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Solve system of linear equations, using matrix method. $2 x+3 y+3 z=5$ ; $x-2 y+z=-4$ ; $3 x-y-2 z=3$
$x=-1, y=2,z=-1$
$x=1, y=-2,z=-1$
$x=1, y=2,z=1$
$x=1, y=2,z=-1$
Solution
The given system of equation can be written in the form of $A X=B$, where
$\begin{aligned}
|A| &=2(4+1)-3(2-3)+3(-1+6) \\
&=2(5)-3(-5)+3(5) \\
&=10+15+15=40 \neq 0
\end{aligned}$
Thus, $A$ is non-singular. Therefore, its inverse exists.
Now,
$A_{11}=5, A_{12}=5, A_{13}=5$
$A_{21}=3, A_{22}=-13, A_{23}=11$
$A_{31}=9, A_{32}=1, A_{33}=-7$
$\therefore A^{-1}=\begin{array}{c}1 \\ |A|\end{array}(a d j A)=\frac{1}{40}\left[\begin{array}{ccc}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{array}\right]$
$\therefore X=A^{-1} B=\frac{1}{40}\left[\begin{array}{ccc}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{array}\right]\left[\begin{array}{c}5 \\ -4 \\ 3\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{40}\left[\begin{array}{c}25-12+27 \\ 25+52+3 \\ 25-44-21\end{array}\right]$
$=\frac{1}{40}\left[\begin{array}{c}40 \\ 80 \\ -40\end{array}\right]$
$=\left[\begin{array}{c}1 \\ 2 \\ -1\end{array}\right]$
Hence, $x=1, y=2$ and $z=-1$
