Step 1: Write the order of matrices. \[ A \; (2\times 3), B \; (3\times 2) \implies AB \; (2\times 2) \] Step 2: Perform multiplication. \[ AB = \begin{bmatrix} 1 & -2 & 3 \\ -4 & 2 & 5 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} (1)(2) + (-2)(4) + (3)(2) & (1)(3) + (-2)(5) + (3)(1) \\ (-4)(2) + (2)(4) + (5)(2) & (-4)(3) + (2)(5) + (5)(1) \end{bmatrix} \] Step 3: Simplify each element. - First row, first column: $2 - 8 + 6 = 0$ - First row, second column: $3 - 10 + 3 = -4$ - Second row, first column: $-8 + 8 + 10 = 10$ - Second row, second column: $-12 + 10 + 5 = 3$ \[ AB = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix} \]
Final Answer: \[ \boxed{AB = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}} \]
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: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]