Answer
a) The matrix corresponding to linear transformation $T$ is given by: $=\left[ {\begin{array}{*{20}{c}} { -1}&{0}\\ 0&{ 1} \end{array}} \right]$
b) Hence, the image of $v=\left[ {\begin{array}{*{20}{c}} 2\\ {-3} \end{array}} \right]$ is $\left[ {\begin{array}{*{20}{c}} { -2}\\ { -3} \end{array}} \right]$
c) See image below.
Work Step by Step
a) Take the standard basis for $R^2,$ which is $\{(1,0),(0,1)\}=\{e_1,e_2\}$.
We are given
$T(x,y)=(-x,y)$
Then,
$T(1,0)=(-1,0)\\ T(0,1)=(0,1)$
Therefore, the matrix corresponding to linear transformation $T$ is given by: $A=[T(e_1)$ $T(e_2)]$ $=\left[ {\begin{array}{*{20}{c}} { -1}&{0}\\ 0&{ 1} \end{array}} \right]$
b) Hence, the image of $v=\left[ {\begin{array}{*{20}{c}} 2\\ {-3} \end{array}} \right]$ is $\begin{array}{l} T(v) = Av\\ \Rightarrow T\left[ {\begin{array}{*{20}{c}} 2\\ {-3} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { -1}&0\\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2\\ {-3} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { -2}\\ { -3} \end{array}} \right] \end{array}$
c)
