Solving a Iwo-step Matrix Equation Given: \[ A=\left[\begin{array}{cc} 1 & -1 \\ -2 & 3 \end{array}\right] \quad B=\left[\begin{array}{cc} 1 & 5 \\ 8 & -8 \end{array}\right] \quad C=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \] Solve: \[ A X+B=C \] \[ X=\left[\begin{array}{ll} \square & \square \\ \square & \square \end{array}\right] \] DONE

To solve the equation \( A X + B = C \), first, isolate \( AX \) by subtracting \( B \) from both sides: \( AX = C - B \). Calculating \( C - B \): \[ C - B = \left[\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}\right] - \left[\begin{array}{cc} 1 & 5 \\ 8 & -8 \end{array}\right] = \left[\begin{array}{cc} 0 & -3 \\ -5 & 12 \end{array}\right] \] Now we need to solve \( AX = \left[\begin{array}{cc} 0 & -3 \\ -5 & 12 \end{array}\right] \). To find \( X \), we can multiply both sides by \( A^{-1} \), the inverse of \( A \). Calculating the inverse \( A^{-1} \) involves finding the determinant of \( A \): \[ \text{Det}(A) = 1(3) - (-1)(-2) = 3 - 2 = 1 \] \[ A^{-1} = \frac{1}{\text{Det}(A)} \begin{bmatrix} 3 & 1 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 2 & 1 \end{bmatrix} \] Multiplying \( A^{-1} \) by \( C - B \): \[ X = A^{-1}(C - B) = \begin{bmatrix} 3 & 1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 0 & -3 \\ -5 & 12 \end{bmatrix} = \begin{bmatrix} -3 & 33 \\ -10 & 24 \end{bmatrix} \] Thus, the solution is: \[ X = \left[\begin{array}{cc} -3 & 33 \\ -10 & 24 \end{array}\right] \]