The given system of equation can be written in the form of AX = B, where
A = $\left[\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right]$ , X = $\left[\begin{array}{c}\mathit{X}\\ \mathit{Y}\\ \mathit{Z}\end{array}\right]$ and B = $\left[\begin{array}{c}7\\ -5\\ 12\end{array}\right]$
Now,
|A| = 1 (12 - 5) + 1 (9 + 10) + 2 (- 3 - 8) = 7 + 19 - 22 = 4 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
Now, ${\mathit{A}}_{11}$ = 7 , ${\mathit{A}}_{12}$ = - 19 , ${\mathit{A}}_{13}$ = - 11
${\mathit{A}}_{21}$ = 1 , ${\mathit{A}}_{22}$ = - 1 , ${\mathit{A}}_{23}$ = - 1
${\mathit{A}}_{31}$ = - 3 , ${\mathit{A}}_{32}$ = 11 , ${\mathit{A}}_{33}$ = 7
∴ ${\mathit{A}}^{-1}$ = $\frac{1}{\left|\mathit{A}\right|}$ (adj A) = $\frac{1}{4}\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]$
OR
Consider the given matrix.
Let A = $\left[\begin{array}{ccc}-1& 1& 2\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]$
We know that, A = ${\mathit{I}}_{\mathit{n}}$ A
Perform sequence of elementary row operations on A on the left hand side and the term ${\mathit{I}}_{\mathit{n}}$ on the right hand side till we obtain the result
${\mathit{I}}_{\mathit{n}}$ = BA
Thus, B = ${\mathit{A}}^{-1}$
Here, ${\mathit{I}}_3$ = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
Thus,we have,
$\left[\begin{array}{ccc}-1& 1& 2\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]$ = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ A
${\mathit{R}}_1$ ↔ ${\mathit{R}}_2$
$\left[\begin{array}{ccc}1& 2& 3\\ -1& 1& 2\\ 3& 1& 1\end{array}\right]$ = $\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$ A
${\mathit{R}}_2$ → ${\mathit{R}}_2+{\mathit{R}}_1$
${\mathit{R}}_3$ → ${\mathit{R}}_3-3{\mathit{R}}_1$
$\left[\begin{array}{ccc}1& 2& 3\\ 0& 3& 5\\ 0& -5& -8\end{array}\right]$ = $\left[\begin{array}{ccc}0& 1& 0\\ 1& 1& 0\\ 0& -3& 1\end{array}\right]$ A
${\mathit{R}}_1$ → ${\mathit{R}}_1+{\mathit{R}}_2$
$\left[\begin{array}{ccc}1& 5& 8\\ 0& 3& 5\\ 0& -5& -8\end{array}\right]$ = $\left[\begin{array}{ccc}1& 2& 0\\ 1& 1& 0\\ 0& -3& 1\end{array}\right]$ A
${\mathit{R}}_1$ → ${\mathit{R}}_1+{\mathit{R}}_3$
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 3& 5\\ 0& -5& -8\end{array}\right]$ = $\left[\begin{array}{ccc}1& -1& 1\\ 1& 1& 0\\ 0& -3& 1\end{array}\right]$ A
${\mathit{R}}_2$ → $\frac{{\mathit{R}}_2}{3}$
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& \frac{5}{3}\\ 0& -5& -8\end{array}\right]$ = $\left[\begin{array}{ccc}1& -1& 1\\ \frac{1}{3}& \frac{1}{3}& 0\\ 0& -3& 1\end{array}\right]$ A
${\mathit{R}}_{32}$ → ${\mathit{R}}_2+5{\mathit{R}}_2$
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& \frac{5}{3}\\ 0& 0& \frac{1}{3}\end{array}\right]$ = $\left[\begin{array}{ccc}1& -1& 1\\ \frac{1}{3}& \frac{1}{3}& 0\\ \frac{5}{3}& -\frac{4}{3}& 1\end{array}\right]$ A
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& \frac{5}{3}\\ 0& 0& 1\end{array}\right]$ = $\left[\begin{array}{ccc}1& -1& 1\\ \frac{1}{3}& \frac{1}{3}& 0\\ 5& -4& 3\end{array}\right]$ A
${\mathit{R}}_2$ → ${\mathit{R}}_2-\frac{5}{3}{\mathit{R}}_3$
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ = $\left[\begin{array}{ccc}1& -1& 1\\ -8& 7& -5\\ 5& -4& 3\end{array}\right]$ A
Thus the inverse of the matrix A is given by
$\left[\begin{array}{ccc}1& -1& 1\\ -8& 7& -5\\ 5& -4& 3\end{array}\right]$