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:
$\dim [Ker(T_2T_1)]+\dim [Rng(T)=\dim P_2(R)\\
1+\dim [Rng(T)]=3\\
\dim [Rng(T)]=2$
$\dim P_1(R)=2=\dim [Rng(T)]\\
\rightarrow Rng=R^2$
$\rightarrow T$ is onto.
Since $T$ is not one-to-one, $T^{-1}$ does not exist.
