Answer
See below
Work Step by Step
Assume $ax^2+bx+c \in P_2(R)$
and $\alpha, \beta, \gamma \in R$
Obtain $ax^2+bx+c=\alpha(x^2-x-3)+\beta(2x+5)+\gamma(6) $
We have the system: $\alpha=a\\
\beta+\gamma =b\\
-\alpha +2\beta=b\\
-3\alpha+5\beta +6\gamma=c\\
\rightarrow \alpha =a\\
\beta =\frac{a+b}{2}\\
\gamma=\frac{a-5b+2c}{12}$
Hence,
$T(ax^2+bx+c)=T[a(x^2-x-3)+\frac{a+b}{2}(2x+5)+\frac{a-5b+2c}{12}(6)]\\
=aT(x^2-x-3)+\frac{a+b}{2}T(2x+5)+\frac{a-5b+2c}{12}T(6)\\
=a\begin{bmatrix}
-2 & 1\\
-4& -1
\end{bmatrix}+\frac{a+b}{2}\begin{bmatrix}
0 & 1\\
2 & -2
\end{bmatrix}+\frac{a-5b+2c}{12}\begin{bmatrix}
12 & 6\\
6 & 8
\end{bmatrix}\\
=\begin{bmatrix}
-2a & a\\
-4a& -a
\end{bmatrix}+\begin{bmatrix}
0 & \frac{a+b}{2}\\
a+b & -a-b
\end{bmatrix}+\begin{bmatrix}
a-5b+2c & \frac{a-5b+2c}{2}\\\frac{a-5b+2c}{2} & \frac{3a-15b+6c}{12}
\end{bmatrix}\\
=\begin{bmatrix}
-a-5b+2c & 2a-2b+c\\\frac{-5a-3b+2c}{2} & \frac{-a-17b+6c}{12}
\end{bmatrix}$
