Solve the following system of equation by matrix method.
$3x-2y+3z=8$
$2x+y-z=1$
$4x-3y+2z=4$

$A=\left[\begin{array}{ccc}3& -2& 3\\ 2& 1& -1\\ 4& -3& 2\end{array}\right],x=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}8\\ 1\\ 4\end{array}\right]$

$|A|=\left[\begin{array}{ccc}3& -2& 3\\ 2& 1& -1\\ 4& -3& 2\end{array}\right]$

$=3\left[\begin{array}{cc}1& -1\\ -3& 2\end{array}\right]-1(-2)\left[\begin{array}{cc}2& -1\\ 4& 2\end{array}\right]+3\left[\begin{array}{cc}2& 1\\ 4& -3\end{array}\right]\Rightarrow 3(2-3)-2(4+4)+3(-6-4)$

$=-3+16-30=-17$

Thus $|A|\ne 0$

$\therefore$ We have unique solution & consistent

$AX=B$

$X=A^{-1}B=\frac{1}{|A|}$adj(A)B.

adj $C=(Aii{)}^T$

$A=\left[\begin{array}{ccc}m& -2& 3\\ 2& 1& -1\\ 4& -3& 2\end{array}\right]$

cofactor of $3=\left[\begin{array}{cc}1& -1\\ 3& 2\end{array}\right]=2-3=-1$

cofactor of $-2=8$

cofactor of $3=-10$

cofactor of $2=5$

cofactor of $1=-6$

cofactor of $-1=-1$

cofactor of $4-1$

cofactor of $-3=-9$

cofactor of $2=7$

$\Rightarrow \left[\begin{array}{ccc}-1& 8& -10\\ 5& -6& -1\\ -1& -9& 7\end{array}\right]$

$\Rightarrow$ adj $A=(aii{)}^T=\left[\begin{array}{ccc}-1& -5& -1\\ -8& -6& +9\\ -10& +1& 7\end{array}\right]$

Now $\Rightarrow x=\frac{-1}{17}\left[\begin{array}{ccc}-1& -5& -1\\ -8& -6& 9\\ -10& 1& 7\end{array}\right]\left[\begin{array}{ccc}8& 1& 4\end{array}\right]=\frac{-1}{17}\left[\begin{array}{ccc}-1\left(8\right)& +(-5)1& +(-1)4\\ -8\left(8\right)& +(-6)1& +9\left(4\right)\\ -10\left(8\right)& +1\left(1\right)& +7\left(4\right)\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{-1}{17}\left[\begin{array}{c}-17\\ -34\\ -51\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$

$\therefore x=1,y=2,z=3$.