Answer
See below
Work Step by Step
Assume $\begin{bmatrix}
a & b \\ c & d
\end{bmatrix} \in M_2(R)$
and $\alpha, \beta, \gamma, \delta \in R$
Obtain $\begin{bmatrix}
a & b \\ c & d
\end{bmatrix} =\alpha \begin{bmatrix}
1& 0 \\ 0 & 1
\end{bmatrix} +\beta \begin{bmatrix}
0 & 1\\ 1 & 0
\end{bmatrix} +\gamma \begin{bmatrix}
1 & 0 \\ 0 & 0
\end{bmatrix} +\delta \begin{bmatrix}
1 & 1 \\ 0 & 0
\end{bmatrix} $
We have the system: $\alpha +\beta+\gamma +\delta=a\\
\beta+\gamma =b\\
\beta=c\\
\alpha =d\\
\rightarrow \alpha =d\\
\beta =c\\
\gamma=a-b+c-d\\
\delta=b-c$
Hence,
$T\begin{bmatrix}
a & b \\ c & d
\end{bmatrix} =T(d\begin{bmatrix}
1 & 0 \\ 0 & 1
\end{bmatrix} +c\begin{bmatrix}
0 & 1 \\ 1 & 0
\end{bmatrix} +(a-b+c-d)\begin{bmatrix}
1 & 0 \\ 0 & 0
\end{bmatrix} +(b-c)\begin{bmatrix}
1 &1 \\ 0 & 0
\end{bmatrix} )\\
=\begin{bmatrix}
1 &1 \\ 0 & 0
\end{bmatrix}+cT\begin{bmatrix}
0 &1 \\ 1 & 0
\end{bmatrix}+(a-b+c-d)T\begin{bmatrix}
1 & 0\\ 0 & 0
\end{bmatrix}+(b-c)T\begin{bmatrix}
1 &1 \\ 0 & 0
\end{bmatrix}\\
=d(2,-5)+c(0,-3)+(a-b+c-d)(1,1)+(b-c)(-6,2)\\
=(2d,-5d)+(0,-3c)+(a-b+c-d,a-b+c-d)+(-6b+6c,2b-2c)\\
=(2d+a-b+c-d-6b+6c,-5d-3c+a-b+c-d+2b-2c)\\
=(a-7b+7c+d,a+b-4c-6d)$
