Answer
$ \left[ {\begin{array}{*{20}{c}}
4\\
{ - 2}\\
{ - 2}
\end{array}} \right]$
Work Step by Step
Take the standard basis for $R^2$, which is
$(1,0),(0,1)=\{e_1,e_2\}$.
Then,
$T(1,0)=(1-0,1+2(0),0)=(1,1,0)\\
T(0,1)=(0-1,0+2(1),1)=(-1,2,1)$
Therefore, the matrix corresponding to linear the transformation $T$ is given by:
$A=[T(e_1)$ $T(e_2)]$
$=\left[ {\begin{array}{*{20}{c}}
1&{ - 1}\\
1&2\\
0&1
\end{array}} \right]$
Hence, the image of
$v=\left[ {\begin{array}{*{20}{c}}
2\\
{ - 2}
\end{array}} \right]$
is
$\begin{array}{l}
T(v) = Av\\
\Rightarrow T\left[ {\begin{array}{*{20}{c}}
2\\
{ - 2}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{ - 1}\\
1&2\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
2\\
{ - 2}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
4\\
{ - 2}\\
{ - 2}
\end{array}} \right]
\end{array}$
