Answer
See below
Work Step by Step
Obtain: $Ker(T)=\{A \in M_2(R): T(A)=0\}\\
=\{\begin{bmatrix}
a & b\\
c & d
\end{bmatrix} \in M_2(R):T( \begin{bmatrix}
a & b\\
c & d
\end{bmatrix})=0\}$
We have the system: $a-b+d=0\\
2a+b=0\\
c=0\\
4a-b+2d=0$
Hence, $\rightarrow b=-2a\\
d=-3a\\
c=0\\
\rightarrow Ker(T)=\{ \begin{bmatrix}
a & -2a\\
0 & -3a
\end{bmatrix}: a \in R\}\\
=\{a \begin{bmatrix}
1 & -2\\
0 & -3
\end{bmatrix}: a\in R\}\\ =span\{ \begin{bmatrix}
1 & -2\\
0 & -3
\end{bmatrix}\}\\
\rightarrow Ker(T)= \{0\}$
$T$ is not one-to-one.
Apply Rank Nullity Theorem:
$\dim [Ker(T_2T_1)]+\dim [Rng(T)=\dim R^3\\
1+\dim [Rng(T)]=4\\
\dim [Rng(T)]=3$
Since, $\dim P_3(R)=4 \ne 3=\dim [Rng(T)]\\
\rightarrow Rng(T) \ne P_3(R)$
$\rightarrow T$ is not onto.
Hence, $T^{-1}$ does not exist.
