%% This document created by Scientific Notebook (R) Version 3.0


\documentclass[12pt,thmsa]{article}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage{sw20jart}

%TCIDATA{TCIstyle=article/art4.lat,jart,sw20jart}

%TCIDATA{OutputFilter=LATEX.DLL}
%TCIDATA{Version=4.10.0.2359}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Monday, February 17, 2003 09:08:07}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{CSTFile=webmath_11pt.cst}
%TCIDATA{PageSetup=72,72,36,36,0}
%TCIDATA{ComputeDefs=
%$x\left( t\right) =c_{1}e^{2t}\left[ 
%\begin{array}{c}
%-2 \\ 
%1 \\ 
%4%
%\end{array}%
%\right] +c_{2}e^{3t}\left[ 
%\begin{array}{c}
%-1 \\ 
%1 \\ 
%4%
%\end{array}%
%\right] +c_{3}e^{t}\left[ 
%\begin{array}{c}
%-1 \\ 
%1 \\ 
%2%
%\end{array}%
%\right] $
%}


\input{tcilatex}

\begin{document}


\section{Ma 227\qquad \qquad \qquad Practice for Exam I - Solutions\qquad
\qquad \qquad \qquad}

$\qquad \qquad \qquad \qquad \qquad $

\subsection{\protect\vspace{1pt}Problem 1 \ }

Find the eigenvalues and eigenvectors of

\vspace{1pt}

\begin{center}
\qquad \qquad \qquad \qquad \qquad $\left[ 
\begin{array}{lll}
1 & 3 & -2 \\ 
0 & 2 & 4 \\ 
0 & 1 & -1%
\end{array}%
\right] \qquad $
\end{center}

$\qquad \det \left[ 
\begin{array}{lll}
1-r & 3 & -2 \\ 
0 & 2-r & 4 \\ 
0 & 1 & -1-r%
\end{array}%
\right] =$ $\left( 1-r\right) \left( -6-r+r^{2}\right) =\allowbreak -\left(
-1+r\right) \left( r+2\right) \left( r-3\right) $,

The system of equations that gives the eigenvectors is%
\begin{eqnarray*}
\left( 1-r\right) x_{1}+3x_{2}-2x_{3} &=&0 \\
0x_{1}+\left( 2-r\right) x_{2}+4x_{3} &=&0 \\
0x_{3}+x_{2}+\left( -1-r\right) x_{3} &=&0
\end{eqnarray*}

\vspace{1pt}

$r=1\qquad \left[ 
\begin{array}{cccc}
0 & 3 & -2 & 0 \\ 
0 & 1 & 4 & 0 \\ 
0 & 1 & -2 & 0%
\end{array}%
\right] $, row echelon form: $\left[ 
\begin{array}{cccc}
0 & 1 & 0 & 0 \\ 
0 & 0 & 1 & 0 \\ 
0 & 0 & 0 & 0%
\end{array}%
\right] $

eigenvectors: $\left\{ \left[ 
\begin{array}{c}
\frac{5}{3} \\ 
-1 \\ 
1%
\end{array}%
\right] \right\} \leftrightarrow -2,\left\{ \left[ 
\begin{array}{c}
1 \\ 
0 \\ 
0%
\end{array}%
\right] \right\} \leftrightarrow 1,\left\{ \left[ 
\begin{array}{c}
5 \\ 
4 \\ 
1%
\end{array}%
\right] \right\} \leftrightarrow 3$

\vspace{1pt}

\subsection{Problem 2}

\vspace{1pt}a) \ Row reduce the matrix $B$ to echelon form, where

\qquad \qquad

\begin{center}
$\vspace{1pt}B=\left[ 
\begin{array}{llll}
2 & 1 & 1 & 2 \\ 
1 & 2 & 1 & 1 \\ 
-1 & 1 & -2 & -1%
\end{array}%
\right] $, row echelon form: $\left[ 
\begin{array}{cccc}
1 & 0 & 0 & 1 \\ 
0 & 1 & 0 & 0 \\ 
0 & 0 & 1 & 0%
\end{array}%
\right] $ .
\end{center}

\vspace{1pt}\qquad \qquad \qquad \qquad

\vspace{1pt}\vspace{1pt}b) \ Find all values of $t$\ for which the system of
equations below will have a non-trivial solution.

\qquad \qquad \qquad 
\begin{eqnarray*}
x-ty+z &=&0 \\
2y+tz &=&0 \\
y+z &=&0
\end{eqnarray*}

$\det \left[ 
\begin{array}{ccc}
1 & -t & 1 \\ 
0 & 2 & t \\ 
0 & 1 & 1%
\end{array}%
\right] =\allowbreak 2-t,$ Thus for $t=2$ the system has a non-trivial
solution.

c) Solve, if possible, the system

\begin{eqnarray*}
x_{1}+2x_{2}-3x_{3}-4x_{4} &=&6 \\
x_{1}+3x_{2}+x_{3}-2x_{4} &=&4 \\
2x_{1}+5x_{2}-2x_{3}-5x_{4} &=&10\smallskip
\end{eqnarray*}

$\left[ 
\begin{array}{ccccc}
1 & 2 & -3 & -4 & 6 \\ 
1 & 3 & 1 & -2 & 4 \\ 
2 & 5 & -2 & -5 & 10%
\end{array}%
\right] $, row echelon form: $\left[ 
\begin{array}{ccccc}
1 & 0 & -11 & 0 & 10 \\ 
0 & 1 & 4 & 0 & -2 \\ 
0 & 0 & 0 & 1 & 0%
\end{array}%
\right] $

Thus%
\begin{eqnarray*}
x_{4} &=&0 \\
x_{2} &=&-4x_{3}-2 \\
x_{1} &=&11x_{3}+10
\end{eqnarray*}

$x_{3}$ arbitrary.

\subsection{\protect\vspace{1pt}Problem 3}

Let

\vspace{1pt}\qquad \qquad \qquad 
\[
B=\left[ 
\begin{array}{lll}
1 & 2 & 1 \\ 
1 & 3 & 2 \\ 
1 & 0 & 1%
\end{array}%
\right] 
\]%
Find $B^{-1}$.\qquad

\vspace{1pt}$\left[ 
\begin{array}{cccccc}
1 & 2 & 1 & 1 & 0 & 0 \\ 
1 & 3 & 2 & 0 & 1 & 0 \\ 
1 & 0 & 1 & 0 & 0 & 1%
\end{array}%
\right] \rightarrow ^{-R_{1}+R_{2};-R_{1}+R_{3}}\left[ 
\begin{array}{cccccc}
1 & 2 & 1 & 1 & 0 & 0 \\ 
0 & 1 & 1 & -1 & 1 & 0 \\ 
0 & -2 & 0 & -1 & 0 & 1%
\end{array}%
\right] \rightarrow ^{2R_{2}+R_{3};-2R_{2}+R_{1}}$

$\left[ 
\begin{array}{cccccc}
1 & 0 & -1 & 3 & -2 & 0 \\ 
0 & 1 & 1 & -1 & 1 & 0 \\ 
0 & 0 & 2 & -3 & 2 & 1%
\end{array}%
\right] \rightarrow ^{\frac{1}{2}R_{3}}\left[ 
\begin{array}{cccccc}
1 & 0 & -1 & 3 & -2 & 0 \\ 
0 & 1 & 1 & -1 & 1 & 0 \\ 
0 & 0 & 1 & -\frac{3}{2} & 1 & \frac{1}{2}%
\end{array}%
\right] \rightarrow ^{-R_{3}+R_{2};R_{3}+R_{1}}$

$\left[ 
\begin{array}{cccccc}
1 & 0 & 0 & \dfrac{3}{2} & -1 & \dfrac{1}{2} \\ 
0 & 1 & 0 & \dfrac{1}{2} & 0 & \dfrac{1}{2} \\ 
0 & 0 & 1 & -\dfrac{3}{2} & 1 & \dfrac{1}{2}%
\end{array}%
\right] $

Check:$\left[ 
\begin{array}{lll}
1 & 2 & 1 \\ 
1 & 3 & 2 \\ 
1 & 0 & 1%
\end{array}%
\right] $, inverse: $\left[ 
\begin{array}{ccc}
\frac{3}{2} & -1 & \frac{1}{2} \\ 
\frac{1}{2} & 0 & -\frac{1}{2} \\ 
-\frac{3}{2} & 1 & \frac{1}{2}%
\end{array}%
\right] $

\subsection{Problem 4 \ }

a) Find the general solution of 
\[
x^{\prime }=Ax 
\]

where 
\[
A=\left[ 
\begin{array}{cc}
1 & -1 \\ 
-2 & 0%
\end{array}%
\right] 
\]

$\left[ 
\begin{array}{cc}
1 & -1 \\ 
-2 & 0%
\end{array}%
\right] $, eigenvectors: $\left\{ \left[ 
\begin{array}{c}
-1 \\ 
1%
\end{array}%
\right] \right\} \leftrightarrow 2,\left\{ \left[ 
\begin{array}{c}
1 \\ 
2%
\end{array}%
\right] \right\} \leftrightarrow -1$

\[
x\left( t\right) =c_{1}e^{2t}\left[ 
\begin{array}{c}
-1 \\ 
1%
\end{array}%
\right] +c_{2}e^{-t}\left[ 
\begin{array}{c}
1 \\ 
2%
\end{array}%
\right] 
\]

b) Write the equation $y^{\prime \prime \prime }-2y^{\prime \prime
}+2y^{\prime }+4y=\sin t$ in \underline{normal} form $(x^{\prime }=Ax+f)$

\vspace{1pt}$\left[ 
\begin{array}{l}
x_{1}^{\prime } \\ 
x_{2}^{\prime } \\ 
x_{3}^{\prime }%
\end{array}%
\right] =\left[ 
\begin{array}{lll}
0 & 1 & 0 \\ 
0 & 0 & 1 \\ 
-4 & -2 & 2%
\end{array}%
\right] \left[ 
\begin{array}{l}
x_{1} \\ 
x_{2} \\ 
x_{3}%
\end{array}%
\right] +\left[ 
\begin{array}{l}
0 \\ 
0 \\ 
\sin t%
\end{array}%
\right] $

c) Solve the initial value problem%
\[
x^{\prime }\left( t\right) =\left[ 
\begin{array}{ccc}
1 & 2 & -1 \\ 
1 & 0 & 1 \\ 
4 & -4 & 5%
\end{array}%
\right] \text{ \ \ \ \ \ }x\left( 0\right) =\left[ 
\begin{array}{c}
-1 \\ 
0 \\ 
0%
\end{array}%
\right] 
\]

\vspace{1pt}$\left[ 
\begin{array}{ccc}
1 & 2 & -1 \\ 
1 & 0 & 1 \\ 
4 & -4 & 5%
\end{array}%
\right] $, eigenvectors: $\left\{ \left[ 
\begin{array}{c}
-2 \\ 
1 \\ 
4%
\end{array}%
\right] \right\} \leftrightarrow 2,\left\{ \left[ 
\begin{array}{c}
-1 \\ 
1 \\ 
4%
\end{array}%
\right] \right\} \leftrightarrow 3,\left\{ \left[ 
\begin{array}{c}
-1 \\ 
1 \\ 
2%
\end{array}%
\right] \right\} \leftrightarrow 1$

\vspace{1pt}

\[
x\left( t\right) =c_{1}e^{2t}\left[ 
\begin{array}{c}
-2 \\ 
1 \\ 
4%
\end{array}%
\right] +c_{2}e^{3t}\left[ 
\begin{array}{c}
-1 \\ 
1 \\ 
4%
\end{array}%
\right] +c_{3}e^{t}\left[ 
\begin{array}{c}
-1 \\ 
1 \\ 
2%
\end{array}%
\right] 
\]

$x\left( 0\right) =\allowbreak \left[ 
\begin{array}{c}
-2c_{1}-c_{2}-c_{3} \\ 
c_{1}+c_{2}+c_{3} \\ 
4c_{1}+4c_{2}+2c_{3}%
\end{array}%
\right] =\left[ 
\begin{array}{c}
-1 \\ 
0 \\ 
0%
\end{array}%
\right] $

\vspace{1pt}To solve for $c_{1},c_{2},c_{3}$ form

$\left[ 
\begin{array}{cccc}
-2 & -1 & -1 & -1 \\ 
1 & 1 & 1 & 0 \\ 
4 & 4 & 2 & 0%
\end{array}%
\right] $, row echelon form: $\left[ 
\begin{array}{cccc}
1 & 0 & 0 & 1 \\ 
0 & 1 & 0 & -1 \\ 
0 & 0 & 1 & 0%
\end{array}%
\right] $

Thus $c_{1}=1,c_{2}=-1,c_{3}=0$ so

\qquad \qquad \qquad $\ \ \ \ \ $%
\[
x\left( t\right) =e^{2t}\left[ 
\begin{array}{c}
-2 \\ 
1 \\ 
4%
\end{array}%
\right] -e^{3t}\left[ 
\begin{array}{c}
-1 \\ 
1 \\ 
4%
\end{array}%
\right] 
\]%
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad
\qquad \ \ \ \ \ \ $

\qquad \qquad \qquad \qquad $\ $

\vspace{1pt}

\end{document}
