STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

$\begin{aligned}
|A| &=2(4+1)-3(2-3)+3(-1+6) \\
&=2(5)-3(-5)+3(5) \\
&=10+15+15=40 \neq 0
\end{aligned}$

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 $z=-1$