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%%%%%%%%%%%%%%%%%%% Start /document/ex2.1_sol_12.tex %%%%%%%%%%%%%%%%%%

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


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%TCIDATA{Created=Monday, February 01, 1999 14:09:24}
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\input{tcilatex}

\begin{document}


\section{Ma 112}

\section{Exercises 2.1:  Solutions}

\vspace{1pt}1. SNB ICE

\vspace{1pt}

Graph the system of equations

\vspace{1pt}

\begin{eqnarray*}
x-3y &=&5 \\
2x+y &=&1
\end{eqnarray*}

\QTP{Body Math}
$\vspace{1pt}y=\frac{1}{3}x-\frac{5}{3}$\FRAME{itbpF}{3in}{2.0003in}{0in}{}{%
}{}{\special{language "Scientific Word";type "MAPLEPLOT";width 3in;height
2.0003in;depth 0in;display "USEDEF";function
\TEXUX{$\frac{1}{3}x-\frac{5}{3}$};linecolor "black";linestyle
1;linethickness 1;pointstyle "point";function \TEXUX{$-2x+1$};linecolor
"black";linestyle 1;linethickness 1;pointstyle "point";xmin "-5";xmax
"5";xviewmin "-5.000000";xviewmax "5.000000";yviewmin "-9.400000";yviewmax
"11.408000";rangeset"X";phi 45;theta 45;plottype 4;numpoints 49;axesstyle
"normal";xis \TEXUX{x};var1name \TEXUX{$x$};}}

\QTP{Body Math}
$y=-2x+1$

Now use SNB to find the solution to the system.

$x-3y=5$

$2x+y=1$, Solution is : $\left\{ y=-\frac{9}{7},x=\frac{8}{7}\right\} $

\QTP{Body Math}
$\vspace{1pt}$

2. Strang, page 30 \#10

\vspace{1pt} \ \ (a)$Ax=2\left[ 
\begin{array}{c}
1 \\ 
-2 \\ 
-4
\end{array}
\right] +2\left[ 
\begin{array}{c}
2 \\ 
3 \\ 
1
\end{array}
\right] +3\left[ 
\begin{array}{c}
4 \\ 
1 \\ 
2
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
18 \\ 
5 \\ 
0
\end{array}
\right] $

\ \ \ \ \ \ \ $Ax=1\left[ 
\begin{array}{c}
2 \\ 
1 \\ 
0 \\ 
0
\end{array}
\right] +1\left[ 
\begin{array}{c}
1 \\ 
2 \\ 
1 \\ 
0
\end{array}
\right] +1\left[ 
\begin{array}{c}
0 \\ 
1 \\ 
2 \\ 
1
\end{array}
\right] +2\left[ 
\begin{array}{c}
0 \\ 
0 \\ 
1 \\ 
2
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
3 \\ 
4 \\ 
5 \\ 
5
\end{array}
\right] $

\ \ (b) $9$ multiplications,

3. Strang, page 30 \#11

\vspace{1pt}\qquad (a) $Ax=4\left[ 
\begin{array}{c}
2 \\ 
5
\end{array}
\right] +2\left[ 
\begin{array}{c}
3 \\ 
1
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
14 \\ 
22
\end{array}
\right] $

\qquad (b) $Ax=2\left[ 
\begin{array}{c}
3 \\ 
6
\end{array}
\right] -1\left[ 
\begin{array}{c}
6 \\ 
12
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
0 \\ 
0
\end{array}
\right] $

\qquad (c) $Ax=3\left[ 
\begin{array}{c}
1 \\ 
2
\end{array}
\right] +1\left[ 
\begin{array}{c}
2 \\ 
0
\end{array}
\right] +1\left[ 
\begin{array}{c}
4 \\ 
1
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
9 \\ 
7
\end{array}
\right] $

4. Strang, page 30 \#12

\vspace{1pt}\qquad (a) $Ax=x\left[ 
\begin{array}{c}
0 \\ 
0 \\ 
1
\end{array}
\right] +y\left[ 
\begin{array}{c}
0 \\ 
1 \\ 
0
\end{array}
\right] +z\left[ 
\begin{array}{c}
1 \\ 
0 \\ 
0
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
z \\ 
y \\ 
x
\end{array}
\right] $

\qquad (b) $Ax=1\left[ 
\begin{array}{c}
2 \\ 
1 \\ 
3
\end{array}
\right] +1\left[ 
\begin{array}{c}
1 \\ 
2 \\ 
3
\end{array}
\right] -1\left[ 
\begin{array}{c}
3 \\ 
3 \\ 
6
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
0 \\ 
0 \\ 
0
\end{array}
\right] $

\qquad (c) $Ax=1\left[ 
\begin{array}{c}
2 \\ 
1 \\ 
3
\end{array}
\right] +1\left[ 
\begin{array}{c}
1 \\ 
2 \\ 
3
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
3 \\ 
3 \\ 
6
\end{array}
\right] $

5. Strang, page 30 \#15

\vspace{1pt}\qquad (a) $\left[ 
\begin{array}{cc}
1 & 0 \\ 
0 & 1
\end{array}
\right] \left[ 
\begin{array}{c}
x \\ 
y
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
x \\ 
y
\end{array}
\right] $

\qquad (b)$\left[ 
\begin{array}{cc}
0 & 1 \\ 
1 & 0
\end{array}
\right] \left[ 
\begin{array}{c}
x \\ 
y
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
y \\ 
x
\end{array}
\right] $

6. Strang, page 30 \#16

\vspace{1pt}$\qquad $(a)$R\left[ 
\begin{array}{c}
x \\ 
y
\end{array}
\right] =\left[ 
\begin{array}{cc}
0 & 1 \\ 
-1 & 0
\end{array}
\right] \left[ 
\begin{array}{c}
x \\ 
y
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
y \\ 
-x
\end{array}
\right] $

\qquad (b)rotate by $180^{0}:R1\left[ 
\begin{array}{c}
x \\ 
y
\end{array}
\right] =\left[ 
\begin{array}{c}
-x \\ 
-y
\end{array}
\right] $, $R1=R^{2}$

\qquad \qquad $\left[ 
\begin{array}{cc}
-1 & 0 \\ 
0 & -1
\end{array}
\right] \left[ 
\begin{array}{c}
x \\ 
y
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
-x \\ 
-y
\end{array}
\right] $

7. Strang, page 30 \#17

\vspace{1pt}\qquad (a) $P=\left[ 
\begin{array}{ccc}
0 & 1 & 0 \\ 
0 & 0 & 1 \\ 
1 & 0 & 0
\end{array}
\right] \left[ 
\begin{array}{c}
x \\ 
y \\ 
z
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
y \\ 
z \\ 
x
\end{array}
\right] $

\qquad (b) $P^{-1}=\left[ 
\begin{array}{ccc}
0 & 0 & 1 \\ 
1 & 0 & 0 \\ 
0 & 1 & 0
\end{array}
\right] \left[ 
\begin{array}{c}
y \\ 
z \\ 
x
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
x \\ 
y \\ 
z
\end{array}
\right] $

8. Strang, page 30 \#18

\vspace{1pt}\qquad $E\left[ 
\begin{array}{c}
3 \\ 
5
\end{array}
\right] =\left[ 
\begin{array}{c}
3 \\ 
2
\end{array}
\right] $ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ $E\left[ 
\begin{array}{c}
3 \\ 
5 \\ 
7
\end{array}
\right] =\left[ 
\begin{array}{c}
3 \\ 
2 \\ 
7
\end{array}
\right] $

\qquad (a) $\left[ 
\begin{array}{cc}
1 & 0 \\ 
-1 & 1
\end{array}
\right] \left[ 
\begin{array}{c}
3 \\ 
5
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
3 \\ 
2
\end{array}
\right] $

\qquad (b) $\left[ 
\begin{array}{ccc}
1 & 0 & 0 \\ 
-1 & 1 & 0 \\ 
0 & 0 & 1
\end{array}
\right] \left[ 
\begin{array}{c}
3 \\ 
5 \\ 
7
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
3 \\ 
2 \\ 
7
\end{array}
\right] $

\qquad There cannot be any other answer because it has to be ``substract the
first component from the second component''.

9. SNB ICE

\vspace{1pt}

Use forward elimination to find the pivots and solve

\vspace{1pt}

\begin{center}
$x+y+2z=-1$

$2x+y+3z=0$

$-2y+z=2$

\begin{eqnarray*}
x+y+2z &=&-1 \\
2x+y+3z &=&0 \\
-2y+z &=&2
\end{eqnarray*}
\end{center}

\vspace{1pt}The first pivot is $1.$ Eliminating $x$ from the second equation
yields:

\begin{eqnarray*}
x+y+2z &=&-1 \\
-y-z &=&2 \\
-2y+z &=&2
\end{eqnarray*}

Thus the second pivot is $-1.$ Eliminating $y$ from the tird equation yields

\vspace{1pt}

\begin{eqnarray*}
x+y+2z &=&-1 \\
-y-z &=&2 \\
3z &=&-2
\end{eqnarray*}

Thus the third pivot is $3.$ The solution to the system is $z=-2/3,$

$y=-2-z=-2+\frac{2}{3}=-\frac{4}{3}$

$x=-y-2z-1$ $=\frac{4}{3}-2\left( -\frac{2}{3}\right) -1$ $=\frac{5}{3}$

\vspace{1pt}

which is easily checked by Maple.

\begin{eqnarray*}
x+y+2z &=&-1 \\
-y-z &=&2 \\
3z &=&-2
\end{eqnarray*}
, Solution is: $\left\{ z=-\frac{2}{3},y=-\frac{4}{3},x=\frac{5}{3}\right\} $

\end{document}

%%%%%%%%%%%%%%%%%%%% End /document/ex2.1_sol_12.tex %%%%%%%%%%%%%%%%%%%