Question
Verify that \( A(B + C) = AB + AC \) for:
\( A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \)
Step 1: Find \( B + C \)
\[ B + C = \begin{bmatrix} 0+1 & 1+(-1) \\ 1+0 & 1+1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} \]Step 2: Compute \( A(B + C) \)
\[ A(B+C) = \begin{bmatrix} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} \] \[ = \begin{bmatrix} (2)(1)+(-1)(1) & (2)(0)+(-1)(2) \\ (1)(1)+(1)(1) & (1)(0)+(1)(2) \\ (-1)(1)+(2)(1) & (-1)(0)+(2)(2) \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 2 & 2 \\ 1 & 4 \end{bmatrix} \]Step 3: Compute \( AB \)
\[ AB = \begin{bmatrix} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} -1 & 1 \\ 1 & 2 \\ 2 & 1 \end{bmatrix} \]Step 4: Compute \( AC \)
\[ AC = \begin{bmatrix} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} 2 & -3 \\ 1 & 0 \\ -1 & 3 \end{bmatrix} \]Step 5: Compute \( AB + AC \)
\[ AB + AC = \begin{bmatrix} -1 & 1 \\ 1 & 2 \\ 2 & 1 \end{bmatrix} + \begin{bmatrix} 2 & -3 \\ 1 & 0 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 2 & 2 \\ 1 & 4 \end{bmatrix} \]Final Result
\[ A(B + C) = \begin{bmatrix} 1 & -2 \\ 2 & 2 \\ 1 & 4 \end{bmatrix} = AB + AC \]Hence Verified: \( A(B + C) = AB + AC \)