2x – 3y + 5 = 11
3x + 2y – 4z = -5
x + y – 2z= -3
System of equations can be written as AX = B
Where, A = $\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]$
X = $\left[\begin{array}{c}\mathit{x}\\ \mathit{y}\\ \mathit{z}\end{array}\right]$ , B = $\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]$
A = $\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]$
|A| = 2(- 4 + 4) + 3 ( - 6 + 4) + 5 (3 - 2)
|A| = - 6 + 5 = - 1 ≠ 0
∴ ${\mathit{A}}^{-1}$ exists and system of equations has a unique solution
${\mathit{A}}^{-1}$ = $\frac{1}{\left|\mathit{A}\right|}$ (adj A)
adj A = $\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]$
${\mathit{A}}^{-1}$ = $\frac{1}{\left|\mathit{A}\right|}\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]$
= $\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]$
X = ${\mathit{A}}^{-1}\mathit{B}$ = $\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]$ $\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]$
X = $\left[\begin{array}{c}-5+6\\ -22-45+69\\ -11-25+39\end{array}\right]$ = $\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$
So x = 1 , y = 2 , z = 3