Answer
See below
Work Step by Step
Obtain: $Ker(T)=\{(a,b,c) \in R^3: T(a,b,c)=0\}\\
=\{(a,b,c) \in R^3: \begin{bmatrix}
-a+3c & a-b-c\\
2a+b & 0
\end{bmatrix}\begin {bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix}\}$
We have the system: $-a+3c=0\\
a-b-c=0\\
2a+b=0$
Hence, $\rightarrow a=b=c=0\\
\rightarrow Ker(T)=\{(0,0,0)\}=\{0\}$
$T$ is one-to-one.
Apply Rank Nullity Theorem:
$\dim [Ker(T_2T_1)]+\dim [Rng(T)=\dim R^3\\
0+\dim [Rng(T)]=3\\
\dim [Rng(T)]=3$
Since, $\dim M_2(R)=4 \ne 3=\dim [Rng(T)]\\
\rightarrow T \ne R^3$
$\rightarrow T$ is onto.
Hence, $T^{-1}$ does not exist.
