Answer
$\left[ {\begin{array}{*{20}{c}}
1\\
4
\end{array}} \right]$
Work Step by Step
Take the standard basis for $R^3$, which is
${(1,0,0),(0,1,0),(0,0,1)}=\{{e_1,e_2,e_3}\}$.
Then,
$T(1,0,0)=(2(1)+0,3(0)-0)=(2,0)\\
T(0,1,0)=(2(0)+1,3(1)-0)=(1,3)\\
T(0,0,1)=(2(0)+0,3(0)-1)=(0,-1) $
Therefore, the matrix corresponding to linear the transformation $T$ is given by:
$A=[T(e_1)$ $T(e_2)$ $T(e_3)]$ $=\left[ {\begin{array}{*{20}{c}} 2&1&0\\ 0&3&{ - 1} \end{array}} \right]$
Hence, the image of
$v = \left[ {\begin{array}{*{20}{c}}
0\\
1\\
{ - 1}
\end{array}} \right]$
is
$T(v) = Av = \left[ {\begin{array}{*{20}{c}}
2&1&0\\
0&3&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0\\
1\\
{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1\\
4
\end{array}} \right]$
