I looked everywhere in the book for answer for this kind of question and they don't have it.
all they have is the easy case where T[1 0] and T[0 1]
it's in the attachment below
Well, solve
$\displaystyle \left ( \begin{matrix} a & b \\ c & d \end{matrix} \right ) \left ( \begin{matrix} 1 \\ 1 \end{matrix} \right ) = \left ( \begin{matrix} 3 \\ -2 \end{matrix} \right )$
and
$\displaystyle \left ( \begin{matrix} a & b \\ c & d \end{matrix} \right ) \left ( \begin{matrix} 1 \\ -1 \end{matrix} \right ) = \left ( \begin{matrix} 1 \\ 5 \end{matrix} \right )$
for a, b, c, and d.
After you get that, see if
$\displaystyle Tx + Ty = T(x + y)$
where x and y are arbitrary column vectors.
-Dan