Answer
T is not a linear transformation.
Work Step by Step
A linear transformation is said to be linear if
$T(av+w)=aT(v)+T(w)$, for all $v,w$ in the given domain and for all scalars $a$.
Here, the above condition is violated.
Take, $a = 1;v = w = \left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]$
Then,
$T\left[ {1\left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]} \right] = T\left( {\left[ {\begin{array}{*{20}{c}}
2&0\\
0&2
\end{array}} \right]} \right) = \left| {\begin{array}{*{20}{c}}
2&0\\
0&2
\end{array}} \right| = 4$
whereas,
$1T\left[ {\left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]} \right] + T\left[ {\left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]} \right] = \left| {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right| = 1 + 1 = 2$
Therefore,
$T\left[ {1\left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]} \right] \ne 1T\left[ {\left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]} \right] + T\left[ {\left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]} \right]$
