Solve system of linear equations, using matrix method.

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

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

Now,

$|A|=2(4+1)-3(-2-3)+3(-1+6)=2(5)-3(-5)+3(5)=10+15+15=40 \neq 0$

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

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

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

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

$\therefore A^{-1}=\frac{1}{|A|}($ adjA $)=\frac{1}{40}\left[\begin{array}{ccc}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{array}\right]$

$\therefore X=A^{-1} B=\frac{1}{40}\left[\begin{array}{ccc}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{array}\right]\left[\begin{array}{l}5 \\ -4 \\ 3\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{40}\left[\begin{array}{l}25-12+27 \\ 25+52+3 \\ 25-44-21\end{array}\right]$

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

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

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