" stroke=\"rgb( 0, 128, 0)\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"2.00\">\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"2.00\">\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"black\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "</g>\n",
+ "<g color=\"black\" fill=\"none\" stroke=\"currentColor\" stroke-linecap=\"butt\" stroke-linejoin=\"miter\" stroke-width=\"0.50\">\n",
+ "</g>\n",
+ "</g>\n",
+ "</svg>"
+ ],
+ "text/plain": [
+ "<IPython.core.display.SVG object>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "% time LU solution vs backslash\n",
+ "t_lu=zeros(100,1);\n",
+ "t_bs=zeros(100,1);\n",
+ "for N=1:100\n",
+ " A=rand(N,N);\n",
+ " y=rand(N,1);\n",
+ " [L,U,P]=lu(A);\n",
+ "\n",
+ " tic; d=inv(L)*y; x=inv(U)*d; t_lu(N)=toc;\n",
+ "\n",
+ " tic; x=inv(A)*y; t_bs(N)=toc;\n",
+ "end\n",
+ "plot([1:100],t_lu,[1:100],t_bs) \n",
+ "legend('LU decomp','Octave \\\\')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "Consider the problem again from the intro to Linear Algebra, 4 masses are connected in series to 4 springs with K=10 N/m. What are the final positions of the masses? \n",
+ "\n",
+ "\n",
+ "\n",
+ "The masses haves the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass:\n",
+ "\n",
+ "$m_{1}g+k(x_{2}-x_{1})-kx_{1}=0$\n",
+ "\n",
+ "$m_{2}g+k(x_{3}-x_{2})-k(x_{2}-x_{1})=0$\n",
+ "\n",
+ "$m_{3}g+k(x_{4}-x_{3})-k(x_{3}-x_{2})=0$\n",
+ "\n",
+ "$m_{4}g-k(x_{4}-x_{3})=0$\n",
+ "\n",
+ "in matrix form:\n",
+ "\n",
+ "$\\left[ \\begin{array}{cccc}\n",
+ "2k & -k & 0 & 0 \\\\\n",
+ "-k & 2k & -k & 0 \\\\\n",
+ "0 & -k & 2k & -k \\\\\n",
+ "0 & 0 & -k & k \\end{array} \\right]\n",
+ "\\left[ \\begin{array}{c}\n",
+ "x_{1} \\\\\n",
+ "x_{2} \\\\\n",
+ "x_{3} \\\\\n",
+ "x_{4} \\end{array} \\right]=\n",
+ "\\left[ \\begin{array}{c}\n",
+ "m_{1}g \\\\\n",
+ "m_{2}g \\\\\n",
+ "m_{3}g \\\\\n",
+ "m_{4}g \\end{array} \\right]$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "K =\n",
+ "\n",
+ " 20 -10 0 0\n",
+ " -10 20 -10 0\n",
+ " 0 -10 20 -10\n",
+ " 0 0 -10 10\n",
+ "\n",
+ "y =\n",
+ "\n",
+ " 9.8100\n",
+ " 19.6200\n",
+ " 29.4300\n",
+ " 39.2400\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "k=10; % N/m\n",
+ "m1=1; % kg\n",
+ "m2=2;\n",
+ "m3=3;\n",
+ "m4=4;\n",
+ "g=9.81; % m/s^2\n",
+ "K=[2*k -k 0 0; -k 2*k -k 0; 0 -k 2*k -k; 0 0 -k k]\n",
+ "y=[m1*g;m2*g;m3*g;m4*g]"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This matrix, K, is symmetric. \n",
+ "\n",
+ "`K(i,j)==K(j,i)`\n",
+ "\n",
+ "Now we can use,\n",
+ "\n",
+ "## Cholesky Factorization\n",
+ "\n",
+ "We can decompose the matrix, K into two matrices, $U$ and $U^{T}$, where\n",
+ "\n",
+ "$K=U^{T}U$\n",
+ "\n",
+ "each of the components of U can be calculated with the following equations:\n",
+ "\n",
+ "$u_{ii}=\\sqrt{a_{ii}-\\sum_{k=1}^{i-1}u_{ki}^{2}}$\n",
+ "\n",
+ "$u_{ij}=\\frac{a_{ij}-\\sum_{k=1}^{i-1}u_{ki}u_{kj}}{u_{ii}}$\n",
+ "\n",
+ "so for K"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "K =\n",
+ "\n",
+ " 20 -10 0 0\n",
+ " -10 20 -10 0\n",
+ " 0 -10 20 -10\n",
+ " 0 0 -10 10\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "K"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "u11 = 4.4721\n",
+ "u12 = -2.2361\n",
+ "u13 = 0\n",
+ "u14 = 0\n",
+ "u22 = 3.8730\n",
+ "u23 = -2.5820\n",
+ "u24 = 0\n",
+ "u33 = 3.6515\n",
+ "u34 = -2.7386\n",
+ "u44 = 1.5811\n",
+ "U =\n",
+ "\n",
+ " 4.47214 -2.23607 0.00000 0.00000\n",
+ " 0.00000 3.87298 -2.58199 0.00000\n",
+ " 0.00000 0.00000 3.65148 -2.73861\n",
+ " 0.00000 0.00000 0.00000 1.58114\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "u11=sqrt(K(1,1))\n",
+ "u12=(K(1,2))/u11\n",
+ "u13=(K(1,3))/u11\n",
+ "u14=(K(1,4))/u11\n",
+ "u22=sqrt(K(2,2)-u12^2)\n",
+ "u23=(K(2,3)-u12*u13)/u22\n",
+ "u24=(K(2,4)-u12*u14)/u22\n",
+ "u33=sqrt(K(3,3)-u13^2-u23^2)\n",
+ "u34=(K(3,4)-u13*u14-u23*u24)/u33\n",
+ "u44=sqrt(K(4,4)-u14^2-u24^2-u34^2)\n",
+ "U=[u11,u12,u13,u14;0,u22,u23,u24;0,0,u33,u34;0,0,0,u44]"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans =\n",
+ "\n",
+ " 1 1 1 1\n",
+ " 1 1 1 1\n",
+ " 1 1 1 1\n",
+ " 1 1 1 1\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "(U'*U)'==U'*U"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "average time spent for Cholesky factored solution = 1.465964e-05+/-9.806001e-06\n",
+ "average time spent for backslash solution = 1.555967e-05+/-1.048561e-05\n"
+ ]
+ }
+ ],
+ "source": [
+ "% time solution for Cholesky vs backslash\n",
+ "t_chol=zeros(1000,1);\n",
+ "t_bs=zeros(1000,1);\n",
+ "for i=1:1000\n",
+ " tic; d=U'*y; x=U\\d; t_chol(i)=toc;\n",
+ " tic; x=K\\y; t_bs(i)=toc;\n",
+ "end\n",
+ "fprintf('average time spent for Cholesky factored solution = %e+/-%e',mean(t_chol),std(t_chol))\n",
+ "\n",
+ "fprintf('average time spent for backslash solution = %e+/-%e',mean(t_bs),std(t_bs))"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Octave",
+ "language": "octave",
+ "name": "octave"
+ },
+ "language_info": {
+ "file_extension": ".m",
+ "help_links": [
+ {
+ "text": "MetaKernel Magics",
+ "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+ }
+ ],
+ "mimetype": "text/x-octave",
+ "name": "octave",
+ "version": "0.19.14"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
diff --git a/lecture_12/LU_naive.m b/lecture_12/LU_naive.m
new file mode 100644
index 0000000..92efde6
--- /dev/null
+++ b/lecture_12/LU_naive.m
@@ -0,0 +1,27 @@
+function [L, U] = LU_naive(A)
+% GaussNaive: naive Gauss elimination
+% x = GaussNaive(A,b): Gauss elimination without pivoting.
+% input:
+% A = coefficient matrix
+% y = right hand side vector
+% output:
+% x = solution vector
+[m,n] = size(A);
+if m~=n, error('Matrix A must be square'); end
+nb = n;
+L=diag(ones(n,1));
+U=A;
+% forward elimination
+for k = 1:n-1
+ for i = k+1:n
+ fik = U(i,k)/U(k,k);
+ L(i,k)=fik;
+ U(i,k:nb) = U(i,k:nb)-fik*U(k,k:nb);
+ end
+end
+%% back substitution
+%x = zeros(n,1);
+%x(n) = Aug(n,nb)/Aug(n,n);
+%for i = n-1:-1:1
+% x(i) = (Aug(i,nb)-Aug(i,i+1:n)*x(i+1:n))/Aug(i,i);
+%end
diff --git a/lecture_12/chol_pre.png b/lecture_12/chol_pre.png
new file mode 100644
index 0000000..67d2493
Binary files /dev/null and b/lecture_12/chol_pre.png differ
diff --git a/lecture_12/det_L.png b/lecture_12/det_L.png
new file mode 100644
index 0000000..aaac5a2
Binary files /dev/null and b/lecture_12/det_L.png differ
diff --git a/lecture_12/lecture_12.aux b/lecture_12/lecture_12.aux
new file mode 100644
index 0000000..4aa3a49
--- /dev/null
+++ b/lecture_12/lecture_12.aux
@@ -0,0 +1,38 @@
+\relax
+\providecommand\hyper@newdestlabel[2]{}
+\providecommand\HyperFirstAtBeginDocument{\AtBeginDocument}
+\HyperFirstAtBeginDocument{\ifx\hyper@anchor\@undefined
+\global\let\oldcontentsline\contentsline
+\gdef\contentsline#1#2#3#4{\oldcontentsline{#1}{#2}{#3}}
+\global\let\oldnewlabel\newlabel
+\gdef\newlabel#1#2{\newlabelxx{#1}#2}
+\gdef\newlabelxx#1#2#3#4#5#6{\oldnewlabel{#1}{{#2}{#3}}}
+\AtEndDocument{\ifx\hyper@anchor\@undefined
+\let\contentsline\oldcontentsline
+\let\newlabel\oldnewlabel
+\fi}
+\fi}
+\global\let\hyper@last\relax
+\gdef\HyperFirstAtBeginDocument#1{#1}
+\providecommand\HyField@AuxAddToFields[1]{}
+\providecommand\HyField@AuxAddToCoFields[2]{}
+\providecommand \oddpage@label [2]{}
+\@writefile{toc}{\contentsline {subsection}{\numberline {0.1}My question from last class}{2}{subsection.0.1}}
+\newlabel{my-question-from-last-class}{{0.1}{2}{My question from last class}{subsection.0.1}{}}
+\@writefile{lof}{\contentsline {figure}{\numberline {1}{\ignorespaces q1\relax }}{2}{figure.caption.1}}
+\@writefile{lof}{\contentsline {figure}{\numberline {2}{\ignorespaces q2\relax }}{3}{figure.caption.2}}
+\@writefile{toc}{\contentsline {subsection}{\numberline {0.2}Your questions from last class}{3}{subsection.0.2}}
+\newlabel{your-questions-from-last-class}{{0.2}{3}{Your questions from last class}{subsection.0.2}{}}
+\@writefile{toc}{\contentsline {section}{\numberline {1}Matrix Inverse and Condition}{3}{section.1}}
+\newlabel{matrix-inverse-and-condition}{{1}{3}{Matrix Inverse and Condition}{section.1}{}}
+\newlabel{note-on-solving-for-a-1-column-1}{{1}{5}{\texorpdfstring {Note on solving for \(A^{-1}\) column 1}{Note on solving for A\^{}\{-1\} column 1}}{section*.3}{}}
+\@writefile{toc}{\contentsline {paragraph}{Note on solving for \(A^{-1}\) column 1}{5}{section*.3}}
+\@writefile{toc}{\contentsline {subsection}{\numberline {1.1}Condition of a matrix}{10}{subsection.1.1}}
+\newlabel{condition-of-a-matrix}{{1.1}{10}{Condition of a matrix}{subsection.1.1}{}}
+\@writefile{toc}{\contentsline {subsubsection}{\numberline {1.1.1}\emph {just checked in to see what condition my condition was in}}{10}{subsubsection.1.1.1}}
+\newlabel{just-checked-in-to-see-what-condition-my-condition-was-in}{{1.1.1}{10}{\texorpdfstring {\emph {just checked in to see what condition my condition was in}}{just checked in to see what condition my condition was in}}{subsubsection.1.1.1}{}}
+\@writefile{toc}{\contentsline {subsubsection}{\numberline {1.1.2}Matrix norms}{10}{subsubsection.1.1.2}}
+\newlabel{matrix-norms}{{1.1.2}{10}{Matrix norms}{subsubsection.1.1.2}{}}
+\@writefile{toc}{\contentsline {subsubsection}{\numberline {1.1.3}Condition of Matrix}{11}{subsubsection.1.1.3}}
+\newlabel{condition-of-matrix}{{1.1.3}{11}{Condition of Matrix}{subsubsection.1.1.3}{}}
+\@writefile{lof}{\contentsline {figure}{\numberline {3}{\ignorespaces Springs-masses\relax }}{13}{figure.caption.4}}
diff --git a/lecture_12/lecture_12.bbl b/lecture_12/lecture_12.bbl
new file mode 100644
index 0000000..e69de29
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diff --git a/lecture_12/lecture_12.ipynb b/lecture_12/lecture_12.ipynb
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+++ b/lecture_12/lecture_12.ipynb
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+{
+ "cells": [
+ {
+ "cell_type": "code",
+ "execution_count": 27,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "%plot --format svg"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 28,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "setdefaults"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A =\n",
+ "\n",
+ " 0.447394 0.357071 0.720915 0.499926\n",
+ " 0.648313 0.323276 0.521677 0.288345\n",
+ " 0.084982 0.581513 0.466420 0.142342\n",
+ " 0.576580 0.658089 0.916987 0.923165\n",
+ "\n",
+ "L =\n",
+ "\n",
+ " 1.00000 0.00000 0.00000 0.00000\n",
+ " 0.13108 1.00000 0.00000 0.00000\n",
+ " 0.69009 0.24851 1.00000 0.00000\n",
+ " 0.88935 0.68736 0.68488 1.00000\n",
+ "\n",
+ "U =\n",
+ "\n",
+ " 0.64831 0.32328 0.52168 0.28834\n",
+ " 0.00000 0.53914 0.39804 0.10455\n",
+ " 0.00000 0.00000 0.26199 0.27496\n",
+ " 0.00000 0.00000 0.00000 0.40655\n",
+ "\n",
+ "P =\n",
+ "\n",
+ "Permutation Matrix\n",
+ "\n",
+ " 0 1 0 0\n",
+ " 0 0 1 0\n",
+ " 1 0 0 0\n",
+ " 0 0 0 1\n",
+ "\n",
+ "ans = 1\n"
+ ]
+ }
+ ],
+ "source": [
+ "A=rand(4,4)\n",
+ "\n",
+ "[L,U,P]=lu(A)\n",
+ "\n",
+ "det(L)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 44,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans =\n",
+ "\n",
+ " 4 4\n",
+ "\n",
+ "ans = 23.586\n",
+ "ans = 35.826\n",
+ "ans = 14.869\n",
+ "C =\n",
+ "\n",
+ " 5.98549 4.28555 4.35707 4.31359\n",
+ " 0.00000 3.63950 1.35005 1.45342\n",
+ " 0.00000 0.00000 3.62851 1.50580\n",
+ " 0.00000 0.00000 0.00000 3.21911\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "A=rand(4,100)';\n",
+ "A=A'*A;\n",
+ "size(A)\n",
+ "min(min(A))\n",
+ "max(max(A))\n",
+ "cond(A)\n",
+ "C=chol(A)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## My question from last class \n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Your questions from last class\n",
+ "\n",
+ "1. Will the exam be more theoretical or problem based?\n",
+ "\n",
+ "2. Writing code is difficult \n",
+ "\n",
+ "3. What format can we expect for the midterm? \n",
+ "\n",
+ "2. Could we go over some example questions for the exam?\n",
+ "\n",
+ "3. Will the use of GitHub be tested on the Midterm exam? Or is it more focused on linear algebra techniques/what was covered in the lectures?\n",
+ "\n",
+ "4. This is not my strong suit, getting a bit overwhelmed with matrix multiplication.\n",
+ "\n",
+ "5. I forgot how much I learned in linear algebra.\n",
+ "\n",
+ "6. What's the most exciting project you've ever worked on with Matlab/Octave?"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Matrix Inverse and Condition\n",
+ "\n",
+ "\n",
+ "Considering the same solution set:\n",
+ "\n",
+ "$y=Ax$\n",
+ "\n",
+ "If we know that $A^{-1}A=I$, then \n",
+ "\n",
+ "$A^{-1}y=A^{-1}Ax=x$\n",
+ "\n",
+ "so \n",
+ "\n",
+ "$x=A^{-1}y$\n",
+ "\n",
+ "Where, $A^{-1}$ is the inverse of matrix $A$.\n",
+ "\n",
+ "$2x_{1}+x_{2}=1$\n",
+ "\n",
+ "$x_{1}+3x_{2}=1$\n",
+ "\n",
+ "$Ax=y$\n",
+ "\n",
+ "$\\left[ \\begin{array}{cc}\n",
+ "2 & 1 \\\\\n",
+ "1 & 3 \\end{array} \\right]\n",
+ "\\left[\\begin{array}{c} \n",
+ "x_{1} \\\\ \n",
+ "x_{2} \\end{array}\\right]=\n",
+ "\\left[\\begin{array}{c} \n",
+ "1 \\\\\n",
+ "1\\end{array}\\right]$\n",
+ "\n",
+ "$A^{-1}=\\frac{1}{2*3-1*1}\\left[ \\begin{array}{cc}\n",
+ "3 & -1 \\\\\n",
+ "-1 & 2 \\end{array} \\right]=\n",
+ "\\left[ \\begin{array}{cc}\n",
+ "3/5 & -1/5 \\\\\n",
+ "-1/5 & 2/5 \\end{array} \\right]$\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 45,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A =\n",
+ "\n",
+ " 2 1\n",
+ " 1 3\n",
+ "\n",
+ "invA =\n",
+ "\n",
+ " 0.60000 -0.20000\n",
+ " -0.20000 0.40000\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " 1.00000 0.00000\n",
+ " 0.00000 1.00000\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " 1.00000 0.00000\n",
+ " 0.00000 1.00000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "A=[2,1;1,3]\n",
+ "invA=1/5*[3,-1;-1,2]\n",
+ "\n",
+ "A*invA\n",
+ "invA*A"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "How did we know the inverse of A? \n",
+ "\n",
+ "for 2$\\times$2 matrices, it is always:\n",
+ "\n",
+ "$A=\\left[ \\begin{array}{cc}\n",
+ "A_{11} & A_{12} \\\\\n",
+ "A_{21} & A_{22} \\end{array} \\right]$\n",
+ "\n",
+ "$A^{-1}=\\frac{1}{det(A)}\\left[ \\begin{array}{cc}\n",
+ "A_{22} & -A_{12} \\\\\n",
+ "-A_{21} & A_{11} \\end{array} \\right]$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$AA^{-1}=\\frac{1}{A_{11}A_{22}-A_{21}A_{12}}\\left[ \\begin{array}{cc}\n",
+ "A_{11}A_{22}-A_{21}A_{12} & -A_{11}A_{12}+A_{12}A_{11} \\\\\n",
+ "A_{21}A_{22}-A_{22}A_{21} & -A_{21}A_{12}+A_{22}A_{11} \\end{array} \\right]\n",
+ "=\\left[ \\begin{array}{cc}\n",
+ "1 & 0 \\\\\n",
+ "0 & 1 \\end{array} \\right]$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "What about bigger matrices?\n",
+ "\n",
+ "We can use the LU-decomposition\n",
+ "\n",
+ "$A=LU$\n",
+ "\n",
+ "$A^{-1}=(LU)^{-1}=U^{-1}L^{-1}$\n",
+ "\n",
+ "if we divide $A^{-1}$ into n-column vectors, $a_{n}$, then\n",
+ "\n",
+ "$Aa_{1}=\\left[\\begin{array}{c} \n",
+ "1 \\\\ \n",
+ "0 \\\\ \n",
+ "\\vdots \\\\\n",
+ "0 \\end{array} \\right]$\n",
+ "$Aa_{2}=\\left[\\begin{array}{c} \n",
+ "0 \\\\ \n",
+ "1 \\\\ \n",
+ "\\vdots \\\\\n",
+ "0 \\end{array} \\right]$\n",
+ "$Aa_{n}=\\left[\\begin{array}{c} \n",
+ "0 \\\\ \n",
+ "0 \\\\ \n",
+ "\\vdots \\\\\n",
+ "1 \\end{array} \\right]$\n",
+ "\n",
+ "\n",
+ "Which we can solve for each $a_{n}$ with LU-decomposition, knowing the lower and upper triangular decompositions, then \n",
+ "\n",
+ "$A^{-1}=\\left[ \\begin{array}{cccc}\n",
+ "| & | & & | \\\\\n",
+ "a_{1} & a_{2} & \\cdots & a_{n} \\\\\n",
+ "| & | & & | \\end{array} \\right]$\n",
+ "\n",
+ "\n",
+ "$Ld_{1}=\\left[\\begin{array}{c} \n",
+ "1 \\\\ \n",
+ "0 \\\\ \n",
+ "\\vdots \\\\\n",
+ "0 \\end{array} \\right]$\n",
+ "$;~Ua_{1}=d_{1}$\n",
+ "\n",
+ "$Ld_{2}=\\left[\\begin{array}{c} \n",
+ "0 \\\\ \n",
+ "1 \\\\ \n",
+ "\\vdots \\\\\n",
+ "0 \\end{array} \\right]$\n",
+ "$;~Ua_{2}=d_{2}$\n",
+ "\n",
+ "$Ld_{n}=\\left[\\begin{array}{c} \n",
+ "0 \\\\ \n",
+ "1 \\\\ \n",
+ "\\vdots \\\\\n",
+ "n \\end{array} \\right]$\n",
+ "$;~Ua_{n}=d_{n}$\n",
+ "\n",
+ "Consider the following matrix:\n",
+ "\n",
+ "$A=\\left[ \\begin{array}{ccc}\n",
+ "2 & -1 & 0\\\\\n",
+ "-1 & 2 & -1\\\\\n",
+ "0 & -1 & 1 \\end{array} \\right]$\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "#### Note on solving for $A^{-1}$ column 1\n",
+ "\n",
+ "$Aa_1=I(:,1)$\n",
+ "\n",
+ "$LUa_1=I(:,1)$\n",
+ "\n",
+ "$(LUa_1-I(:,1))=0$\n",
+ "\n",
+ "$L(Ua_1-d_1)=0$\n",
+ "\n",
+ "$I(:,1)=Ld_1$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 56,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A =\n",
+ "\n",
+ " 2 -1 0\n",
+ " -1 2 -1\n",
+ " 0 -1 1\n",
+ "\n",
+ "U =\n",
+ "\n",
+ " 2.00000 -1.00000 0.00000\n",
+ " 0.00000 1.50000 -1.00000\n",
+ " 0.00000 -1.00000 1.00000\n",
+ "\n",
+ "L =\n",
+ "\n",
+ " 1.00000 0.00000 0.00000\n",
+ " -0.50000 1.00000 0.00000\n",
+ " 0.00000 0.00000 1.00000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "A=[2,-1,0;-1,2,-1;0,-1,1]\n",
+ "U=A;\n",
+ "L=eye(3,3);\n",
+ "U(2,:)=U(2,:)-U(2,1)/U(1,1)*U(1,:)\n",
+ "L(2,1)=A(2,1)/A(1,1)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 57,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "L =\n",
+ "\n",
+ " 1.00000 0.00000 0.00000\n",
+ " -0.50000 1.00000 0.00000\n",
+ " 0.00000 -0.66667 1.00000\n",
+ "\n",
+ "U =\n",
+ "\n",
+ " 2.00000 -1.00000 0.00000\n",
+ " 0.00000 1.50000 -1.00000\n",
+ " 0.00000 0.00000 0.33333\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "L(3,2)=U(3,2)/U(2,2)\n",
+ "U(3,:)=U(3,:)-U(3,2)/U(2,2)*U(2,:)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Now solve for $d_1$ then $a_1$, $d_2$ then $a_2$, and $d_3$ then $a_{3}$\n",
+ "\n",
+ "$Ld_{1}=\\left[\\begin{array}{c} \n",
+ "1 \\\\ \n",
+ "0 \\\\ \n",
+ "\\vdots \\\\\n",
+ "0 \\end{array} \\right]=\n",
+ "\\left[\\begin{array}{ccc} \n",
+ "1 & 0 & 0 \\\\ \n",
+ "-1/2 & 1 & 0 \\\\\n",
+ "0 & -2/3 & 1 \\end{array} \\right]\\left[\\begin{array}{c} \n",
+ "d1(1) \\\\ \n",
+ "d1(2) \\\\ \n",
+ "d1(3)\\end{array} \\right]=\\left[\\begin{array}{c} \n",
+ "1 \\\\ \n",
+ "0 \\\\ \n",
+ "0 \\end{array} \\right]\n",
+ ";~Ua_{1}=d_{1}$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 58,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "d1 =\n",
+ "\n",
+ " 1.00000\n",
+ " 0.50000\n",
+ " 0.33333\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "d1=zeros(3,1);\n",
+ "d1(1)=1;\n",
+ "d1(2)=0-L(2,1)*d1(1);\n",
+ "d1(3)=0-L(3,1)*d1(1)-L(3,2)*d1(2)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 59,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "a1 =\n",
+ "\n",
+ " 1.00000\n",
+ " 1.00000\n",
+ " 1.00000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "a1=zeros(3,1);\n",
+ "a1(3)=d1(3)/U(3,3);\n",
+ "a1(2)=1/U(2,2)*(d1(2)-U(2,3)*a1(3));\n",
+ "a1(1)=1/U(1,1)*(d1(1)-U(1,2)*a1(2)-U(1,3)*a1(3))"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 60,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "d2 =\n",
+ "\n",
+ " 0.00000\n",
+ " 1.00000\n",
+ " 0.66667\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "d2=zeros(3,1);\n",
+ "d2(1)=0;\n",
+ "d2(2)=1-L(2,1)*d2(1);\n",
+ "d2(3)=0-L(3,1)*d2(1)-L(3,2)*d2(2)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 61,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "a2 =\n",
+ "\n",
+ " 1.0000\n",
+ " 2.0000\n",
+ " 2.0000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "a2=zeros(3,1);\n",
+ "a2(3)=d2(3)/U(3,3);\n",
+ "a2(2)=1/U(2,2)*(d2(2)-U(2,3)*a2(3));\n",
+ "a2(1)=1/U(1,1)*(d2(1)-U(1,2)*a2(2)-U(1,3)*a2(3))"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 62,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "d3 =\n",
+ "\n",
+ " 0\n",
+ " 0\n",
+ " 1\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "d3=zeros(3,1);\n",
+ "d3(1)=0;\n",
+ "d3(2)=0-L(2,1)*d3(1);\n",
+ "d3(3)=1-L(3,1)*d3(1)-L(3,2)*d3(2)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 63,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "a3 =\n",
+ "\n",
+ " 1.00000\n",
+ " 2.00000\n",
+ " 3.00000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "a3=zeros(3,1);\n",
+ "a3(3)=d3(3)/U(3,3);\n",
+ "a3(2)=1/U(2,2)*(d3(2)-U(2,3)*a3(3));\n",
+ "a3(1)=1/U(1,1)*(d3(1)-U(1,2)*a3(2)-U(1,3)*a3(3))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Final solution for $A^{-1}$ is $[a_{1}~a_{2}~a_{3}]$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 69,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "invA =\n",
+ "\n",
+ " 1.00000 1.00000 1.00000\n",
+ " 1.00000 2.00000 2.00000\n",
+ " 1.00000 2.00000 3.00000\n",
+ "\n",
+ "I_app =\n",
+ "\n",
+ " 1.00000 0.00000 0.00000\n",
+ " 0.00000 1.00000 -0.00000\n",
+ " -0.00000 -0.00000 1.00000\n",
+ "\n",
+ "ans = -4.4409e-16\n",
+ "ans = 2.2204e-16\n",
+ "ans = 0.0039062\n"
+ ]
+ }
+ ],
+ "source": [
+ "invA=[a1,a2,a3]\n",
+ "I_app=A*invA\n",
+ "I_app(2,3)\n",
+ "eps\n",
+ "\n",
+ "2^-8"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Now the solution of $x$ to $Ax=y$ is $x=A^{-1}y$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 70,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "y =\n",
+ "\n",
+ " 1\n",
+ " 2\n",
+ " 3\n",
+ "\n",
+ "x =\n",
+ "\n",
+ " 6.0000\n",
+ " 11.0000\n",
+ " 14.0000\n",
+ "\n",
+ "xbs =\n",
+ "\n",
+ " 6.0000\n",
+ " 11.0000\n",
+ " 14.0000\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " -3.5527e-15\n",
+ " -8.8818e-15\n",
+ " -1.0658e-14\n",
+ "\n",
+ "ans = 2.2204e-16\n"
+ ]
+ }
+ ],
+ "source": [
+ "y=[1;2;3]\n",
+ "x=invA*y\n",
+ "xbs=A\\y\n",
+ "x-xbs\n",
+ "eps"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 71,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "data": {
+ "image/svg+xml": [
+ "<svg height=\"420px\" viewBox=\"0 0 560 420\" width=\"560px\" xmlns=\"http://www.w3.org/2000/svg\" xmlns:xlink=\"http://www.w3.org/1999/xlink\">\n",
+ "\n",
+ "<title>Gnuplot</title>\n",
+ "<desc>Produced by GNUPLOT 5.0 patchlevel 3 </desc>\n",
+ "\n",
+ "<g id=\"gnuplot_canvas\">\n",
+ "\n",
+ "<rect fill=\"none\" height=\"420\" width=\"560\" x=\"0\" y=\"0\"/>\n",
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+ "\t<g id=\"gnuplot_plot_1a\"><title>inversion</title>\n",
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+ "N=100;\n",
+ "n=[1:N];\n",
+ "t_inv=zeros(N,1);\n",
+ "t_bs=zeros(N,1);\n",
+ "t_mult=zeros(N,1);\n",
+ "for i=1:N\n",
+ " A=rand(i,i);\n",
+ " tic\n",
+ " invA=inv(A);\n",
+ " t_inv(i)=toc;\n",
+ " b=rand(i,1);\n",
+ " tic;\n",
+ " x=A\\b;\n",
+ " t_bs(i)=toc;\n",
+ " tic;\n",
+ " x=invA*b;\n",
+ " t_mult(i)=toc;\n",
+ "end\n",
+ "plot(n,t_inv,n,t_bs,n,t_mult)\n",
+ "axis([0 100 0 0.002])\n",
+ "legend('inversion','backslash','multiplication','Location','NorthWest')"
+ ]
+ },
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+ "source": [
+ "## Condition of a matrix \n",
+ "### *just checked in to see what condition my condition was in*\n",
+ "### Matrix norms\n",
+ "\n",
+ "The Euclidean norm of a vector is measure of the magnitude (in 3D this would be: $|x|=\\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$) in general the equation is:\n",
+ "\n",
+ "$||x||_{e}=\\sqrt{\\sum_{i=1}^{n}x_{i}^{2}}$\n",
+ "\n",
+ "For a matrix, A, the same norm is called the Frobenius norm:\n",
+ "\n",
+ "$||A||_{f}=\\sqrt{\\sum_{i=1}^{n}\\sum_{i=1}^{m}A_{i,j}^{2}}$\n",
+ "\n",
+ "In general we can calculate any $p$-norm where\n",
+ "\n",
+ "$||A||_{p}=\\sqrt{\\sum_{i=1}^{n}\\sum_{i=1}^{m}A_{i,j}^{p}}$\n",
+ "\n",
+ "so the p=1, 1-norm is \n",
+ "\n",
+ "$||A||_{1}=\\sqrt{\\sum_{i=1}^{n}\\sum_{i=1}^{m}A_{i,j}^{1}}=\\sum_{i=1}^{n}\\sum_{i=1}^{m}|A_{i,j}|$\n",
+ "\n",
+ "$||A||_{\\infty}=\\sqrt{\\sum_{i=1}^{n}\\sum_{i=1}^{m}A_{i,j}^{\\infty}}=\\max_{1\\le i \\le n}\\sum_{j=1}^{m}|A_{i,j}|$\n",
+ "\n",
+ "### Condition of Matrix\n",
+ "\n",
+ "The matrix condition is the product of \n",
+ "\n",
+ "$Cond(A) = ||A||\\cdot||A^{-1}||$ \n",
+ "\n",
+ "So each norm will have a different condition number, but the limit is $Cond(A)\\ge 1$\n",
+ "\n",
+ "An estimate of the rounding error is based on the condition of A:\n",
+ "\n",
+ "$\\frac{||\\Delta x||}{x} \\le Cond(A) \\frac{||\\Delta A||}{||A||}$\n",
+ "\n",
+ "So if the coefficients of A have accuracy to $10^{-t}\n",
+ "\n",
+ "and the condition of A, $Cond(A)=10^{c}$\n",
+ "\n",
+ "then the solution for x can have rounding errors up to $10^{c-t}$\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 72,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A =\n",
+ "\n",
+ " 1.00000 0.50000 0.33333\n",
+ " 0.50000 0.33333 0.25000\n",
+ " 0.33333 0.25000 0.20000\n",
+ "\n",
+ "L =\n",
+ "\n",
+ " 1.00000 0.00000 0.00000\n",
+ " 0.50000 1.00000 0.00000\n",
+ " 0.33333 1.00000 1.00000\n",
+ "\n",
+ "U =\n",
+ "\n",
+ " 1.00000 0.50000 0.33333\n",
+ " 0.00000 0.08333 0.08333\n",
+ " 0.00000 -0.00000 0.00556\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "A=[1,1/2,1/3;1/2,1/3,1/4;1/3,1/4,1/5]\n",
+ "[L,U]=LU_naive(A)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "Then, $A^{-1}=(LU)^{-1}=U^{-1}L^{-1}$\n",
+ "\n",
+ "$Ld_{1}=\\left[\\begin{array}{c}\n",
+ "1 \\\\\n",
+ "0 \\\\\n",
+ "0 \\end{array}\\right]$, $Ux_{1}=d_{1}$ ..."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 75,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "invA =\n",
+ "\n",
+ " 9.0000 -36.0000 30.0000\n",
+ " -36.0000 192.0000 -180.0000\n",
+ " 30.0000 -180.0000 180.0000\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " 1.0000e+00 3.5527e-15 2.9976e-15\n",
+ " -1.3249e-14 1.0000e+00 -9.1038e-15\n",
+ " 8.5117e-15 7.1054e-15 1.0000e+00\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "invA=zeros(3,3);\n",
+ "d1=L\\[1;0;0];\n",
+ "d2=L\\[0;1;0];\n",
+ "d3=L\\[0;0;1];\n",
+ "invA(:,1)=U\\d1;\n",
+ "invA(:,2)=U\\d2;\n",
+ "invA(:,3)=U\\d3\n",
+ "invA*A"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Find the condition of A, $cond(A)$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 74,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "normf_A = 1.4136\n",
+ "normf_invA = 372.21\n",
+ "cond_f_A = 526.16\n",
+ "ans = 1.4136\n",
+ "norm1_A = 1.8333\n",
+ "norm1_invA = 30.000\n",
+ "ans = 1.8333\n",
+ "cond_1_A = 55.000\n",
+ "norminf_A = 1.8333\n",
+ "norminf_invA = 30.000\n",
+ "ans = 1.8333\n",
+ "cond_inf_A = 55.000\n"
+ ]
+ }
+ ],
+ "source": [
+ "% Frobenius norm\n",
+ "normf_A = sqrt(sum(sum(A.^2)))\n",
+ "normf_invA = sqrt(sum(sum(invA.^2)))\n",
+ "\n",
+ "cond_f_A = normf_A*normf_invA\n",
+ "\n",
+ "norm(A,'fro')\n",
+ "\n",
+ "% p=1, column sum norm\n",
+ "norm1_A = max(sum(A,2))\n",
+ "norm1_invA = max(sum(invA,2))\n",
+ "norm(A,1)\n",
+ "\n",
+ "cond_1_A=norm1_A*norm1_invA\n",
+ "\n",
+ "% p=inf, row sum norm\n",
+ "norminf_A = max(sum(A,1))\n",
+ "norminf_invA = max(sum(invA,1))\n",
+ "norm(A,inf)\n",
+ "\n",
+ "cond_inf_A=norminf_A*norminf_invA\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "Consider the problem again from the intro to Linear Algebra, 4 masses are connected in series to 4 springs with spring constants $K_{i}$. What does a high condition number mean for this problem? \n",
+ "\n",
+ "\n",
+ "\n",
+ "The masses haves the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass:\n",
+ "\n",
+ "$m_{1}g+k_{2}(x_{2}-x_{1})-k_{1}x_{1}=0$\n",
+ "\n",
+ "$m_{2}g+k_{3}(x_{3}-x_{2})-k_{2}(x_{2}-x_{1})=0$\n",
+ "\n",
+ "$m_{3}g+k_{4}(x_{4}-x_{3})-k_{3}(x_{3}-x_{2})=0$\n",
+ "\n",
+ "$m_{4}g-k_{4}(x_{4}-x_{3})=0$\n",
+ "\n",
+ "in matrix form:\n",
+ "\n",
+ "$\\left[ \\begin{array}{cccc}\n",
+ "k_{1}+k_{2} & -k_{2} & 0 & 0 \\\\\n",
+ "-k_{2} & k_{2}+k_{3} & -k_{3} & 0 \\\\\n",
+ "0 & -k_{3} & k_{3}+k_{4} & -k_{4} \\\\\n",
+ "0 & 0 & -k_{4} & k_{4} \\end{array} \\right]\n",
+ "\\left[ \\begin{array}{c}\n",
+ "x_{1} \\\\\n",
+ "x_{2} \\\\\n",
+ "x_{3} \\\\\n",
+ "x_{4} \\end{array} \\right]=\n",
+ "\\left[ \\begin{array}{c}\n",
+ "m_{1}g \\\\\n",
+ "m_{2}g \\\\\n",
+ "m_{3}g \\\\\n",
+ "m_{4}g \\end{array} \\right]$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 21,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "K =\n",
+ "\n",
+ " 100010 -100000 0 0\n",
+ " -100000 100010 -10 0\n",
+ " 0 -10 11 -1\n",
+ " 0 0 -1 1\n",
+ "\n",
+ "y =\n",
+ "\n",
+ " 9.8100\n",
+ " 19.6200\n",
+ " 29.4300\n",
+ " 39.2400\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "k1=10; % N/m\n",
+ "k2=100000;\n",
+ "k3=10;\n",
+ "k4=1;\n",
+ "m1=1; % kg\n",
+ "m2=2;\n",
+ "m3=3;\n",
+ "m4=4;\n",
+ "g=9.81; % m/s^2\n",
+ "K=[k1+k2 -k2 0 0; -k2 k2+k3 -k3 0; 0 -k3 k3+k4 -k4; 0 0 -k4 k4]\n",
+ "y=[m1*g;m2*g;m3*g;m4*g]"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans = 3.2004e+05\n",
+ "ans = 3.2004e+05\n",
+ "ans = 2.5925e+05\n",
+ "ans = 2.5293e+05\n"
+ ]
+ }
+ ],
+ "source": [
+ "cond(K,inf)\n",
+ "cond(K,1)\n",
+ "cond(K,'fro')\n",
+ "cond(K,2)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "e =\n",
+ "\n",
+ " 7.9078e-01\n",
+ " 3.5881e+00\n",
+ " 1.7621e+01\n",
+ " 2.0001e+05\n",
+ "\n",
+ "ans = 2.5293e+05\n"
+ ]
+ }
+ ],
+ "source": [
+ "e=eig(K)\n",
+ "max(e)/min(e)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Octave",
+ "language": "octave",
+ "name": "octave"
+ },
+ "language_info": {
+ "file_extension": ".m",
+ "help_links": [
+ {
+ "text": "MetaKernel Magics",
+ "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+ }
+ ],
+ "mimetype": "text/x-octave",
+ "name": "octave",
+ "version": "0.19.14"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
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diff --git a/lecture_12/lecture_12.md b/lecture_12/lecture_12.md
new file mode 100644
index 0000000..82f9bea
--- /dev/null
+++ b/lecture_12/lecture_12.md
@@ -0,0 +1,772 @@
+
+
+```octave
+%plot --format svg
+```
+
+
+```octave
+setdefaults
+```
+
+
+```octave
+A=rand(4,4)
+
+[L,U,P]=lu(A)
+
+det(L)
+```
+
+ A =
+
+ 0.447394 0.357071 0.720915 0.499926
+ 0.648313 0.323276 0.521677 0.288345
+ 0.084982 0.581513 0.466420 0.142342
+ 0.576580 0.658089 0.916987 0.923165
+
+ L =
+
+ 1.00000 0.00000 0.00000 0.00000
+ 0.13108 1.00000 0.00000 0.00000
+ 0.69009 0.24851 1.00000 0.00000
+ 0.88935 0.68736 0.68488 1.00000
+
+ U =
+
+ 0.64831 0.32328 0.52168 0.28834
+ 0.00000 0.53914 0.39804 0.10455
+ 0.00000 0.00000 0.26199 0.27496
+ 0.00000 0.00000 0.00000 0.40655
+
+ P =
+
+ Permutation Matrix
+
+ 0 1 0 0
+ 0 0 1 0
+ 1 0 0 0
+ 0 0 0 1
+
+ ans = 1
+
+
+
+```octave
+A=rand(4,100)';
+A=A'*A;
+size(A)
+min(min(A))
+max(max(A))
+cond(A)
+C=chol(A)
+```
+
+ ans =
+
+ 4 4
+
+ ans = 23.586
+ ans = 35.826
+ ans = 14.869
+ C =
+
+ 5.98549 4.28555 4.35707 4.31359
+ 0.00000 3.63950 1.35005 1.45342
+ 0.00000 0.00000 3.62851 1.50580
+ 0.00000 0.00000 0.00000 3.21911
+
+
+
+## My question from last class
+
+
+
+
+
+
+## Your questions from last class
+
+1. Will the exam be more theoretical or problem based?
+
+2. Writing code is difficult
+
+3. What format can we expect for the midterm?
+
+2. Could we go over some example questions for the exam?
+
+3. Will the use of GitHub be tested on the Midterm exam? Or is it more focused on linear algebra techniques/what was covered in the lectures?
+
+4. This is not my strong suit, getting a bit overwhelmed with matrix multiplication.
+
+5. I forgot how much I learned in linear algebra.
+
+6. What's the most exciting project you've ever worked on with Matlab/Octave?
+
+# Matrix Inverse and Condition
+
+
+Considering the same solution set:
+
+$y=Ax$
+
+If we know that $A^{-1}A=I$, then
+
+$A^{-1}y=A^{-1}Ax=x$
+
+so
+
+$x=A^{-1}y$
+
+Where, $A^{-1}$ is the inverse of matrix $A$.
+
+$2x_{1}+x_{2}=1$
+
+$x_{1}+3x_{2}=1$
+
+$Ax=y$
+
+$\left[ \begin{array}{cc}
+2 & 1 \\
+1 & 3 \end{array} \right]
+\left[\begin{array}{c}
+x_{1} \\
+x_{2} \end{array}\right]=
+\left[\begin{array}{c}
+1 \\
+1\end{array}\right]$
+
+$A^{-1}=\frac{1}{2*3-1*1}\left[ \begin{array}{cc}
+3 & -1 \\
+-1 & 2 \end{array} \right]=
+\left[ \begin{array}{cc}
+3/5 & -1/5 \\
+-1/5 & 2/5 \end{array} \right]$
+
+
+
+```octave
+A=[2,1;1,3]
+invA=1/5*[3,-1;-1,2]
+
+A*invA
+invA*A
+```
+
+ A =
+
+ 2 1
+ 1 3
+
+ invA =
+
+ 0.60000 -0.20000
+ -0.20000 0.40000
+
+ ans =
+
+ 1.00000 0.00000
+ 0.00000 1.00000
+
+ ans =
+
+ 1.00000 0.00000
+ 0.00000 1.00000
+
+
+
+How did we know the inverse of A?
+
+for 2$\times$2 matrices, it is always:
+
+$A=\left[ \begin{array}{cc}
+A_{11} & A_{12} \\
+A_{21} & A_{22} \end{array} \right]$
+
+$A^{-1}=\frac{1}{det(A)}\left[ \begin{array}{cc}
+A_{22} & -A_{12} \\
+-A_{21} & A_{11} \end{array} \right]$
+
+$AA^{-1}=\frac{1}{A_{11}A_{22}-A_{21}A_{12}}\left[ \begin{array}{cc}
+A_{11}A_{22}-A_{21}A_{12} & -A_{11}A_{12}+A_{12}A_{11} \\
+A_{21}A_{22}-A_{22}A_{21} & -A_{21}A_{12}+A_{22}A_{11} \end{array} \right]
+=\left[ \begin{array}{cc}
+1 & 0 \\
+0 & 1 \end{array} \right]$
+
+What about bigger matrices?
+
+We can use the LU-decomposition
+
+$A=LU$
+
+$A^{-1}=(LU)^{-1}=U^{-1}L^{-1}$
+
+if we divide $A^{-1}$ into n-column vectors, $a_{n}$, then
+
+$Aa_{1}=\left[\begin{array}{c}
+1 \\
+0 \\
+\vdots \\
+0 \end{array} \right]$
+$Aa_{2}=\left[\begin{array}{c}
+0 \\
+1 \\
+\vdots \\
+0 \end{array} \right]$
+$Aa_{n}=\left[\begin{array}{c}
+0 \\
+0 \\
+\vdots \\
+1 \end{array} \right]$
+
+
+Which we can solve for each $a_{n}$ with LU-decomposition, knowing the lower and upper triangular decompositions, then
+
+$A^{-1}=\left[ \begin{array}{cccc}
+| & | & & | \\
+a_{1} & a_{2} & \cdots & a_{n} \\
+| & | & & | \end{array} \right]$
+
+
+$Ld_{1}=\left[\begin{array}{c}
+1 \\
+0 \\
+\vdots \\
+0 \end{array} \right]$
+$;~Ua_{1}=d_{1}$
+
+$Ld_{2}=\left[\begin{array}{c}
+0 \\
+1 \\
+\vdots \\
+0 \end{array} \right]$
+$;~Ua_{2}=d_{2}$
+
+$Ld_{n}=\left[\begin{array}{c}
+0 \\
+1 \\
+\vdots \\
+n \end{array} \right]$
+$;~Ua_{n}=d_{n}$
+
+Consider the following matrix:
+
+$A=\left[ \begin{array}{ccc}
+2 & -1 & 0\\
+-1 & 2 & -1\\
+0 & -1 & 1 \end{array} \right]$
+
+
+#### Note on solving for $A^{-1}$ column 1
+
+$Aa_1=I(:,1)$
+
+$LUa_1=I(:,1)$
+
+$(LUa_1-I(:,1))=0$
+
+$L(Ua_1-d_1)=0$
+
+$I(:,1)=Ld_1$
+
+
+```octave
+A=[2,-1,0;-1,2,-1;0,-1,1]
+U=A;
+L=eye(3,3);
+U(2,:)=U(2,:)-U(2,1)/U(1,1)*U(1,:)
+L(2,1)=A(2,1)/A(1,1)
+```
+
+ A =
+
+ 2 -1 0
+ -1 2 -1
+ 0 -1 1
+
+ U =
+
+ 2.00000 -1.00000 0.00000
+ 0.00000 1.50000 -1.00000
+ 0.00000 -1.00000 1.00000
+
+ L =
+
+ 1.00000 0.00000 0.00000
+ -0.50000 1.00000 0.00000
+ 0.00000 0.00000 1.00000
+
+
+
+
+```octave
+L(3,2)=U(3,2)/U(2,2)
+U(3,:)=U(3,:)-U(3,2)/U(2,2)*U(2,:)
+
+```
+
+ L =
+
+ 1.00000 0.00000 0.00000
+ -0.50000 1.00000 0.00000
+ 0.00000 -0.66667 1.00000
+
+ U =
+
+ 2.00000 -1.00000 0.00000
+ 0.00000 1.50000 -1.00000
+ 0.00000 0.00000 0.33333
+
+
+
+Now solve for $d_1$ then $a_1$, $d_2$ then $a_2$, and $d_3$ then $a_{3}$
+
+$Ld_{1}=\left[\begin{array}{c}
+1 \\
+0 \\
+\vdots \\
+0 \end{array} \right]=
+\left[\begin{array}{ccc}
+1 & 0 & 0 \\
+-1/2 & 1 & 0 \\
+0 & -2/3 & 1 \end{array} \right]\left[\begin{array}{c}
+d1(1) \\
+d1(2) \\
+d1(3)\end{array} \right]=\left[\begin{array}{c}
+1 \\
+0 \\
+0 \end{array} \right]
+;~Ua_{1}=d_{1}$
+
+
+```octave
+d1=zeros(3,1);
+d1(1)=1;
+d1(2)=0-L(2,1)*d1(1);
+d1(3)=0-L(3,1)*d1(1)-L(3,2)*d1(2)
+```
+
+ d1 =
+
+ 1.00000
+ 0.50000
+ 0.33333
+
+
+
+
+```octave
+a1=zeros(3,1);
+a1(3)=d1(3)/U(3,3);
+a1(2)=1/U(2,2)*(d1(2)-U(2,3)*a1(3));
+a1(1)=1/U(1,1)*(d1(1)-U(1,2)*a1(2)-U(1,3)*a1(3))
+```
+
+ a1 =
+
+ 1.00000
+ 1.00000
+ 1.00000
+
+
+
+
+```octave
+d2=zeros(3,1);
+d2(1)=0;
+d2(2)=1-L(2,1)*d2(1);
+d2(3)=0-L(3,1)*d2(1)-L(3,2)*d2(2)
+```
+
+ d2 =
+
+ 0.00000
+ 1.00000
+ 0.66667
+
+
+
+
+```octave
+a2=zeros(3,1);
+a2(3)=d2(3)/U(3,3);
+a2(2)=1/U(2,2)*(d2(2)-U(2,3)*a2(3));
+a2(1)=1/U(1,1)*(d2(1)-U(1,2)*a2(2)-U(1,3)*a2(3))
+```
+
+ a2 =
+
+ 1.0000
+ 2.0000
+ 2.0000
+
+
+
+
+```octave
+d3=zeros(3,1);
+d3(1)=0;
+d3(2)=0-L(2,1)*d3(1);
+d3(3)=1-L(3,1)*d3(1)-L(3,2)*d3(2)
+```
+
+ d3 =
+
+ 0
+ 0
+ 1
+
+
+
+
+```octave
+a3=zeros(3,1);
+a3(3)=d3(3)/U(3,3);
+a3(2)=1/U(2,2)*(d3(2)-U(2,3)*a3(3));
+a3(1)=1/U(1,1)*(d3(1)-U(1,2)*a3(2)-U(1,3)*a3(3))
+```
+
+ a3 =
+
+ 1.00000
+ 2.00000
+ 3.00000
+
+
+
+Final solution for $A^{-1}$ is $[a_{1}~a_{2}~a_{3}]$
+
+
+```octave
+invA=[a1,a2,a3]
+I_app=A*invA
+I_app(2,3)
+eps
+
+2^-8
+```
+
+ invA =
+
+ 1.00000 1.00000 1.00000
+ 1.00000 2.00000 2.00000
+ 1.00000 2.00000 3.00000
+
+ I_app =
+
+ 1.00000 0.00000 0.00000
+ 0.00000 1.00000 -0.00000
+ -0.00000 -0.00000 1.00000
+
+ ans = -4.4409e-16
+ ans = 2.2204e-16
+ ans = 0.0039062
+
+
+Now the solution of $x$ to $Ax=y$ is $x=A^{-1}y$
+
+
+```octave
+y=[1;2;3]
+x=invA*y
+xbs=A\y
+x-xbs
+eps
+```
+
+ y =
+
+ 1
+ 2
+ 3
+
+ x =
+
+ 6.0000
+ 11.0000
+ 14.0000
+
+ xbs =
+
+ 6.0000
+ 11.0000
+ 14.0000
+
+ ans =
+
+ -3.5527e-15
+ -8.8818e-15
+ -1.0658e-14
+
+ ans = 2.2204e-16
+
+
+
+```octave
+N=100;
+n=[1:N];
+t_inv=zeros(N,1);
+t_bs=zeros(N,1);
+t_mult=zeros(N,1);
+for i=1:N
+ A=rand(i,i);
+ tic
+ invA=inv(A);
+ t_inv(i)=toc;
+ b=rand(i,1);
+ tic;
+ x=A\b;
+ t_bs(i)=toc;
+ tic;
+ x=invA*b;
+ t_mult(i)=toc;
+end
+plot(n,t_inv,n,t_bs,n,t_mult)
+axis([0 100 0 0.002])
+legend('inversion','backslash','multiplication','Location','NorthWest')
+```
+
+
+
+
+
+## Condition of a matrix
+### *just checked in to see what condition my condition was in*
+### Matrix norms
+
+The Euclidean norm of a vector is measure of the magnitude (in 3D this would be: $|x|=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$) in general the equation is:
+
+$||x||_{e}=\sqrt{\sum_{i=1}^{n}x_{i}^{2}}$
+
+For a matrix, A, the same norm is called the Frobenius norm:
+
+$||A||_{f}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{2}}$
+
+In general we can calculate any $p$-norm where
+
+$||A||_{p}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{p}}$
+
+so the p=1, 1-norm is
+
+$||A||_{1}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{1}}=\sum_{i=1}^{n}\sum_{i=1}^{m}|A_{i,j}|$
+
+$||A||_{\infty}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{\infty}}=\max_{1\le i \le n}\sum_{j=1}^{m}|A_{i,j}|$
+
+### Condition of Matrix
+
+The matrix condition is the product of
+
+$Cond(A) = ||A||\cdot||A^{-1}||$
+
+So each norm will have a different condition number, but the limit is $Cond(A)\ge 1$
+
+An estimate of the rounding error is based on the condition of A:
+
+$\frac{||\Delta x||}{x} \le Cond(A) \frac{||\Delta A||}{||A||}$
+
+So if the coefficients of A have accuracy to $10^{-t}
+
+and the condition of A, $Cond(A)=10^{c}$
+
+then the solution for x can have rounding errors up to $10^{c-t}$
+
+
+
+```octave
+A=[1,1/2,1/3;1/2,1/3,1/4;1/3,1/4,1/5]
+[L,U]=LU_naive(A)
+```
+
+ A =
+
+ 1.00000 0.50000 0.33333
+ 0.50000 0.33333 0.25000
+ 0.33333 0.25000 0.20000
+
+ L =
+
+ 1.00000 0.00000 0.00000
+ 0.50000 1.00000 0.00000
+ 0.33333 1.00000 1.00000
+
+ U =
+
+ 1.00000 0.50000 0.33333
+ 0.00000 0.08333 0.08333
+ 0.00000 -0.00000 0.00556
+
+
+
+Then, $A^{-1}=(LU)^{-1}=U^{-1}L^{-1}$
+
+$Ld_{1}=\left[\begin{array}{c}
+1 \\
+0 \\
+0 \end{array}\right]$, $Ux_{1}=d_{1}$ ...
+
+
+```octave
+invA=zeros(3,3);
+d1=L\[1;0;0];
+d2=L\[0;1;0];
+d3=L\[0;0;1];
+invA(:,1)=U\d1;
+invA(:,2)=U\d2;
+invA(:,3)=U\d3
+invA*A
+```
+
+ invA =
+
+ 9.0000 -36.0000 30.0000
+ -36.0000 192.0000 -180.0000
+ 30.0000 -180.0000 180.0000
+
+ ans =
+
+ 1.0000e+00 3.5527e-15 2.9976e-15
+ -1.3249e-14 1.0000e+00 -9.1038e-15
+ 8.5117e-15 7.1054e-15 1.0000e+00
+
+
+
+Find the condition of A, $cond(A)$
+
+
+```octave
+% Frobenius norm
+normf_A = sqrt(sum(sum(A.^2)))
+normf_invA = sqrt(sum(sum(invA.^2)))
+
+cond_f_A = normf_A*normf_invA
+
+norm(A,'fro')
+
+% p=1, column sum norm
+norm1_A = max(sum(A,2))
+norm1_invA = max(sum(invA,2))
+norm(A,1)
+
+cond_1_A=norm1_A*norm1_invA
+
+% p=inf, row sum norm
+norminf_A = max(sum(A,1))
+norminf_invA = max(sum(invA,1))
+norm(A,inf)
+
+cond_inf_A=norminf_A*norminf_invA
+
+```
+
+ normf_A = 1.4136
+ normf_invA = 372.21
+ cond_f_A = 526.16
+ ans = 1.4136
+ norm1_A = 1.8333
+ norm1_invA = 30.000
+ ans = 1.8333
+ cond_1_A = 55.000
+ norminf_A = 1.8333
+ norminf_invA = 30.000
+ ans = 1.8333
+ cond_inf_A = 55.000
+
+
+Consider the problem again from the intro to Linear Algebra, 4 masses are connected in series to 4 springs with spring constants $K_{i}$. What does a high condition number mean for this problem?
+
+
+
+The masses haves the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass:
+
+$m_{1}g+k_{2}(x_{2}-x_{1})-k_{1}x_{1}=0$
+
+$m_{2}g+k_{3}(x_{3}-x_{2})-k_{2}(x_{2}-x_{1})=0$
+
+$m_{3}g+k_{4}(x_{4}-x_{3})-k_{3}(x_{3}-x_{2})=0$
+
+$m_{4}g-k_{4}(x_{4}-x_{3})=0$
+
+in matrix form:
+
+$\left[ \begin{array}{cccc}
+k_{1}+k_{2} & -k_{2} & 0 & 0 \\
+-k_{2} & k_{2}+k_{3} & -k_{3} & 0 \\
+0 & -k_{3} & k_{3}+k_{4} & -k_{4} \\
+0 & 0 & -k_{4} & k_{4} \end{array} \right]
+\left[ \begin{array}{c}
+x_{1} \\
+x_{2} \\
+x_{3} \\
+x_{4} \end{array} \right]=
+\left[ \begin{array}{c}
+m_{1}g \\
+m_{2}g \\
+m_{3}g \\
+m_{4}g \end{array} \right]$
+
+
+```octave
+k1=10; % N/m
+k2=100000;
+k3=10;
+k4=1;
+m1=1; % kg
+m2=2;
+m3=3;
+m4=4;
+g=9.81; % m/s^2
+K=[k1+k2 -k2 0 0; -k2 k2+k3 -k3 0; 0 -k3 k3+k4 -k4; 0 0 -k4 k4]
+y=[m1*g;m2*g;m3*g;m4*g]
+```
+
+ K =
+
+ 100010 -100000 0 0
+ -100000 100010 -10 0
+ 0 -10 11 -1
+ 0 0 -1 1
+
+ y =
+
+ 9.8100
+ 19.6200
+ 29.4300
+ 39.2400
+
+
+
+
+```octave
+cond(K,inf)
+cond(K,1)
+cond(K,'fro')
+cond(K,2)
+```
+
+ ans = 3.2004e+05
+ ans = 3.2004e+05
+ ans = 2.5925e+05
+ ans = 2.5293e+05
+
+
+
+```octave
+e=eig(K)
+max(e)/min(e)
+```
+
+ e =
+
+ 7.9078e-01
+ 3.5881e+00
+ 1.7621e+01
+ 2.0001e+05
+
+ ans = 2.5293e+05
+
+
+
+```octave
+
+```
diff --git a/lecture_12/lecture_12.out b/lecture_12/lecture_12.out
new file mode 100644
index 0000000..e7b5af8
--- /dev/null
+++ b/lecture_12/lecture_12.out
@@ -0,0 +1,7 @@
+\BOOKMARK [2][-]{subsection.0.1}{My question from last class}{}% 1
+\BOOKMARK [2][-]{subsection.0.2}{Your questions from last class}{}% 2
+\BOOKMARK [1][-]{section.1}{Matrix Inverse and Condition}{}% 3
+\BOOKMARK [2][-]{subsection.1.1}{Condition of a matrix}{section.1}% 4
+\BOOKMARK [3][-]{subsubsection.1.1.1}{just checked in to see what condition my condition was in}{subsection.1.1}% 5
+\BOOKMARK [3][-]{subsubsection.1.1.2}{Matrix norms}{subsection.1.1}% 6
+\BOOKMARK [3][-]{subsubsection.1.1.3}{Condition of Matrix}{subsection.1.1}% 7
diff --git a/lecture_12/lecture_12.pdf b/lecture_12/lecture_12.pdf
new file mode 100644
index 0000000..e0b4e6c
Binary files /dev/null and b/lecture_12/lecture_12.pdf differ
diff --git a/lecture_12/lecture_12.tex b/lecture_12/lecture_12.tex
new file mode 100644
index 0000000..1265a06
--- /dev/null
+++ b/lecture_12/lecture_12.tex
@@ -0,0 +1,1015 @@
+
+% Default to the notebook output style
+
+
+
+
+% Inherit from the specified cell style.
+
+
+
+
+
+\documentclass[11pt]{article}
+
+
+
+ \usepackage[T1]{fontenc}
+ % Nicer default font (+ math font) than Computer Modern for most use cases
+ \usepackage{mathpazo}
+
+ % Basic figure setup, for now with no caption control since it's done
+ % automatically by Pandoc (which extracts  syntax from Markdown).
+ \usepackage{graphicx}
+ % We will generate all images so they have a width \maxwidth. This means
+ % that they will get their normal width if they fit onto the page, but
+ % are scaled down if they would overflow the margins.
+ \makeatletter
+ \def\maxwidth{\ifdim\Gin@nat@width>\linewidth\linewidth
+ \else\Gin@nat@width\fi}
+ \makeatother
+ \let\Oldincludegraphics\includegraphics
+ % Set max figure width to be 80% of text width, for now hardcoded.
+ \renewcommand{\includegraphics}[1]{\Oldincludegraphics[width=.8\maxwidth]{#1}}
+ % Ensure that by default, figures have no caption (until we provide a
+ % proper Figure object with a Caption API and a way to capture that
+ % in the conversion process - todo).
+ \usepackage{caption}
+ \DeclareCaptionLabelFormat{nolabel}{}
+ \captionsetup{labelformat=nolabel}
+
+ \usepackage{adjustbox} % Used to constrain images to a maximum size
+ \usepackage{xcolor} % Allow colors to be defined
+ \usepackage{enumerate} % Needed for markdown enumerations to work
+ \usepackage{geometry} % Used to adjust the document margins
+ \usepackage{amsmath} % Equations
+ \usepackage{amssymb} % Equations
+ \usepackage{textcomp} % defines textquotesingle
+ % Hack from http://tex.stackexchange.com/a/47451/13684:
+ \AtBeginDocument{%
+ \def\PYZsq{\textquotesingle}% Upright quotes in Pygmentized code
+ }
+ \usepackage{upquote} % Upright quotes for verbatim code
+ \usepackage{eurosym} % defines \euro
+ \usepackage[mathletters]{ucs} % Extended unicode (utf-8) support
+ \usepackage[utf8x]{inputenc} % Allow utf-8 characters in the tex document
+ \usepackage{fancyvrb} % verbatim replacement that allows latex
+ \usepackage{grffile} % extends the file name processing of package graphics
+ % to support a larger range
+ % The hyperref package gives us a pdf with properly built
+ % internal navigation ('pdf bookmarks' for the table of contents,
+ % internal cross-reference links, web links for URLs, etc.)
+ \usepackage{hyperref}
+ \usepackage{longtable} % longtable support required by pandoc >1.10
+ \usepackage{booktabs} % table support for pandoc > 1.12.2
+ \usepackage[inline]{enumitem} % IRkernel/repr support (it uses the enumerate* environment)
+ \usepackage[normalem]{ulem} % ulem is needed to support strikethroughs (\sout)
+ % normalem makes italics be italics, not underlines
+
+
+
+
+ % Colors for the hyperref package
+ \definecolor{urlcolor}{rgb}{0,.145,.698}
+ \definecolor{linkcolor}{rgb}{.71,0.21,0.01}
+ \definecolor{citecolor}{rgb}{.12,.54,.11}
+
+ % ANSI colors
+ \definecolor{ansi-black}{HTML}{3E424D}
+ \definecolor{ansi-black-intense}{HTML}{282C36}
+ \definecolor{ansi-red}{HTML}{E75C58}
+ \definecolor{ansi-red-intense}{HTML}{B22B31}
+ \definecolor{ansi-green}{HTML}{00A250}
+ \definecolor{ansi-green-intense}{HTML}{007427}
+ \definecolor{ansi-yellow}{HTML}{DDB62B}
+ \definecolor{ansi-yellow-intense}{HTML}{B27D12}
+ \definecolor{ansi-blue}{HTML}{208FFB}
+ \definecolor{ansi-blue-intense}{HTML}{0065CA}
+ \definecolor{ansi-magenta}{HTML}{D160C4}
+ \definecolor{ansi-magenta-intense}{HTML}{A03196}
+ \definecolor{ansi-cyan}{HTML}{60C6C8}
+ \definecolor{ansi-cyan-intense}{HTML}{258F8F}
+ \definecolor{ansi-white}{HTML}{C5C1B4}
+ \definecolor{ansi-white-intense}{HTML}{A1A6B2}
+
+ % commands and environments needed by pandoc snippets
+ % extracted from the output of `pandoc -s`
+ \providecommand{\tightlist}{%
+ \setlength{\itemsep}{0pt}\setlength{\parskip}{0pt}}
+ \DefineVerbatimEnvironment{Highlighting}{Verbatim}{commandchars=\\\{\}}
+ % Add ',fontsize=\small' for more characters per line
+ \newenvironment{Shaded}{}{}
+ \newcommand{\KeywordTok}[1]{\textcolor[rgb]{0.00,0.44,0.13}{\textbf{{#1}}}}
+ \newcommand{\DataTypeTok}[1]{\textcolor[rgb]{0.56,0.13,0.00}{{#1}}}
+ \newcommand{\DecValTok}[1]{\textcolor[rgb]{0.25,0.63,0.44}{{#1}}}
+ \newcommand{\BaseNTok}[1]{\textcolor[rgb]{0.25,0.63,0.44}{{#1}}}
+ \newcommand{\FloatTok}[1]{\textcolor[rgb]{0.25,0.63,0.44}{{#1}}}
+ \newcommand{\CharTok}[1]{\textcolor[rgb]{0.25,0.44,0.63}{{#1}}}
+ \newcommand{\StringTok}[1]{\textcolor[rgb]{0.25,0.44,0.63}{{#1}}}
+ \newcommand{\CommentTok}[1]{\textcolor[rgb]{0.38,0.63,0.69}{\textit{{#1}}}}
+ \newcommand{\OtherTok}[1]{\textcolor[rgb]{0.00,0.44,0.13}{{#1}}}
+ \newcommand{\AlertTok}[1]{\textcolor[rgb]{1.00,0.00,0.00}{\textbf{{#1}}}}
+ \newcommand{\FunctionTok}[1]{\textcolor[rgb]{0.02,0.16,0.49}{{#1}}}
+ \newcommand{\RegionMarkerTok}[1]{{#1}}
+ \newcommand{\ErrorTok}[1]{\textcolor[rgb]{1.00,0.00,0.00}{\textbf{{#1}}}}
+ \newcommand{\NormalTok}[1]{{#1}}
+
+ % Additional commands for more recent versions of Pandoc
+ \newcommand{\ConstantTok}[1]{\textcolor[rgb]{0.53,0.00,0.00}{{#1}}}
+ \newcommand{\SpecialCharTok}[1]{\textcolor[rgb]{0.25,0.44,0.63}{{#1}}}
+ \newcommand{\VerbatimStringTok}[1]{\textcolor[rgb]{0.25,0.44,0.63}{{#1}}}
+ \newcommand{\SpecialStringTok}[1]{\textcolor[rgb]{0.73,0.40,0.53}{{#1}}}
+ \newcommand{\ImportTok}[1]{{#1}}
+ \newcommand{\DocumentationTok}[1]{\textcolor[rgb]{0.73,0.13,0.13}{\textit{{#1}}}}
+ \newcommand{\AnnotationTok}[1]{\textcolor[rgb]{0.38,0.63,0.69}{\textbf{\textit{{#1}}}}}
+ \newcommand{\CommentVarTok}[1]{\textcolor[rgb]{0.38,0.63,0.69}{\textbf{\textit{{#1}}}}}
+ \newcommand{\VariableTok}[1]{\textcolor[rgb]{0.10,0.09,0.49}{{#1}}}
+ \newcommand{\ControlFlowTok}[1]{\textcolor[rgb]{0.00,0.44,0.13}{\textbf{{#1}}}}
+ \newcommand{\OperatorTok}[1]{\textcolor[rgb]{0.40,0.40,0.40}{{#1}}}
+ \newcommand{\BuiltInTok}[1]{{#1}}
+ \newcommand{\ExtensionTok}[1]{{#1}}
+ \newcommand{\PreprocessorTok}[1]{\textcolor[rgb]{0.74,0.48,0.00}{{#1}}}
+ \newcommand{\AttributeTok}[1]{\textcolor[rgb]{0.49,0.56,0.16}{{#1}}}
+ \newcommand{\InformationTok}[1]{\textcolor[rgb]{0.38,0.63,0.69}{\textbf{\textit{{#1}}}}}
+ \newcommand{\WarningTok}[1]{\textcolor[rgb]{0.38,0.63,0.69}{\textbf{\textit{{#1}}}}}
+
+
+ % Define a nice break command that doesn't care if a line doesn't already
+ % exist.
+ \def\br{\hspace*{\fill} \\* }
+ % Math Jax compatability definitions
+ \def\gt{>}
+ \def\lt{<}
+ % Document parameters
+ \title{lecture\_12}
+
+
+
+
+ % Pygments definitions
+
+\makeatletter
+\def\PY@reset{\let\PY@it=\relax \let\PY@bf=\relax%
+ \let\PY@ul=\relax \let\PY@tc=\relax%
+ \let\PY@bc=\relax \let\PY@ff=\relax}
+\def\PY@tok#1{\csname PY@tok@#1\endcsname}
+\def\PY@toks#1+{\ifx\relax#1\empty\else%
+ \PY@tok{#1}\expandafter\PY@toks\fi}
+\def\PY@do#1{\PY@bc{\PY@tc{\PY@ul{%
+ \PY@it{\PY@bf{\PY@ff{#1}}}}}}}
+\def\PY#1#2{\PY@reset\PY@toks#1+\relax+\PY@do{#2}}
+
+\expandafter\def\csname PY@tok@gd\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.63,0.00,0.00}{##1}}}
+\expandafter\def\csname PY@tok@gu\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.50,0.00,0.50}{##1}}}
+\expandafter\def\csname PY@tok@gt\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.00,0.27,0.87}{##1}}}
+\expandafter\def\csname PY@tok@gs\endcsname{\let\PY@bf=\textbf}
+\expandafter\def\csname PY@tok@gr\endcsname{\def\PY@tc##1{\textcolor[rgb]{1.00,0.00,0.00}{##1}}}
+\expandafter\def\csname PY@tok@cm\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
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+\expandafter\def\csname PY@tok@vi\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
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+\expandafter\def\csname PY@tok@cs\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
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+\expandafter\def\csname PY@tok@nc\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.00,1.00}{##1}}}
+\expandafter\def\csname PY@tok@nd\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.67,0.13,1.00}{##1}}}
+\expandafter\def\csname PY@tok@ne\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.82,0.25,0.23}{##1}}}
+\expandafter\def\csname PY@tok@nf\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.00,0.00,1.00}{##1}}}
+\expandafter\def\csname PY@tok@si\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.73,0.40,0.53}{##1}}}
+\expandafter\def\csname PY@tok@s2\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
+\expandafter\def\csname PY@tok@nt\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@nv\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
+\expandafter\def\csname PY@tok@s1\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
+\expandafter\def\csname PY@tok@ch\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
+\expandafter\def\csname PY@tok@m\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
+\expandafter\def\csname PY@tok@gp\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.00,0.50}{##1}}}
+\expandafter\def\csname PY@tok@sh\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
+\expandafter\def\csname PY@tok@ow\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.67,0.13,1.00}{##1}}}
+\expandafter\def\csname PY@tok@sx\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@bp\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@c1\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
+\expandafter\def\csname PY@tok@o\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
+\expandafter\def\csname PY@tok@kc\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@c\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
+\expandafter\def\csname PY@tok@mf\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
+\expandafter\def\csname PY@tok@err\endcsname{\def\PY@bc##1{\setlength{\fboxsep}{0pt}\fcolorbox[rgb]{1.00,0.00,0.00}{1,1,1}{\strut ##1}}}
+\expandafter\def\csname PY@tok@mb\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
+\expandafter\def\csname PY@tok@ss\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
+\expandafter\def\csname PY@tok@sr\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.40,0.53}{##1}}}
+\expandafter\def\csname PY@tok@mo\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
+\expandafter\def\csname PY@tok@kd\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@mi\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
+\expandafter\def\csname PY@tok@kn\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@cpf\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
+\expandafter\def\csname PY@tok@kr\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@s\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
+\expandafter\def\csname PY@tok@kp\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@w\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.73,0.73}{##1}}}
+\expandafter\def\csname PY@tok@kt\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.69,0.00,0.25}{##1}}}
+\expandafter\def\csname PY@tok@sc\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
+\expandafter\def\csname PY@tok@sb\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
+\expandafter\def\csname PY@tok@k\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
+\expandafter\def\csname PY@tok@se\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.73,0.40,0.13}{##1}}}
+\expandafter\def\csname PY@tok@sd\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
+
+\def\PYZbs{\char`\\}
+\def\PYZus{\char`\_}
+\def\PYZob{\char`\{}
+\def\PYZcb{\char`\}}
+\def\PYZca{\char`\^}
+\def\PYZam{\char`\&}
+\def\PYZlt{\char`\<}
+\def\PYZgt{\char`\>}
+\def\PYZsh{\char`\#}
+\def\PYZpc{\char`\%}
+\def\PYZdl{\char`\$}
+\def\PYZhy{\char`\-}
+\def\PYZsq{\char`\'}
+\def\PYZdq{\char`\"}
+\def\PYZti{\char`\~}
+% for compatibility with earlier versions
+\def\PYZat{@}
+\def\PYZlb{[}
+\def\PYZrb{]}
+\makeatother
+
+
+ % Exact colors from NB
+ \definecolor{incolor}{rgb}{0.0, 0.0, 0.5}
+ \definecolor{outcolor}{rgb}{0.545, 0.0, 0.0}
+
+
+
+
+ % Prevent overflowing lines due to hard-to-break entities
+ \sloppy
+ % Setup hyperref package
+ \hypersetup{
+ breaklinks=true, % so long urls are correctly broken across lines
+ colorlinks=true,
+ urlcolor=urlcolor,
+ linkcolor=linkcolor,
+ citecolor=citecolor,
+ }
+ % Slightly bigger margins than the latex defaults
+
+ \geometry{verbose,tmargin=1in,bmargin=1in,lmargin=1in,rmargin=1in}
+
+
+
+ \begin{document}
+
+
+ \maketitle
+
+
+
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}27}]:} \PY{c}{\PYZpc{}plot \PYZhy{}\PYZhy{}format svg}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}28}]:} \PY{n}{setdefaults}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}29}]:} \PY{n}{A}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{)}
+
+ \PY{p}{[}\PY{n}{L}\PY{p}{,}\PY{n}{U}\PY{p}{,}\PY{n}{P}\PY{p}{]}\PY{p}{=}\PY{n+nb}{lu}\PY{p}{(}\PY{n}{A}\PY{p}{)}
+
+ \PY{n+nb}{det}\PY{p}{(}\PY{n}{L}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+A =
+
+ 0.447394 0.357071 0.720915 0.499926
+ 0.648313 0.323276 0.521677 0.288345
+ 0.084982 0.581513 0.466420 0.142342
+ 0.576580 0.658089 0.916987 0.923165
+
+L =
+
+ 1.00000 0.00000 0.00000 0.00000
+ 0.13108 1.00000 0.00000 0.00000
+ 0.69009 0.24851 1.00000 0.00000
+ 0.88935 0.68736 0.68488 1.00000
+
+U =
+
+ 0.64831 0.32328 0.52168 0.28834
+ 0.00000 0.53914 0.39804 0.10455
+ 0.00000 0.00000 0.26199 0.27496
+ 0.00000 0.00000 0.00000 0.40655
+
+P =
+
+Permutation Matrix
+
+ 0 1 0 0
+ 0 0 1 0
+ 1 0 0 0
+ 0 0 0 1
+
+ans = 1
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}44}]:} \PY{n}{A}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{100}\PY{p}{)}\PY{o}{\PYZsq{}}\PY{p}{;}
+ \PY{n}{A}\PY{p}{=}\PY{n}{A}\PY{o}{\PYZsq{}}\PY{o}{*}\PY{n}{A}\PY{p}{;}
+ \PY{n+nb}{size}\PY{p}{(}\PY{n}{A}\PY{p}{)}
+ \PY{n+nb}{min}\PY{p}{(}\PY{n+nb}{min}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{)}
+ \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{max}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{)}
+ \PY{n+nb}{cond}\PY{p}{(}\PY{n}{A}\PY{p}{)}
+ \PY{n}{C}\PY{p}{=}\PY{n+nb}{chol}\PY{p}{(}\PY{n}{A}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+ans =
+
+ 4 4
+
+ans = 23.586
+ans = 35.826
+ans = 14.869
+C =
+
+ 5.98549 4.28555 4.35707 4.31359
+ 0.00000 3.63950 1.35005 1.45342
+ 0.00000 0.00000 3.62851 1.50580
+ 0.00000 0.00000 0.00000 3.21911
+
+
+ \end{Verbatim}
+
+ \subsection{My question from last
+class}\label{my-question-from-last-class}
+
+\begin{figure}[htbp]
+\centering
+\includegraphics{det_L.png}
+\caption{q1}
+\end{figure}
+
+\begin{figure}[htbp]
+\centering
+\includegraphics{chol_pre.png}
+\caption{q2}
+\end{figure}
+
+\subsection{Your questions from last
+class}\label{your-questions-from-last-class}
+
+\begin{enumerate}
+\def\labelenumi{\arabic{enumi}.}
+\item
+ Will the exam be more theoretical or problem based?
+\item
+ Writing code is difficult
+\item
+ What format can we expect for the midterm?
+\item
+ Could we go over some example questions for the exam?
+\item
+ Will the use of GitHub be tested on the Midterm exam? Or is it more
+ focused on linear algebra techniques/what was covered in the lectures?
+\item
+ This is not my strong suit, getting a bit overwhelmed with matrix
+ multiplication.
+\item
+ I forgot how much I learned in linear algebra.
+\item
+ What's the most exciting project you've ever worked on with
+ Matlab/Octave?
+\end{enumerate}
+
+ \section{Matrix Inverse and
+Condition}\label{matrix-inverse-and-condition}
+
+Considering the same solution set:
+
+\(y=Ax\)
+
+If we know that \(A^{-1}A=I\), then
+
+\(A^{-1}y=A^{-1}Ax=x\)
+
+so
+
+\(x=A^{-1}y\)
+
+Where, \(A^{-1}\) is the inverse of matrix \(A\).
+
+\(2x_{1}+x_{2}=1\)
+
+\(x_{1}+3x_{2}=1\)
+
+\(Ax=y\)
+
+\(\left[ \begin{array}{cc} 2 & 1 \\ 1 & 3 \end{array} \right] \left[\begin{array}{c} x_{1} \\ x_{2} \end{array}\right]= \left[\begin{array}{c} 1 \\ 1\end{array}\right]\)
+
+\(A^{-1}=\frac{1}{2*3-1*1}\left[ \begin{array}{cc} 3 & -1 \\ -1 & 2 \end{array} \right]= \left[ \begin{array}{cc} 3/5 & -1/5 \\ -1/5 & 2/5 \end{array} \right]\)
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}45}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{]}
+ \PY{n}{invA}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{5}\PY{o}{*}\PY{p}{[}\PY{l+m+mi}{3}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{]}
+
+ \PY{n}{A}\PY{o}{*}\PY{n}{invA}
+ \PY{n}{invA}\PY{o}{*}\PY{n}{A}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+A =
+
+ 2 1
+ 1 3
+
+invA =
+
+ 0.60000 -0.20000
+ -0.20000 0.40000
+
+ans =
+
+ 1.00000 0.00000
+ 0.00000 1.00000
+
+ans =
+
+ 1.00000 0.00000
+ 0.00000 1.00000
+
+
+ \end{Verbatim}
+
+ How did we know the inverse of A?
+
+for 2$\times$2 matrices, it is always:
+
+$A=\left[ \begin{array}{cc} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array} \right]$
+
+$A^{-1}=\frac{1}{det(A)}\left[ \begin{array}{cc} A_{22} & -A_{12} \\ -A_{21} & A_{11} \end{array} \right]$
+
+ $AA^{-1}=\frac{1}{A_{11}A_{22}-A_{21}A_{12}}\left[ \begin{array}{cc} A_{11}A_{22}-A_{21}A_{12} & -A_{11}A_{12}+A_{12}A_{11} \\ A_{21}A_{22}-A_{22}A_{21} & -A_{21}A_{12}+A_{22}A_{11} \end{array} \right] =\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$
+
+ What about bigger matrices?
+
+We can use the LU-decomposition
+
+\(A=LU\)
+
+\(A^{-1}=(LU)^{-1}=U^{-1}L^{-1}\)
+
+if we divide \(A^{-1}\) into n-column vectors, \(a_{n}\), then
+
+\(Aa_{1}=\left[\begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right]\)
+\(Aa_{2}=\left[\begin{array}{c} 0 \\ 1 \\ \vdots \\ 0 \end{array} \right]\)
+\(Aa_{n}=\left[\begin{array}{c} 0 \\ 0 \\ \vdots \\ 1 \end{array} \right]\)
+
+Which we can solve for each \(a_{n}\) with LU-decomposition, knowing the
+lower and upper triangular decompositions, then
+
+\(A^{-1}=\left[ \begin{array}{cccc} | & | & & | \\ a_{1} & a_{2} & \cdots & a_{n} \\ | & | & & | \end{array} \right]\)
+
+\(Ld_{1}=\left[\begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right]\)
+\(;~Ua_{1}=d_{1}\)
+
+\(Ld_{2}=\left[\begin{array}{c} 0 \\ 1 \\ \vdots \\ 0 \end{array} \right]\)
+\(;~Ua_{2}=d_{2}\)
+
+\(Ld_{n}=\left[\begin{array}{c} 0 \\ 1 \\ \vdots \\ n \end{array} \right]\)
+\(;~Ua_{n}=d_{n}\)
+
+Consider the following matrix:
+
+\(A=\left[ \begin{array}{ccc} 2 & -1 & 0\\ -1 & 2 & -1\\ 0 & -1 & 1 \end{array} \right]\)
+
+ \paragraph{\texorpdfstring{Note on solving for \(A^{-1}\) column
+1}{Note on solving for A\^{}\{-1\} column 1}}\label{note-on-solving-for-a-1-column-1}
+
+\(Aa_1=I(:,1)\)
+
+\(LUa_1=I(:,1)\)
+
+\((LUa_1-I(:,1))=0\)
+
+\(L(Ua_1-d_1)=0\)
+
+\(I(:,1)=Ld_1\)
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}56}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{2}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{]}
+ \PY{n}{U}\PY{p}{=}\PY{n}{A}\PY{p}{;}
+ \PY{n}{L}\PY{p}{=}\PY{n+nb}{eye}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;}
+ \PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{p}{:}\PY{p}{)}
+ \PY{n}{L}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{/}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+A =
+
+ 2 -1 0
+ -1 2 -1
+ 0 -1 1
+
+U =
+
+ 2.00000 -1.00000 0.00000
+ 0.00000 1.50000 -1.00000
+ 0.00000 -1.00000 1.00000
+
+L =
+
+ 1.00000 0.00000 0.00000
+ -0.50000 1.00000 0.00000
+ 0.00000 0.00000 1.00000
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}57}]:} \PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}
+ \PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+L =
+
+ 1.00000 0.00000 0.00000
+ -0.50000 1.00000 0.00000
+ 0.00000 -0.66667 1.00000
+
+U =
+
+ 2.00000 -1.00000 0.00000
+ 0.00000 1.50000 -1.00000
+ 0.00000 0.00000 0.33333
+
+
+ \end{Verbatim}
+
+ Now solve for \(d_1\) then \(a_1\), \(d_2\) then \(a_2\), and \(d_3\)
+then \(a_{3}\)
+
+\(Ld_{1}=\left[\begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right]= \left[\begin{array}{ccc} 1 & 0 & 0 \\ -1/2 & 1 & 0 \\ 0 & -2/3 & 1 \end{array} \right]\left[\begin{array}{c} d1(1) \\ d1(2) \\ d1(3)\end{array} \right]=\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right] ;~Ua_{1}=d_{1}\)
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}58}]:} \PY{n}{d1}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{d1}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;}
+ \PY{n}{d1}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{d1}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{d1}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{d1}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{d1}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+d1 =
+
+ 1.00000
+ 0.50000
+ 0.33333
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}59}]:} \PY{n}{a1}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{a1}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{n}{d1}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;}
+ \PY{n}{a1}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{d1}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{a1}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)}\PY{p}{;}
+ \PY{n}{a1}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{d1}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{a1}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{a1}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+a1 =
+
+ 1.00000
+ 1.00000
+ 1.00000
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}60}]:} \PY{n}{d2}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{p}{;}
+ \PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+d2 =
+
+ 0.00000
+ 1.00000
+ 0.66667
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}61}]:} \PY{n}{a2}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;}
+ \PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)}\PY{p}{;}
+ \PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+a2 =
+
+ 1.0000
+ 2.0000
+ 2.0000
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}62}]:} \PY{n}{d3}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{p}{;}
+ \PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+d3 =
+
+ 0
+ 0
+ 1
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}63}]:} \PY{n}{a3}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;}
+ \PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)}\PY{p}{;}
+ \PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+a3 =
+
+ 1.00000
+ 2.00000
+ 3.00000
+
+
+ \end{Verbatim}
+
+ Final solution for \(A^{-1}\) is \([a_{1}~a_{2}~a_{3}]\)
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}69}]:} \PY{n}{invA}\PY{p}{=}\PY{p}{[}\PY{n}{a1}\PY{p}{,}\PY{n}{a2}\PY{p}{,}\PY{n}{a3}\PY{p}{]}
+ \PY{n}{I\PYZus{}app}\PY{p}{=}\PY{n}{A}\PY{o}{*}\PY{n}{invA}
+ \PY{n}{I\PYZus{}app}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}
+ \PY{n+nb}{eps}
+
+ \PY{l+m+mi}{2}\PYZca{}\PY{o}{\PYZhy{}}\PY{l+m+mi}{8}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+invA =
+
+ 1.00000 1.00000 1.00000
+ 1.00000 2.00000 2.00000
+ 1.00000 2.00000 3.00000
+
+I\_app =
+
+ 1.00000 0.00000 0.00000
+ 0.00000 1.00000 -0.00000
+ -0.00000 -0.00000 1.00000
+
+ans = -4.4409e-16
+ans = 2.2204e-16
+ans = 0.0039062
+
+ \end{Verbatim}
+
+ Now the solution of \(x\) to \(Ax=y\) is \(x=A^{-1}y\)
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}70}]:} \PY{n}{y}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{2}\PY{p}{;}\PY{l+m+mi}{3}\PY{p}{]}
+ \PY{n}{x}\PY{p}{=}\PY{n}{invA}\PY{o}{*}\PY{n}{y}
+ \PY{n}{xbs}\PY{p}{=}\PY{n}{A}\PY{o}{\PYZbs{}}\PY{n}{y}
+ \PY{n}{x}\PY{o}{\PYZhy{}}\PY{n}{xbs}
+ \PY{n+nb}{eps}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+y =
+
+ 1
+ 2
+ 3
+
+x =
+
+ 6.0000
+ 11.0000
+ 14.0000
+
+xbs =
+
+ 6.0000
+ 11.0000
+ 14.0000
+
+ans =
+
+ -3.5527e-15
+ -8.8818e-15
+ -1.0658e-14
+
+ans = 2.2204e-16
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}71}]:} \PY{n}{N}\PY{p}{=}\PY{l+m+mi}{100}\PY{p}{;}
+ \PY{n}{n}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{:}\PY{n}{N}\PY{p}{]}\PY{p}{;}
+ \PY{n}{t\PYZus{}inv}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{n}{N}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{t\PYZus{}bs}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{n}{N}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n}{t\PYZus{}mult}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{n}{N}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{k}{for} \PY{n}{i}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{:}\PY{n}{N}
+ \PY{n}{A}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{n}{i}\PY{p}{,}\PY{n}{i}\PY{p}{)}\PY{p}{;}
+ \PY{n+nb}{tic}
+ \PY{n}{invA}\PY{p}{=}\PY{n+nb}{inv}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{;}
+ \PY{n}{t\PYZus{}inv}\PY{p}{(}\PY{n}{i}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;}
+ \PY{n}{b}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{n}{i}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
+ \PY{n+nb}{tic}\PY{p}{;}
+ \PY{n}{x}\PY{p}{=}\PY{n}{A}\PY{o}{\PYZbs{}}\PY{n}{b}\PY{p}{;}
+ \PY{n}{t\PYZus{}bs}\PY{p}{(}\PY{n}{i}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;}
+ \PY{n+nb}{tic}\PY{p}{;}
+ \PY{n}{x}\PY{p}{=}\PY{n}{invA}\PY{o}{*}\PY{n}{b}\PY{p}{;}
+ \PY{n}{t\PYZus{}mult}\PY{p}{(}\PY{n}{i}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;}
+ \PY{k}{end}
+ \PY{n+nb}{plot}\PY{p}{(}\PY{n}{n}\PY{p}{,}\PY{n}{t\PYZus{}inv}\PY{p}{,}\PY{n}{n}\PY{p}{,}\PY{n}{t\PYZus{}bs}\PY{p}{,}\PY{n}{n}\PY{p}{,}\PY{n}{t\PYZus{}mult}\PY{p}{)}
+ \PY{n+nb}{axis}\PY{p}{(}\PY{p}{[}\PY{l+m+mi}{0} \PY{l+m+mi}{100} \PY{l+m+mi}{0} \PY{l+m+mf}{0.002}\PY{p}{]}\PY{p}{)}
+ \PY{n+nb}{legend}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{inversion\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{backslash\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{multiplication\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{Location\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{NorthWest\PYZsq{}}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{center}
+ \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_12_files/lecture_12_24_0.pdf}
+ \end{center}
+ { \hspace*{\fill} \\}
+
+ \subsection{Condition of a matrix}\label{condition-of-a-matrix}
+
+\subsubsection{\texorpdfstring{\emph{just checked in to see what
+condition my condition was
+in}}{just checked in to see what condition my condition was in}}\label{just-checked-in-to-see-what-condition-my-condition-was-in}
+
+\subsubsection{Matrix norms}\label{matrix-norms}
+
+The Euclidean norm of a vector is measure of the magnitude (in 3D this
+would be: \(|x|=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}\)) in general the
+equation is:
+
+\(||x||_{e}=\sqrt{\sum_{i=1}^{n}x_{i}^{2}}\)
+
+For a matrix, A, the same norm is called the Frobenius norm:
+
+\(||A||_{f}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{2}}\)
+
+In general we can calculate any \(p\)-norm where
+
+\(||A||_{p}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{p}}\)
+
+so the p=1, 1-norm is
+
+\(||A||_{1}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{1}}=\sum_{i=1}^{n}\sum_{i=1}^{m}|A_{i,j}|\)
+
+\(||A||_{\infty}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{\infty}}=\max_{1\le i \le n}\sum_{j=1}^{m}|A_{i,j}|\)
+
+\subsubsection{Condition of Matrix}\label{condition-of-matrix}
+
+The matrix condition is the product of
+
+\(Cond(A) = ||A||\cdot||A^{-1}||\)
+
+So each norm will have a different condition number, but the limit is
+\(Cond(A)\ge 1\)
+
+An estimate of the rounding error is based on the condition of A:
+
+\(\frac{||\Delta x||}{x} \le Cond(A) \frac{||\Delta A||}{||A||}\)
+
+So if the coefficients of A have accuracy to \$10\^{}\{-t\}
+
+and the condition of A, \(Cond(A)=10^{c}\)
+
+then the solution for x can have rounding errors up to \(10^{c-t}\)
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}72}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{4}\PY{p}{;}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{5}\PY{p}{]}
+ \PY{p}{[}\PY{n}{L}\PY{p}{,}\PY{n}{U}\PY{p}{]}\PY{p}{=}\PY{n}{LU\PYZus{}naive}\PY{p}{(}\PY{n}{A}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+A =
+
+ 1.00000 0.50000 0.33333
+ 0.50000 0.33333 0.25000
+ 0.33333 0.25000 0.20000
+
+L =
+
+ 1.00000 0.00000 0.00000
+ 0.50000 1.00000 0.00000
+ 0.33333 1.00000 1.00000
+
+U =
+
+ 1.00000 0.50000 0.33333
+ 0.00000 0.08333 0.08333
+ 0.00000 -0.00000 0.00556
+
+
+ \end{Verbatim}
+
+ Then, \(A^{-1}=(LU)^{-1}=U^{-1}L^{-1}\)
+
+\(Ld_{1}=\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right]\),
+\(Ux_{1}=d_{1}\) ...
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}75}]:} \PY{n}{invA}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;}
+ \PY{n}{d1}\PY{p}{=}\PY{n}{L}\PY{o}{\PYZbs{}}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{;}
+ \PY{n}{d2}\PY{p}{=}\PY{n}{L}\PY{o}{\PYZbs{}}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{;}
+ \PY{n}{d3}\PY{p}{=}\PY{n}{L}\PY{o}{\PYZbs{}}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{;}
+ \PY{n}{invA}\PY{p}{(}\PY{p}{:}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZbs{}}\PY{n}{d1}\PY{p}{;}
+ \PY{n}{invA}\PY{p}{(}\PY{p}{:}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZbs{}}\PY{n}{d2}\PY{p}{;}
+ \PY{n}{invA}\PY{p}{(}\PY{p}{:}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZbs{}}\PY{n}{d3}
+ \PY{n}{invA}\PY{o}{*}\PY{n}{A}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+invA =
+
+ 9.0000 -36.0000 30.0000
+ -36.0000 192.0000 -180.0000
+ 30.0000 -180.0000 180.0000
+
+ans =
+
+ 1.0000e+00 3.5527e-15 2.9976e-15
+ -1.3249e-14 1.0000e+00 -9.1038e-15
+ 8.5117e-15 7.1054e-15 1.0000e+00
+
+
+ \end{Verbatim}
+
+ Find the condition of A, \(cond(A)\)
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}74}]:} \PY{c}{\PYZpc{} Frobenius norm}
+ \PY{n}{normf\PYZus{}A} \PY{p}{=} \PY{n+nb}{sqrt}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{A}\PY{o}{.\PYZca{}}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{)}
+ \PY{n}{normf\PYZus{}invA} \PY{p}{=} \PY{n+nb}{sqrt}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{invA}\PY{o}{.\PYZca{}}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{)}
+
+ \PY{n}{cond\PYZus{}f\PYZus{}A} \PY{p}{=} \PY{n}{normf\PYZus{}A}\PY{o}{*}\PY{n}{normf\PYZus{}invA}
+
+ \PY{n+nb}{norm}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{fro\PYZsq{}}\PY{p}{)}
+
+ \PY{c}{\PYZpc{} p=1, column sum norm}
+ \PY{n}{norm1\PYZus{}A} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}
+ \PY{n}{norm1\PYZus{}invA} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{invA}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}
+ \PY{n+nb}{norm}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}
+
+ \PY{n}{cond\PYZus{}1\PYZus{}A}\PY{p}{=}\PY{n}{norm1\PYZus{}A}\PY{o}{*}\PY{n}{norm1\PYZus{}invA}
+
+ \PY{c}{\PYZpc{} p=inf, row sum norm}
+ \PY{n}{norminf\PYZus{}A} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{)}
+ \PY{n}{norminf\PYZus{}invA} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{invA}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{)}
+ \PY{n+nb}{norm}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{n+nb}{inf}\PY{p}{)}
+
+ \PY{n}{cond\PYZus{}inf\PYZus{}A}\PY{p}{=}\PY{n}{norminf\PYZus{}A}\PY{o}{*}\PY{n}{norminf\PYZus{}invA}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+normf\_A = 1.4136
+normf\_invA = 372.21
+cond\_f\_A = 526.16
+ans = 1.4136
+norm1\_A = 1.8333
+norm1\_invA = 30.000
+ans = 1.8333
+cond\_1\_A = 55.000
+norminf\_A = 1.8333
+norminf\_invA = 30.000
+ans = 1.8333
+cond\_inf\_A = 55.000
+
+ \end{Verbatim}
+
+ Consider the problem again from the intro to Linear Algebra, 4 masses
+are connected in series to 4 springs with spring constants \(K_{i}\).
+What does a high condition number mean for this problem?
+
+\begin{figure}[htbp]
+\centering
+\includegraphics{../lecture_09/mass_springs.png}
+\caption{Springs-masses}
+\end{figure}
+
+The masses haves the following amounts, 1, 2, 3, and 4 kg for masses
+1-4. Using a FBD for each mass:
+
+\(m_{1}g+k_{2}(x_{2}-x_{1})-k_{1}x_{1}=0\)
+
+\(m_{2}g+k_{3}(x_{3}-x_{2})-k_{2}(x_{2}-x_{1})=0\)
+
+\(m_{3}g+k_{4}(x_{4}-x_{3})-k_{3}(x_{3}-x_{2})=0\)
+
+\(m_{4}g-k_{4}(x_{4}-x_{3})=0\)
+
+in matrix form:
+
+\(\left[ \begin{array}{cccc} k_{1}+k_{2} & -k_{2} & 0 & 0 \\ -k_{2} & k_{2}+k_{3} & -k_{3} & 0 \\ 0 & -k_{3} & k_{3}+k_{4} & -k_{4} \\ 0 & 0 & -k_{4} & k_{4} \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right]= \left[ \begin{array}{c} m_{1}g \\ m_{2}g \\ m_{3}g \\ m_{4}g \end{array} \right]\)
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}21}]:} \PY{n}{k1}\PY{p}{=}\PY{l+m+mi}{10}\PY{p}{;} \PY{c}{\PYZpc{} N/m}
+ \PY{n}{k2}\PY{p}{=}\PY{l+m+mi}{100000}\PY{p}{;}
+ \PY{n}{k3}\PY{p}{=}\PY{l+m+mi}{10}\PY{p}{;}
+ \PY{n}{k4}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;}
+ \PY{n}{m1}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} \PY{c}{\PYZpc{} kg}
+ \PY{n}{m2}\PY{p}{=}\PY{l+m+mi}{2}\PY{p}{;}
+ \PY{n}{m3}\PY{p}{=}\PY{l+m+mi}{3}\PY{p}{;}
+ \PY{n}{m4}\PY{p}{=}\PY{l+m+mi}{4}\PY{p}{;}
+ \PY{n}{g}\PY{p}{=}\PY{l+m+mf}{9.81}\PY{p}{;} \PY{c}{\PYZpc{} m/s\PYZca{}2}
+ \PY{n}{K}\PY{p}{=}\PY{p}{[}\PY{n}{k1}\PY{o}{+}\PY{n}{k2} \PY{o}{\PYZhy{}}\PY{n}{k2} \PY{l+m+mi}{0} \PY{l+m+mi}{0}\PY{p}{;} \PY{o}{\PYZhy{}}\PY{n}{k2} \PY{n}{k2}\PY{o}{+}\PY{n}{k3} \PY{o}{\PYZhy{}}\PY{n}{k3} \PY{l+m+mi}{0}\PY{p}{;} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k3} \PY{n}{k3}\PY{o}{+}\PY{n}{k4} \PY{o}{\PYZhy{}}\PY{n}{k4}\PY{p}{;} \PY{l+m+mi}{0} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k4} \PY{n}{k4}\PY{p}{]}
+ \PY{n}{y}\PY{p}{=}\PY{p}{[}\PY{n}{m1}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m2}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m3}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m4}\PY{o}{*}\PY{n}{g}\PY{p}{]}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+K =
+
+ 100010 -100000 0 0
+ -100000 100010 -10 0
+ 0 -10 11 -1
+ 0 0 -1 1
+
+y =
+
+ 9.8100
+ 19.6200
+ 29.4300
+ 39.2400
+
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}25}]:} \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{n+nb}{inf}\PY{p}{)}
+ \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}
+ \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{fro\PYZsq{}}\PY{p}{)}
+ \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+ans = 3.2004e+05
+ans = 3.2004e+05
+ans = 2.5925e+05
+ans = 2.5293e+05
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor}26}]:} \PY{n+nb}{e}\PY{p}{=}\PY{n+nb}{eig}\PY{p}{(}\PY{n}{K}\PY{p}{)}
+ \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{e}\PY{p}{)}\PY{o}{/}\PY{n+nb}{min}\PY{p}{(}\PY{n+nb}{e}\PY{p}{)}
+\end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+e =
+
+ 7.9078e-01
+ 3.5881e+00
+ 1.7621e+01
+ 2.0001e+05
+
+ans = 2.5293e+05
+
+ \end{Verbatim}
+
+ \begin{Verbatim}[commandchars=\\\{\}]
+{\color{incolor}In [{\color{incolor} }]:}
+\end{Verbatim}
+
+
+ % Add a bibliography block to the postdoc
+
+
+
+ \end{document}
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+
+<title>Gnuplot</title>
+<desc>Produced by GNUPLOT 5.0 patchlevel 3 </desc>
+
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+
+<rect fill="none" height="420" width="560" x="0" y="0"/>
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+
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+ </g>
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+ </g>
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+ <text><tspan font-family="{}">0</tspan></text>
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+ <text><tspan font-family="{}">20</tspan></text>
+ </g>
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+ <text><tspan font-family="{}">40</tspan></text>
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+ <text><tspan font-family="{}">60</tspan></text>
+ </g>
+</g>
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+ <text><tspan font-family="{}">80</tspan></text>
+ </g>
+</g>
+<g color="black" fill="none" stroke="currentColor" stroke-linecap="butt" stroke-linejoin="miter" stroke-width="0.50">
+ <path d="M535.0,384.0 L535.0,371.5 M535.0,16.7 L535.0,29.2 " stroke="black"/> <g fill="rgb(0,0,0)" font-family="{}" font-size="16.00" stroke="none" text-anchor="middle" transform="translate(535.0,408.0)">
+ <text><tspan font-family="{}">100</tspan></text>
+ </g>
+</g>
+<g color="black" fill="none" stroke="currentColor" stroke-linecap="butt" stroke-linejoin="miter" stroke-width="0.50">
+</g>
+<g color="black" fill="none" stroke="currentColor" stroke-linecap="butt" stroke-linejoin="miter" stroke-width="0.50">
+ <path d="M70.5,16.7 L70.5,384.0 L535.0,384.0 L535.0,16.7 L70.5,16.7 Z " stroke="black"/></g>
+<g color="black" fill="none" stroke="currentColor" stroke-linecap="butt" stroke-linejoin="miter" stroke-width="0.50">
+</g>
+<g color="black" fill="none" stroke="black" stroke-linecap="butt" stroke-linejoin="miter" stroke-width="1.00">
+</g>
+<g color="black" fill="none" stroke="currentColor" stroke-linecap="butt" stroke-linejoin="miter" stroke-width="1.00">
+ <path d="M78.8,169.7 L78.8,25.7 L311.8,25.7 L311.8,169.7 L78.8,169.7 Z " stroke="black"/></g>
+ <g id="gnuplot_plot_1a"><title>inversion</title>
+<g color="white" fill="none" stroke="black" stroke-linecap="butt" stroke-linejoin="miter" stroke-width="3.00">
+</g>
+<g color="black" fill="none" stroke="currentColor" stroke-linecap="butt" stroke-linejoin="miter" stroke-width="3.00">
+ <g fill="rgb(0,0,0)" font-family="{}" font-size="16.00" stroke="none" text-anchor="end" transform="translate(311.8,55.7)">
+ <text><tspan font-family="{}">inversion</tspan></text>
+ </g>
+</g>
+<g color="black" fill="none" stroke="currentColor" stroke-linecap="butt" stroke-linejoin="miter" stroke-width="3.00">
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+ </g>
+ <g id="gnuplot_plot_2a"><title>backslash</title>
+<g color="black" fill="none" stroke="currentColor" stroke-linecap="butt" stroke-linejoin="miter" stroke-width="3.00">
+ <g fill="rgb(0,0,0)" font-family="{}" font-size="16.00" stroke="none" text-anchor="end" transform="translate(311.8,103.7)">
+ <text><tspan font-family="{}">backslash</tspan></text>
+ </g>
+</g>
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+ </g>
+ <g id="gnuplot_plot_3a"><title>multiplication</title>
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+ <text><tspan font-family="{}">multiplication</tspan></text>
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\ No newline at end of file
diff --git a/lecture_12/nohup.out b/lecture_12/nohup.out
new file mode 100644
index 0000000..f0d30e1
--- /dev/null
+++ b/lecture_12/nohup.out
@@ -0,0 +1,2 @@
+
+(evince:8926): Gtk-WARNING **: gtk_widget_size_allocate(): attempt to allocate widget with width -71 and height 20
diff --git a/lecture_12/octave-workspace b/lecture_12/octave-workspace
new file mode 100644
index 0000000..41ef164
Binary files /dev/null and b/lecture_12/octave-workspace differ
diff --git a/lecture_13/.ipynb_checkpoints/lecture_12-checkpoint.ipynb b/lecture_13/.ipynb_checkpoints/lecture_12-checkpoint.ipynb
new file mode 100644
index 0000000..6a5dc68
--- /dev/null
+++ b/lecture_13/.ipynb_checkpoints/lecture_12-checkpoint.ipynb
@@ -0,0 +1,737 @@
+{
+ "cells": [
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "%plot --format svg"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "setdefaults"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## My question from last class \n",
+ "\n",
+ "\n",
+ "## Your questions from last class\n",
+ "\n",
+ "1. Need more linear algebra review\n",
+ " \n",
+ " -We will keep doing Linear Algebra, try the practice problems in 'linear_algebra'\n",
+ "\n",
+ "2. How do I do HW3? \n",
+ " \n",
+ " -demo today\n",
+ "\n",
+ "3. For hw4 is the spring constant (K) suppose to be given? \n",
+ " \n",
+ " -yes, its 30 N/m\n",
+ " \n",
+ "4. Deapool or Joker?"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# LU Decomposition\n",
+ "### efficient storage of matrices for solutions\n",
+ "\n",
+ "Considering the same solution set:\n",
+ "\n",
+ "$y=Ax$\n",
+ "\n",
+ "Assume that we can perform Gauss elimination and achieve this formula:\n",
+ "\n",
+ "$Ux=d$ \n",
+ "\n",
+ "Where, $U$ is an upper triangular matrix that we derived from Gauss elimination and $d$ is the set of dependent variables after Gauss elimination. \n",
+ "\n",
+ "Assume there is a lower triangular matrix, $L$, with ones on the diagonal and same dimensions of $U$ and the following is true:\n",
+ "\n",
+ "$L(Ux-d)=Ax-y=0$\n",
+ "\n",
+ "Now, $Ax=LUx$, so $A=LU$, and $y=Ld$.\n",
+ "\n",
+ "$2x_{1}+x_{2}=1$\n",
+ "\n",
+ "$x_{1}+3x_{2}=1$\n",
+ "\n",
+ "\n",
+ "$\\left[ \\begin{array}{cc}\n",
+ "2 & 1 \\\\\n",
+ "1 & 3 \\end{array} \\right]\n",
+ "\\left[\\begin{array}{c} \n",
+ "x_{1} \\\\ \n",
+ "x_{2} \\end{array}\\right]=\n",
+ "\\left[\\begin{array}{c} \n",
+ "1 \\\\\n",
+ "1\\end{array}\\right]$\n",
+ "\n",
+ "f21=0.5\n",
+ "\n",
+ "A(2,1)=1-1 = 0 \n",
+ "\n",
+ "A(2,2)=3-0.5=2.5\n",
+ "\n",
+ "y(2)=1-0.5=0.5\n",
+ "\n",
+ "$L(Ux-d)=\n",
+ "\\left[ \\begin{array}{cc}\n",
+ "1 & 0 \\\\\n",
+ "0.5 & 1 \\end{array} \\right]\n",
+ "\\left(\\left[ \\begin{array}{cc}\n",
+ "2 & 1 \\\\\n",
+ "0 & 2.5 \\end{array} \\right]\n",
+ "\\left[\\begin{array}{c} \n",
+ "x_{1} \\\\ \n",
+ "x_{2} \\end{array}\\right]-\n",
+ "\\left[\\begin{array}{c} \n",
+ "1 \\\\\n",
+ "0.5\\end{array}\\right]\\right)=0$\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A =\n",
+ "\n",
+ " 2 1\n",
+ " 1 3\n",
+ "\n",
+ "L =\n",
+ "\n",
+ " 1.00000 0.00000\n",
+ " 0.50000 1.00000\n",
+ "\n",
+ "U =\n",
+ "\n",
+ " 2.00000 1.00000\n",
+ " 0.00000 2.50000\n",
+ "\n",
+ "ans =\n",
+ "\n",
+ " 2 1\n",
+ " 1 3\n",
+ "\n",
+ "d =\n",
+ "\n",
+ " 1.00000\n",
+ " 0.50000\n",
+ "\n",
+ "y =\n",
+ "\n",
+ " 1\n",
+ " 1\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "A=[2,1;1,3]\n",
+ "L=[1,0;0.5,1]\n",
+ "U=[2,1;0,2.5]\n",
+ "L*U\n",
+ "\n",
+ "d=[1;0.5]\n",
+ "y=L*d"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans = 5.0000\n",
+ "ans = 1\n",
+ "ans = 5\n",
+ "ans = 5\n",
+ "ans = 5.0000\n"
+ ]
+ }
+ ],
+ "source": [
+ "% what is the determinant of L, U and A?\n",
+ "\n",
+ "det(A)\n",
+ "det(L)\n",
+ "det(U)\n",
+ "det(L)*det(U)\n",
+ "det(L*U)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Pivoting for LU factorization\n",
+ "\n",
+ "LU factorization uses the same method as Gauss elimination so it is also necessary to perform partial pivoting when creating the lower and upper triangular matrices. \n",
+ "\n",
+ "Matlab and Octave use pivoting in the command \n",
+ "\n",
+ "`[L,U,P]=lu(A)`\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "'lu' is a built-in function from the file libinterp/corefcn/lu.cc\n",
+ "\n",
+ " -- Built-in Function: [L, U] = lu (A)\n",
+ " -- Built-in Function: [L, U, P] = lu (A)\n",
+ " -- Built-in Function: [L, U, P, Q] = lu (S)\n",
+ " -- Built-in Function: [L, U, P, Q, R] = lu (S)\n",
+ " -- Built-in Function: [...] = lu (S, THRES)\n",
+ " -- Built-in Function: Y = lu (...)\n",
+ " -- Built-in Function: [...] = lu (..., \"vector\")\n",
+ " Compute the LU decomposition of A.\n",
+ "\n",
+ " If A is full subroutines from LAPACK are used and if A is sparse\n",
+ " then UMFPACK is used.\n",
+ "\n",
+ " The result is returned in a permuted form, according to the\n",
+ " optional return value P. For example, given the matrix 'a = [1, 2;\n",
+ " 3, 4]',\n",
+ "\n",
+ " [l, u, p] = lu (A)\n",
+ "\n",
+ " returns\n",
+ "\n",
+ " l =\n",
+ "\n",
+ " 1.00000 0.00000\n",
+ " 0.33333 1.00000\n",
+ "\n",
+ " u =\n",
+ "\n",
+ " 3.00000 4.00000\n",
+ " 0.00000 0.66667\n",
+ "\n",
+ " p =\n",
+ "\n",
+ " 0 1\n",
+ " 1 0\n",
+ "\n",
+ " The matrix is not required to be square.\n",
+ "\n",
+ " When called with two or three output arguments and a spare input\n",
+ " matrix, 'lu' does not attempt to perform sparsity preserving column\n",
+ " permutations. Called with a fourth output argument, the sparsity\n",
+ " preserving column transformation Q is returned, such that 'P * A *\n",
+ " Q = L * U'.\n",
+ "\n",
+ " Called with a fifth output argument and a sparse input matrix, 'lu'\n",
+ " attempts to use a scaling factor R on the input matrix such that 'P\n",
+ " * (R \\ A) * Q = L * U'. This typically leads to a sparser and more\n",
+ " stable factorization.\n",
+ "\n",
+ " An additional input argument THRES, that defines the pivoting\n",
+ " threshold can be given. THRES can be a scalar, in which case it\n",
+ " defines the UMFPACK pivoting tolerance for both symmetric and\n",
+ " unsymmetric cases. If THRES is a 2-element vector, then the first\n",
+ " element defines the pivoting tolerance for the unsymmetric UMFPACK\n",
+ " pivoting strategy and the second for the symmetric strategy. By\n",
+ " default, the values defined by 'spparms' are used ([0.1, 0.001]).\n",
+ "\n",
+ " Given the string argument \"vector\", 'lu' returns the values of P\n",
+ " and Q as vector values, such that for full matrix, 'A (P,:) = L *\n",
+ " U', and 'R(P,:) * A (:, Q) = L * U'.\n",
+ "\n",
+ " With two output arguments, returns the permuted forms of the upper\n",
+ " and lower triangular matrices, such that 'A = L * U'. With one\n",
+ " output argument Y, then the matrix returned by the LAPACK routines\n",
+ " is returned. If the input matrix is sparse then the matrix L is\n",
+ " embedded into U to give a return value similar to the full case.\n",
+ " For both full and sparse matrices, 'lu' loses the permutation\n",
+ " information.\n",
+ "\n",
+ " See also: luupdate, ilu, chol, hess, qr, qz, schur, svd.\n",
+ "\n",
+ "Additional help for built-in functions and operators is\n",
+ "available in the online version of the manual. Use the command\n",
+ "'doc <topic>' to search the manual index.\n",
+ "\n",
+ "Help and information about Octave is also available on the WWW\n",
+ "at http://www.octave.org and via the help@octave.org\n",
+ "mailing list.\n"
+ ]
+ }
+ ],
+ "source": [
+ "help lu"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
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+ "source": [
+ "% time LU solution vs backslash\n",
+ "t_lu=zeros(100,1);\n",
+ "t_bs=zeros(100,1);\n",
+ "for N=1:100\n",
+ " A=rand(N,N);\n",
+ " y=rand(N,1);\n",
+ " [L,U,P]=lu(A);\n",
+ "\n",
+ " tic; d=inv(L)*y; x=inv(U)*d; t_lu(N)=toc;\n",
+ "\n",
+ " tic; x=inv(A)*y; t_bs(N)=toc;\n",
+ "end\n",
+ "plot([1:100],t_lu,[1:100],t_bs) \n",
+ "legend('LU decomp','Octave \\\\')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "Consider the problem again from the intro to Linear Algebra, 4 masses are connected in series to 4 springs with K=10 N/m. What are the final positions of the masses? \n",
+ "\n",
+ "\n",
+ "\n",
+ "The masses haves the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass:\n",
+ "\n",
+ "$m_{1}g+k(x_{2}-x_{1})-kx_{1}=0$\n",
+ "\n",
+ "$m_{2}g+k(x_{3}-x_{2})-k(x_{2}-x_{1})=0$\n",
+ "\n",
+ "$m_{3}g+k(x_{4}-x_{3})-k(x_{3}-x_{2})=0$\n",
+ "\n",
+ "$m_{4}g-k(x_{4}-x_{3})=0$\n",
+ "\n",
+ "in matrix form:\n",
+ "\n",
+ "$\\left[ \\begin{array}{cccc}\n",
+ "2k & -k & 0 & 0 \\\\\n",
+ "-k & 2k & -k & 0 \\\\\n",
+ "0 & -k & 2k & -k \\\\\n",
+ "0 & 0 & -k & k \\end{array} \\right]\n",
+ "\\left[ \\begin{array}{c}\n",
+ "x_{1} \\\\\n",
+ "x_{2} \\\\\n",
+ "x_{3} \\\\\n",
+ "x_{4} \\end{array} \\right]=\n",
+ "\\left[ \\begin{array}{c}\n",
+ "m_{1}g \\\\\n",
+ "m_{2}g \\\\\n",
+ "m_{3}g \\\\\n",
+ "m_{4}g \\end{array} \\right]$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "K =\n",
+ "\n",
+ " 20 -10 0 0\n",
+ " -10 20 -10 0\n",
+ " 0 -10 20 -10\n",
+ " 0 0 -10 10\n",
+ "\n",
+ "y =\n",
+ "\n",
+ " 9.8100\n",
+ " 19.6200\n",
+ " 29.4300\n",
+ " 39.2400\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "k=10; % N/m\n",
+ "m1=1; % kg\n",
+ "m2=2;\n",
+ "m3=3;\n",
+ "m4=4;\n",
+ "g=9.81; % m/s^2\n",
+ "K=[2*k -k 0 0; -k 2*k -k 0; 0 -k 2*k -k; 0 0 -k k]\n",
+ "y=[m1*g;m2*g;m3*g;m4*g]"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This matrix, K, is symmetric. \n",
+ "\n",
+ "`K(i,j)==K(j,i)`\n",
+ "\n",
+ "Now we can use,\n",
+ "\n",
+ "## Cholesky Factorization\n",
+ "\n",
+ "We can decompose the matrix, K into two matrices, $U$ and $U^{T}$, where\n",
+ "\n",
+ "$K=U^{T}U$\n",
+ "\n",
+ "each of the components of U can be calculated with the following equations:\n",
+ "\n",
+ "$u_{ii}=\\sqrt{a_{ii}-\\sum_{k=1}^{i-1}u_{ki}^{2}}$\n",
+ "\n",
+ "$u_{ij}=\\frac{a_{ij}-\\sum_{k=1}^{i-1}u_{ki}u_{kj}}{u_{ii}}$\n",
+ "\n",
+ "so for K"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "K =\n",
+ "\n",
+ " 20 -10 0 0\n",
+ " -10 20 -10 0\n",
+ " 0 -10 20 -10\n",
+ " 0 0 -10 10\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "K"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "u11 = 4.4721\n",
+ "u12 = -2.2361\n",
+ "u13 = 0\n",
+ "u14 = 0\n",
+ "u22 = 3.8730\n",
+ "u23 = -2.5820\n",
+ "u24 = 0\n",
+ "u33 = 3.6515\n",
+ "u34 = -2.7386\n",
+ "u44 = 1.5811\n",
+ "U =\n",
+ "\n",
+ " 4.47214 -2.23607 0.00000 0.00000\n",
+ " 0.00000 3.87298 -2.58199 0.00000\n",
+ " 0.00000 0.00000 3.65148 -2.73861\n",
+ " 0.00000 0.00000 0.00000 1.58114\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "u11=sqrt(K(1,1))\n",
+ "u12=(K(1,2))/u11\n",
+ "u13=(K(1,3))/u11\n",
+ "u14=(K(1,4))/u11\n",
+ "u22=sqrt(K(2,2)-u12^2)\n",
+ "u23=(K(2,3)-u12*u13)/u22\n",
+ "u24=(K(2,4)-u12*u14)/u22\n",
+ "u33=sqrt(K(3,3)-u13^2-u23^2)\n",
+ "u34=(K(3,4)-u13*u14-u23*u24)/u33\n",
+ "u44=sqrt(K(4,4)-u14^2-u24^2-u34^2)\n",
+ "U=[u11,u12,u13,u14;0,u22,u23,u24;0,0,u33,u34;0,0,0,u44]"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans =\n",
+ "\n",
+ " 1 1 1 1\n",
+ " 1 1 1 1\n",
+ " 1 1 1 1\n",
+ " 1 1 1 1\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "(U'*U)'==U'*U"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "average time spent for Cholesky factored solution = 1.465964e-05+/-9.806001e-06\n",
+ "average time spent for backslash solution = 1.555967e-05+/-1.048561e-05\n"
+ ]
+ }
+ ],
+ "source": [
+ "% time solution for Cholesky vs backslash\n",
+ "t_chol=zeros(1000,1);\n",
+ "t_bs=zeros(1000,1);\n",
+ "for i=1:1000\n",
+ " tic; d=U'*y; x=U\\d; t_chol(i)=toc;\n",
+ " tic; x=K\\y; t_bs(i)=toc;\n",
+ "end\n",
+ "fprintf('average time spent for Cholesky factored solution = %e+/-%e',mean(t_chol),std(t_chol))\n",
+ "\n",
+ "fprintf('average time spent for backslash solution = %e+/-%e',mean(t_bs),std(t_bs))"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Octave",
+ "language": "octave",
+ "name": "octave"
+ },
+ "language_info": {
+ "file_extension": ".m",
+ "help_links": [
+ {
+ "text": "MetaKernel Magics",
+ "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+ }
+ ],
+ "mimetype": "text/x-octave",
+ "name": "octave",
+ "version": "0.19.14"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
diff --git a/linear_algebra/LU_suggested.html b/linear_algebra/LU_suggested.html
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--- /dev/null
+++ b/linear_algebra/LU_suggested.html
@@ -0,0 +1,39 @@
+<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
+<html xmlns="http://www.w3.org/1999/xhtml">
+<head>
+ <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
+ <meta http-equiv="Content-Style-Type" content="text/css" />
+ <meta name="generator" content="pandoc" />
+ <title></title>
+ <style type="text/css">code{white-space: pre;}</style>
+ <script src="https://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML" type="text/javascript"></script>
+</head>
+<body>
+<h1 id="linear-algebra-review">Linear Algebra Review</h1>
+<h2 id="gauss-elimination-suggested-problems">(Gauss Elimination) Suggested problems</h2>
+<h3 id="no-due-date">No due date</h3>
+<ol style="list-style-type: decimal">
+<li><p>Determine the lower (L) and upper (U) triangular matrices with LU-decomposition for the following matrices:</p>
+<ol style="list-style-type: lower-alpha">
+<li><p><span class="math inline">\(A=\left[ \begin{array}{cc} 1 & 3 \\ 2 & 1 \end{array} \right]\)</span></p></li>
+<li><p><span class="math inline">\(A=\left[ \begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array} \right]\)</span></p></li>
+<li><p><span class="math inline">\(A=\left[ \begin{array}{cc} 1 & 1 \\ 2 & -2 \end{array} \right]\)</span></p></li>
+<li><p><span class="math inline">\(A=\left[ \begin{array}{ccc} 1 & 3 & 1 \\ -4 & -9 & 2 \\ 0 & 3 & 6\end{array} \right]\)</span></p></li>
+<li><p><span class="math inline">\(A=\left[ \begin{array}{ccc} 1 & 3 & 1 \\ -4 & -9 & 2 \\ 0 & 3 & 6\end{array} \right]\)</span></p></li>
+<li><p><span class="math inline">\(A=\left[ \begin{array}{ccc} 1 & 3 & -5 \\ 1 & 4 & -8 \\ -3 & -7 & 9\end{array} \right]\)</span></p></li>
+<li><p><span class="math inline">\(A=\left[ \begin{array}{ccc} 1 & 2 & -1 \\ 2 & 2 & 2 \\ 1 & -1 & 2\end{array} \right]\)</span></p></li>
+</ol></li>
+<li><p>Calculate the determinant of A from 1a-g.</p></li>
+<li><p>Determine the Cholesky factorization, C, of the following matrices, where</p>
+<p><span class="math inline">\(C_{ii}=\sqrt{a_{ii}-\sum_{k=1}^{i-1}C_{ki}^{2}}\)</span></p>
+<p><span class="math inline">\(C_{ij}=\frac{a_{ij}-\sum_{k=1}^{i-1}C_{ki}C_{kj}}{C_{ii}}\)</span>.</p>
+<ol style="list-style-type: lower-alpha">
+<li><p>A=<span class="math inline">\(\left[ \begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array} \right]\)</span></p></li>
+<li><p>A=<span class="math inline">\(\left[ \begin{array}{cc} 10 & 5 \\ 5 & 20 \end{array} \right]\)</span></p></li>
+<li><p>A=<span class="math inline">\(\left[ \begin{array}{ccc} 10 & -10 & 20 \\ -10 & 20 & 10 \\ 20 & 10 & 30 \end{array} \right]\)</span></p></li>
+<li><p>A=<span class="math inline">\(\left[ \begin{array}{cccc} 21 & -1 & 0 & 0 \\ -1 & 21 & -1 & 0 \\ 0 & -1 & 21 & -1 \\ 0 & 0 & -1 & 1 \end{array} \right]\)</span></p></li>
+</ol></li>
+<li><p>Verify that <span class="math inline">\(C^{T}C=A\)</span> for 3a-d</p></li>
+</ol>
+</body>
+</html>
diff --git a/linear_algebra/LU_suggested.md b/linear_algebra/LU_suggested.md
new file mode 100644
index 0000000..f809813
--- /dev/null
+++ b/linear_algebra/LU_suggested.md
@@ -0,0 +1,68 @@
+# Linear Algebra Review
+## (Gauss Elimination) Suggested problems
+### No due date
+
+1. Determine the lower (L) and upper (U) triangular matrices with LU-decomposition for the
+following matrices:
+
+ a. $A=\left[ \begin{array}{cc}
+ 1 & 3 \\
+ 2 & 1 \end{array} \right]$
+
+ a. $A=\left[ \begin{array}{cc}
+ 1 & 1 \\
+ 2 & 3 \end{array} \right]$
+
+ a. $A=\left[ \begin{array}{cc}
+ 1 & 1 \\
+ 2 & -2 \end{array} \right]$
+
+ b. $A=\left[ \begin{array}{ccc}
+ 1 & 3 & 1 \\
+ -4 & -9 & 2 \\
+ 0 & 3 & 6\end{array} \right]$
+
+ c. $A=\left[ \begin{array}{ccc}
+ 1 & 3 & 1 \\
+ -4 & -9 & 2 \\
+ 0 & 3 & 6\end{array} \right]$
+
+ d. $A=\left[ \begin{array}{ccc}
+ 1 & 3 & -5 \\
+ 1 & 4 & -8 \\
+ -3 & -7 & 9\end{array} \right]$
+
+ d. $A=\left[ \begin{array}{ccc}
+ 1 & 2 & -1 \\
+ 2 & 2 & 2 \\
+ 1 & -1 & 2\end{array} \right]$
+
+2. Calculate the determinant of A from 1a-g.
+
+3. Determine the Cholesky factorization, C, of the following matrices, where
+
+ $C_{ii}=\sqrt{a_{ii}-\sum_{k=1}^{i-1}C_{ki}^{2}}$
+
+ $C_{ij}=\frac{a_{ij}-\sum_{k=1}^{i-1}C_{ki}C_{kj}}{C_{ii}}$.
+
+ a. A=$\left[ \begin{array}{cc}
+ 3 & 2 \\
+ 2 & 1 \end{array} \right]$
+
+ a. A=$\left[ \begin{array}{cc}
+ 10 & 5 \\
+ 5 & 20 \end{array} \right]$
+
+ a. A=$\left[ \begin{array}{ccc}
+ 10 & -10 & 20 \\
+ -10 & 20 & 10 \\
+ 20 & 10 & 30 \end{array} \right]$
+
+ a. A=$\left[ \begin{array}{cccc}
+ 21 & -1 & 0 & 0 \\
+ -1 & 21 & -1 & 0 \\
+ 0 & -1 & 21 & -1 \\
+ 0 & 0 & -1 & 1 \end{array} \right]$
+
+4. Verify that $C^{T}C=A$ for 3a-d
+
diff --git a/linear_algebra/gauss_suggested.md b/linear_algebra/gauss_suggested.md
new file mode 100644
index 0000000..6b1f6b6
--- /dev/null
+++ b/linear_algebra/gauss_suggested.md
@@ -0,0 +1,71 @@
+# Linear Algebra Review
+## (Gauss Elimination) Suggested problems
+### No due date
+
+1. Solve for x when Ax=b for the following problems:
+
+ a. $A=\left[ \begin{array}{cc}
+ 1 & 3 \\
+ 2 & 1 \end{array} \right]
+ b=
+ \left[\begin{array}{c}
+ 1 \\
+ 1\end{array}\right]$
+
+ a. $A=\left[ \begin{array}{cc}
+ 1 & 1 \\
+ 2 & 3 \end{array} \right]
+ b=
+ \left[\begin{array}{c}
+ 3 \\
+ 4\end{array}\right]$
+
+ a. $A=\left[ \begin{array}{cc}
+ 1 & 1 \\
+ 2 & -2 \end{array} \right]
+ b=
+ \left[\begin{array}{c}
+ 4 \\
+ 2\end{array}\right]$
+
+ b. $A=\left[ \begin{array}{ccc}
+ 1 & 3 & 1 \\
+ -4 & -9 & 2 \\
+ 0 & 3 & 6\end{array} \right]
+ b=
+ \left[\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0\end{array}\right]$
+
+ c. $A=\left[ \begin{array}{ccc}
+ 1 & 3 & 1 \\
+ -4 & -9 & 2 \\
+ 0 & 3 & 6\end{array} \right]
+ b=
+ \left[\begin{array}{c}
+ 1 \\
+ -1 \\
+ -3\end{array}\right]$
+
+ d. $A=\left[ \begin{array}{ccc}
+ 1 & 3 & -5 \\
+ 1 & 4 & -8 \\
+ -3 & -7 & 9\end{array} \right]
+ b=
+ \left[\begin{array}{c}
+ 1 \\
+ -1 \\
+ -3\end{array}\right]$
+
+ d. $A=\left[ \begin{array}{ccc}
+ 1 & 2 & -1 \\
+ 2 & 2 & 2 \\
+ 1 & -1 & 2\end{array} \right]
+ b=
+ \left[\begin{array}{c}
+ 2 \\
+ 12 \\
+ 5\end{array}\right]$
+
+2. Calculate the determinant of A from 1a-g.
diff --git a/linear_algebra/gauss_suggested.pdf b/linear_algebra/gauss_suggested.pdf
new file mode 100644
index 0000000..a78b277
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