Answer
$T_3^{-1}(x,y)=\frac{1}{14}(11x-16y+z),\frac{1}{14}(-6x+10y+2z),\frac{1}{14}(3x+y-z)$
Work Step by Step
The inverse linear transformation is:
$A^{-1}=\begin{bmatrix}
1 & 1 & 3\\
0 & 1 & 2 \\
3 & 5 & -1
\end{bmatrix}^{-1}=\frac{1}{\det A}\begin{bmatrix}
-11 & 6 & -3\\
16 & -10 & -2 \\
-1 & -2 & 1
\end{bmatrix}=-\frac{1}{14}\begin{bmatrix}
-11 & 16 & -1\\
6 & -10 & -2\\
-3 & -2 & 1
\end{bmatrix}=\frac{1}{14}\begin{bmatrix}
11 & -16 & 1\\
-6 & 10 & 2\\
3 & 2 & -1
\end{bmatrix}$
Hence, $T_3^{-1}(x,y)=\frac{1}{14}(11x-16y+z),\frac{1}{14}(-6x+10y+2z),\frac{1}{14}(3x+y-z)$
