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