Step 1: Find $(A + B)$. \[ A + B = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} (2+1) & (4+3) \\ (3+(-2)) & (2+5) \end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix} \] Step 2: Find $(A - B)$. \[ A - B = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} (2-1) & (4-3) \\ (3-(-2)) & (2-5) \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix} \]
Final Answer: \[ A + B = \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix}, A - B = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix} \]
If \[ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \] then find \( A^{-1} \). Hence, solve the system of linear equations: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]