Answer
$ \left[ {\begin{array}{*{20}{c}}
0\\
6\\
6\\
{-6}
\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+0,1-0,2(1),2(0))=(1,1,2,0)\\
T(0,1)=(0+1,0-1,2(0),2(1))=(1,-1,0,2) $
Therefore, the matrix corresponding to the linear transformation $T$ is given by:
$A=[T(e_1)$ $T(e_2)]$ $=\left[ {\begin{array}{*{20}{c}} 1&1\\ 1&{ - 1}\\2&0\\0&2 \end{array}} \right]$
Hence, the image of
$v = \left[ {\begin{array}{*{20}{c}}
3\\
{ - 3}
\end{array}} \right]$
is
$T(v) = Av = \left[ {\begin{array}{*{20}{c}} 1&1\\ 1&{ - 1}\\2&0\\0&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}}
3\\
{ - 3}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0\\
6\\
6\\
{-6}
\end{array}} \right]$
