Answer
$A=T(e_1,e_2)=\begin{bmatrix}
-\frac{5}{3}& -\frac{2}{3}\\
\frac{1}{3} & \frac{1}{3} \\
\frac{5}{3} & -\frac{1}{3} \\
-1 & 1
\end{bmatrix}$
Work Step by Step
We are given:
$T(-1,1)=(1,0,-2,2) \\
T(1,2)=(-3,1,1,1)$
The standard basis vectors in $R^2$ are:
$e_1=(1,0)=-\frac{2}{3}(-1,1)+\frac{1}{3}(1,2) \\
e_2=\frac{1}{3}(-1,1)+\frac{1}{3}(1,2) $
Consequently,
$T(e_1)=T(1,0)=T(-\frac{2}{3}(-1,1)+\frac{1}{3}(1,2)) \\
=-\frac{2}{3}T(-1,1)+\frac{1}{3}T(1,2) \\= -\frac{2}{3}(1,0,-2,2)+\frac{1}{3}(-3,1,1,1) \\
=(-\frac{2}{3},0,\frac{4}{3},-\frac{4}{3})+(-1,\frac{1}{3},\frac{1}{3},\frac{1}{3})\\
=(-\frac{5}{3},\frac{1}{3},\frac{5}{3},-1)$
$T(e_2)=T(0,2)=T(\frac{1}{3}(-1,1)+\frac{1}{3}(1,2)) \\
=\frac{1}{3}T(-1,1)+\frac{1}{3}T(1,2) \\
= \frac{1}{3}(1,0,-2,2)+\frac{1}{3}(-3,1,1,1) \\
=(\frac{1}{3},0,-\frac{2}{3},-\frac{2}{3})+(-1,\frac{1}{3},\frac{1}{3},\frac{1}{3})\\
=(-\frac{2}{3},\frac{1}{3},-\frac{1}{3},1)$
The matrix of the given transformation is:
$A=T(e_1,e_2)=\begin{bmatrix}
-\frac{5}{3}& -\frac{2}{3}\\
\frac{1}{3} & \frac{1}{3} \\
\frac{5}{3} & -\frac{1}{3} \\
-1 & 1
\end{bmatrix}$
