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\begin{document}

\begin{flushright}
Kyle Daling

MATH 204

Assignment 13
\end{flushright}
\begin{large}
Section 1.8
\end{large}
\begin{enumerate}
\item[17.]
$T(3\vec{u})=3T(\vec{u})=3\begin{bmatrix}
2\\1\\
\end{bmatrix}=\begin{bmatrix}
6\\3\\
\end{bmatrix}
$, 
$T(2\vec{v})=2T(\vec{v})=2\begin{bmatrix}
-1\\3\\
\end{bmatrix}=\begin{bmatrix}
-2\\6\\
\end{bmatrix}
$, and 
$T(3\vec{u}+2\vec{v})=3T(\vec{u})+2T(\vec{v})=\begin{bmatrix}
6\\3\\
\end{bmatrix}+\begin{bmatrix}
-2\\6\\
\end{bmatrix}=\begin{bmatrix}
4\\9\\
\end{bmatrix}
$
\item[19.]
Let $A_{19}$ be a $2\times 2$ matrix.

$A_{19}\begin{bmatrix}1\\0\\\end{bmatrix}=\begin{bmatrix}2\\5\\\end{bmatrix}$ and $A_{19}\begin{bmatrix}0\\1\\\end{bmatrix}=\begin{bmatrix}-1\\6\\\end{bmatrix},\ 
\therefore A_{19}=\begin{bmatrix}
2 & -1 \\
5 & 6 \\
\end{bmatrix}
$

$A_{19}\begin{bmatrix}
5\\-3\\
\end{bmatrix}=5\begin{bmatrix}2\\5\\\end{bmatrix}-3\begin{bmatrix}-1\\6\\\end{bmatrix}=\begin{bmatrix}
13\\7\\
\end{bmatrix}
$

$A_{19}\begin{bmatrix}x_1\\x_2\end{bmatrix}=x_1\begin{bmatrix}2\\5\\\end{bmatrix}+x_2\begin{bmatrix}-1\\6\\\end{bmatrix}=
\begin{bmatrix}
2x_1-x_2\\5x_1+6x_2\\
\end{bmatrix}
$
\item[32.]
Let $A_{32}$ =
$\begin{bmatrix}
4 & -2\\
0 & |x_2|\\
\end{bmatrix}
$
The absolute value of $x_2$ is not a linear function. Therefore the transformation cannot be a linear transformation.
\item[33.]
$
T(x_1,x_2)=\begin{bmatrix}
2x_1 -3x_2\\x_1+4\\5x_2\\
\end{bmatrix},
T(0,0) = \begin{bmatrix}
0-0\\0+4\\0\\
\end{bmatrix}
$ Transforming the origin does not give the origin. Therefore this transformation cannot be a linear transformation.
\end{enumerate}
\begin{large}
Section 1.9
\end{large}
\begin{enumerate}
\item[3.]
Let $A_3=
\begin{bmatrix}
0 & 1 \\
-1 & 0 \\
\end{bmatrix},
\begin{bmatrix}
0 & 1 \\
-1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
1\\0\\
\end{bmatrix}=
\begin{bmatrix}
0 \\ -1 \\
\end{bmatrix},
\begin{bmatrix}
0 & 1 \\
-1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
0 \\ 1\\
\end{bmatrix}=
\begin{bmatrix}
1\\0\\
\end{bmatrix}
$
\item[5.]
Let $A_5=\begin{bmatrix}
1 & 0\\
-2 & 1\\
\end{bmatrix}, \begin{bmatrix}
1 & 0\\
-2 & 1\\
\end{bmatrix}\begin{bmatrix}1\\0\\\end{bmatrix}=\begin{bmatrix}1\\-2\\\end{bmatrix},
\begin{bmatrix}
1 & 0\\
-2 & 1\\
\end{bmatrix}\begin{bmatrix}0\\1\\\end{bmatrix}=\begin{bmatrix}0\\1\\\end{bmatrix}
$
\item[8.]
Let $A_8=
\begin{bmatrix}
0 & -1\\
1 & 1\\
\end{bmatrix},$ proof
$
\begin{bmatrix}
0 & -1\\
1 & 1\\
\end{bmatrix}
\begin{bmatrix}
1\\0\\
\end{bmatrix}=
\begin{bmatrix}
0\\1\\
\end{bmatrix}
$ and 
$
\begin{bmatrix}
0 & 1\\
1 & -1\\
\end{bmatrix}
\begin{bmatrix}
0 \\ 1\\
\end{bmatrix}=
\begin{bmatrix}
-1\\1\\
\end{bmatrix}
$
\item[10.]
\item[13.]
\begin{tikzpicture}
\draw [thick] (-3,0)--(5,0) node[right]{$x_1$};
\draw [thick] (0,-.5)--(0,3) node[above]{$x_2$};
\draw [->](0,0) -- node[above] {$T(\vec{e_1})$}(-2,1);
\draw [->](0,0) -- node[above] {$T(\vec{e_2})$}(5,1);
\draw [->](0,0) -- node[right] {$T(2,1)$}(1,3);
\end{tikzpicture}
\item[14.]
\begin{tikzpicture}
\draw [thick] (-.5,0)--(2,0) node[right]{$x_1$};
\draw [thick] (0,-2)--(0,4) node[above]{$x_2$};
\draw [->](0,0)--(2,-1) node[below]{$\vec{a_1}$};
\draw [->](0,0)--(1,1) node[above]{$\vec{a_2}$};
\draw [->](0,0)--(1,4);
\end{tikzpicture}
\item[20.]
$T(x_1,x_2,x_3,x_4)=
\begin{bmatrix}
2 & 0 & 3 & -4\\
\end{bmatrix}
$
\item[22.]
$T(x_1,x_2)
\Rightarrow
\begin{bmatrix}
1 & -2 \\
-1 & 3 \\
3 & -2 \\
\end{bmatrix}\vec{x}
=
\begin{bmatrix}
-1\\4\\9\\
\end{bmatrix}
\Rightarrow
\begin{bmatrix}
1 & -2 & -1\\
-1 & 3 & 4\\
3 & -2 & 9\\
\end{bmatrix}
\xrightarrow{rref}
\begin{bmatrix}
1 & 0 & 5 \\
0 & 1 & 3 \\
0 & 0 & 0 \\
\end{bmatrix}
\Rightarrow
x_1=5$ and $x_2=3$
\item[28.]
\begin{enumerate}
\item
The transformation in Exercise 14 is {\bf one-to-one} because the two vectors are linearly independent.

\item
The transformation in Exercise 14 is {\bf onto} because the two vectors span all of $\mathbb{R}^2$ because they are linearly independent.

\end{enumerate}
\end{enumerate}
\begin{large}
Section 1.10
\end{large}
\begin{enumerate}
\item[9.]
\item[12.]
\end{enumerate}
\begin{large}
Supplementary Exercises
\end{large}
\begin{enumerate}
\item[1.]
\begin{enumerate}
\item[e.]
\item[f.]
\item[g.]
\item[h.]
\item[j.]
\item[k.]
\end{enumerate}
\end{enumerate}
\begin{large}
Handout
\end{large}
\begin{enumerate}
\item[3.]
\end{enumerate}
\end{document}
