Answer
$T$ is a linear transformation.
Work Step by Step
A 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 $a$.
Let $A,B$ be arbitrary elements of the domain $M_{3,3}$ and $k$ be any arbitrary scalar. Then
$\begin{array}{l}
T\left( {kA + B} \right)\\
= \left[ {\begin{array}{*{20}{c}}
0&0&1\\
0&1&0\\
1&0&0
\end{array}} \right](kA + B)\\
= \left[ {\begin{array}{*{20}{c}}
0&0&1\\
0&1&0\\
1&0&0
\end{array}} \right](kA) + \left[ {\begin{array}{*{20}{c}}
0&0&1\\
0&1&0\\
1&0&0
\end{array}} \right](B)\\
= k\left[ {\begin{array}{*{20}{c}}
0&0&1\\
0&1&0\\
1&0&0
\end{array}} \right]A + \left[ {\begin{array}{*{20}{c}}
0&0&1\\
0&1&0\\
1&0&0
\end{array}} \right]B\\
= kT(A) + T(B)
\end{array}$
Hence, $T$ is a linear transformation.
Note: here we have used distributive properties of matrices.
1) For all matrices $A,B$ and $C$, $A(B+C)=AB+AC$
2) For all matrices $A$ and $B$ and for all scalors $k$, $(kAB)=k(AB)$
