Answer
$A=\left(\begin{array}{cc}0 & -1 \\ -1 & 2\end{array}\right)$
Work Step by Step
\[
T(\mathbf{x})=A \mathbf{x}=A_{2} A_{1} \mathbf{x}
\]
$T$ can be viewed as the composition of two transformations thus let's split up the standard matrix $A=A_{2} A_{1}$
\[
\begin{array}{l}
T_{1}\left(\mathbf{e}_{1}\right)=\mathbf{e}_{1} \\
T_{1}\left(\mathbf{e}_{2}\right)=\mathbf{e}_{2}-\mathbf{e}_{1}
\end{array}
\]
The first transformation $T_{1},$ leaves e $_{1}$ unchanged and maps $\mathbf{e}_{2}$ into $\mathbf{e}_{2}-\mathbf{e}_{1}$
\[
A_{1}=\left(\begin{array}{cc}
1 & -2 \\
0 & 1
\end{array}\right)
\]
Thus the first column of $A_{1}$ is $\mathbf{e}_{1},$ and the second column is $\mathbf{e}_{2}-\mathbf{e}_{1}$
\[
\begin{array}{l}
T_{2}\left(\mathbf{e}_{1}\right)=-\mathbf{e}_{2} \\
T_{2}\left(\mathbf{e}_{2}\right)=-\mathbf{e}_{1}
\end{array}
\]
The second transformation $T_{2},$ reflects all points through the line $x_{2}=-x_{1},$ or $f(x)=-x$ for an alternative notation
\[
A_{2}=\left(\begin{array}{cc}
0 & -1 \\
-1 & 0
\end{array}\right)
\]So the first column of $A_{2}$ is $-\mathbf{e}_{2},$ and the second column is $-\mathbf{e}_{1}$
\[
\begin{array}{l}
T(\mathbf{x})=T_{2}\left(T_{1}(\mathbf{x})\right)=A_{2} A_{1} \mathbf{x} \\
=\left(\begin{array}{cc}
0 & -1 \\
-1 & 0
\end{array}\right)\left(\begin{array}{cc}
1 & -2 \\
0 & 1
\end{array}\right) \mathbf{x}
\end{array}
\]
Now form $T$ with $T_{1,2}$ by using matrix multiplication to get $A$
