Answer
$A=T(e_2,e_3,e_4)=\begin{bmatrix}
3 & 2 & -6 & 3\\
-2 & 3 & -1 & 2
\end{bmatrix}$
Work Step by Step
We are given:
$T(1,0,0,0)=(3,-2) \\
T(1,1,0,0)=(5,1) \\
T(1,1,1,0)=(-1,0)\\
T(1,1,1,1)=(2,2)$
The standard basis vectors in $R^2$ are:
$e_2=(0,1,0,0)=-1(1,0,0,0)+1(1,1,0,0)+0(1,1,1,0)+0(1,1,1,1) \\
e_3=(0,0,1,0)=0(1,0,0,0)+(-1)(1,1,0,0)+1(1,1,1,0)+0(1,1,1,1)\\
e_4=(0,0,0,1)=0(1,0,0,0)+0(1,1,0,0)+(-1)(1,1,1,0)+1(1,1,1,1)$
Consequently,
$T(e_2)=T(0,1,0,0)=T((-1)(1,0,0,0)+1(1,1,0,0)+0(1,1,1,0)+0(1,1,1,1)) \\
=-T(1,0,0,0)+T(1,1,0,0)+0T(1,1,1,0)+0T(1,1,1,1)\\
=-1(3,-2)+1(5,1)+0(-1,0)+0(2,2) \\
=(-3,2)+(5,1)\\
=(2,3)$
$T(e_3)=T(0,1,0,0)=T(0(1,0,0,0)+(-1)(1,1,0,0)+1(1,1,1,0)+0(1,1,1,1)) \\
=0T(1,0,0,0)+(-T)(1,1,0,0)+T(1,1,1,0)+0T(1,1,1,1)\\
=0(3,-2)+(-1)(5,1)+1(-1,0)+0(2,2) \\
=(-5,-1)+(-1,0)\\
=(-6,-1)$
$T(e_4)=T(0,0,0,1)=T(0(1,0,0,0)+0(1,1,0,0)+(-1)(1,1,1,0)+1(1,1,1,1)) \\
=0T(1,0,0,0)+0T(1,1,0,0)+(-T)(1,1,1,0)+T(1,1,1,1)\\
=0(3,-2)+0(5,1)+(-1)(-1,0)+1(2,2) \\
=(1,0)+(2,2)\\
=(3,2)$
The matrix of the given transformation is:
$A=T(e_2,e_3,e_4)=\begin{bmatrix}
3 & 2 & -6 & 3\\
-2 & 3 & -1 & 2
\end{bmatrix}$
