If, find A−1. Using A−1 solve the system of equations

Solution:

$A=\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]$

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

Now, $A_{11}=0, A_{12}=2, A_{13}=1$

$A_{21}=-1, A_{22}=-9, A_{23}=-5$

$A_{31}=2, A_{32}=23, A_{33}=13$

$\therefore A^{-1}=\frac{1}{|A|}($ adjA $)=-\left[\begin{array}{ccc}0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right]$   ....(1)

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

$A=\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}11 \\ -5 \\ -3\end{array}\right]$

The solution of the system of equations is given by $X=A^{-1} B$.

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right]\left[\begin{array}{l}11 \\ -5 \\ -3\end{array}\right] \quad[$ Using (1) $]$

$=\left[\begin{array}{c}0-5+6 \\ -22-45+69 \\ -11-25+39\end{array}\right]$

$=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$

Hence, $x=1, y=2$, and $z=3$.