Answer
$A=T(e_1,e_2,e_3)=\begin{bmatrix}
-4 & 3 & 0\\
-5 & 2 & -3 \\
15 & -7 & 6
\end{bmatrix}$
Work Step by Step
We are given:
$T(1,2,0)=(2,-1,1) \\
T(0,1,1)=(3,-1,-1) \\
T(0,2,3)=(6,-5,4)$
The standard basis vectors in $R^3$ are:
$e_1=(1,0,0)=1(1,2,0)+(-6)(0,1,1)+2(0,2,3)\\
e_2=(0,1,0)=0(1,2,0)+3(0,1,1)+(-1)(0,2,3)\\
e_3=(0,0,1)=0(1,2,0)+(-2)(0,1,1)+1(0,2,3)$
Consequently,
$T(e_1)=T(1,0,0)=T(1(1,2,0)+(-6)(0,1,1)+2(0,2,3))\\
=T(1,2,0)+(-6T)(0,1,1,)+2T(0,2,3)\\
=1(2,-1,1)+(-6)(3,-1,-1)+2(6,-5,4) \\
=(2,-1,1)+(-18,6,6)+(12,-10,8)\\
=(-4,-5,15)$
$T(e_2)=T(0,1,0)=T(0(1,2,0)+3(0,1,1,)+(-1)(0,2,3))\\
=0T(1,2,0)+3T(0,1,1)+(-T)(0,2,3)\\
=0(2,-1,1)+3(3,-1,-1)+(-1)(6,-5,4) \\
=(0,0,0)+(9,-3,-3)+(-6,5,-4)\\
=(3,2,-7)$
$T(e_3)=T(0,0,1)=T(0(1,2,0)+(-2)(0,1,1)+1(0,2,3))\\
=0T(1,2,0)+(-2T)(0,1,1,)+1T(0,2,3)\\
=0(2,-1,1)+(-2)(3,-1,-1)+1(6,-5,4) \\
=(0,0,0)+(-6,2,2)+(6,-5,4)\\
=(0,-3,6)$
The matrix of the given transformation is:
$A=T(e_1,e_2,e_3)=\begin{bmatrix}
-4 & 3 & 0\\
-5 & 2 & -3 \\
15 & -7 & 6
\end{bmatrix}$
