Solve by matrix method: $2x+3y+3z=5$
$x-2y+z=-4$
$3x-y-2z=3$

Given system of linear equations are

$2x+3y+3z=5$
$x-2y+z=-4$
$3x-y-2z=3$.

Represent it in matrix form

$\left[\begin{array}{ccc}2& 3& 3\\ 1& -2& 1\\ 3& -1& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}5\\ -4\\ 3\end{array}\right]$ which is in the form of $AX=B$

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

$|A|=10+15+15=40\ne 0$

$\therefore A^{-1}$ exists

To find adjoint of $A$

$A_{11}=5,A_{12}=5,A_{13}=5$

$A_{21}=3,A_{22}=-13,A_{23}=11$

$A_{31}=9,A_{32}=1,A_{33}=-7$

$Adj\left(A\right)=\text{co-factor}\left[\begin{array}{ccc}5& 5& 5\\ 3& -13& 11\\ 9& 1& -7\end{array}\right]$

$=\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]$

$A^{-1}=\frac{1}{|A|}Adj\left(A\right)$

$=\frac{1}{40}\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]$

$X=A^{-1}B$

$=\frac{1}{40}\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]\left[\begin{array}{c}5\\ -4\\ 3\end{array}\right]$

$X=\frac{1}{40}\left[\begin{array}{c}25-12+27\\ 25+52+3\\ 25-44-21\end{array}\right]$

$X=\frac{1}{40}\left[\begin{array}{c}40\\ 80\\ -40\end{array}\right]$

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

Hence, $x=1,y=2$ and $z=-1$