Solve system of linear equations, using matrix method.
$x-y+2 z=7$
$3 x+4 y-5 z=-5$
$2 x-y+3 z=12$
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 $z=3$.
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