Solve system of linear equations, using matrix method.

The given system of equations can be written in the form of AX = B, where

$A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}7 \\ -5 \\ 12\end{array}\right]$

Now,

$|A|=1(12-5)+1(9+10)+2(-3-8)=7+19-22=4 \neq 0$

Thus, A is non-singular. Therefore, its inverse exists.

Now, $A_{11}=7, A_{12}=-19, A_{13}=-11$

$A_{21}=1, A_{22}=-1, A_{23}=-1$

$A_{31}=-3, A_{32}=11, A_{33}=7$

$\therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{4}\left[\begin{array}{ccc}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{array}\right]$

$\therefore X=A^{-1} B=\frac{1}{4}\left[\begin{array}{ccc}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{array}\right]\left[\begin{array}{l}7 \\ -5 \\ 12\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}49-5-36 \\ -133+5+132 \\ -77+5+84\end{array}\right]$

$=\frac{1}{4}\left[\begin{array}{c}8 \\ 4 \\ 12\end{array}\right]=\left[\begin{array}{l}2 \\ 1 \\ 3\end{array}\right]$

Hence, $x=2, y=1$, and $z=3$.