Answer
See below
Work Step by Step
Obtain: $Ker(T)=\{A \in V: T(A)=0\}\\
=\{\begin{bmatrix}
a & b \\ b & c
\end{bmatrix}: T(\begin{bmatrix}
a & b \\ b & c
\end{bmatrix})=0\}\\
=\{\begin{bmatrix}
a & b \\ b & c
\end{bmatrix}:ax^2+bx+c=0\\ \rightarrow a=b=c\\
\rightarrow Ker(T)=\{\begin{bmatrix}
0 & 0 \\ 0 & 0
\end{bmatrix}\}\\
=\{0\}$
Hence, $T$ is one-to-one.
Apply Rank Nullity Theorem:
$\dim [Ker(T_2T_1)]+\dim [Rng(T)]=\dim P_2(R)\\
0+\dim [Rng(T)]=3\\
\dim [Rng(T)]=3$
$\dim P_1(R)=3=\dim [Rng(T)]\\
\rightarrow Rng=P_2(R)$
$\rightarrow T$ is onto.
Since $T$ is both onto one-to-one, $T^{-1}$ does exist.
We have: $T^{-1}: P_2(R) \rightarrow V$
With $A \in V$, then $T^{-1}T(A)=A\\
T^{-1}(T(\begin{bmatrix}
a & b \\ c & d
\end{bmatrix}))=\begin{bmatrix}
a & b \\ c & d
\end{bmatrix}\\
T^{-1}(ax^2+bx+c)=\begin{bmatrix}
a & b \\ c & d
\end{bmatrix}$
