The system of equations can be written in matrix form 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 = $\begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$

Step 1: Find |A| |A| = 2[(2)(-2) - (-4)(1)] - (-3)[(3)(-2) - (-4)(1)] + 5[(3)(1) - (2)(1)] = 2(0) + 3(-2) + 5(1) = -1 $\neq$ 0 Since |A| $\neq$ 0, $A^{-1}$ exists.

Step 2: Find cofactors 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

Step 3: Find adj(A) and $A^{-1}$ adj(A) = $\begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}$ $A^{-1}$ = (1/|A|) × adj(A) = $\begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$

Step 4: Solve for X X = $A^{-1}$B = $\begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} = \begin{bmatrix} 0-5+6 \\ -22-45+69 \\ -11-25+39 \end{bmatrix}$ = $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

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