Answer
$T_3^{-1}(x,y)=(8x-29y+3z,-5x+19y-2z,2x-8y+z)$
Work Step by Step
The inverse linear transformation is:
$A^{-1}=\begin{bmatrix}
3 & 5 & 1\\
1 & 2 & 1 \\
2 & 6 & 7
\end{bmatrix}^{-1}=\frac{1}{\det A}\begin{bmatrix}
8 & -5 & 2\\
-29 & 19 & 8 \\
3 & -2 & 1
\end{bmatrix}=\frac{1}{1}\begin{bmatrix}
8 & -5 & 2\\
-29 & 19 & 8 \\
3 & -2 & 1
\end{bmatrix}=\begin{bmatrix}
8 & -29 & 3\\
-5 & 19 & -2 \\
2 & 8 & 1
\end{bmatrix}$
Hence, $T_3^{-1}(x,y)=(8x-29y+3z,-5x+19y-2z,2x-8y+z)$
