Question:
Verify the distributive property of matrix multiplication: \[ A(B + C) = AB + AC \] where
\[ A = \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} -1 & 0 \\ 2 & 1 \end{bmatrix}, \quad C = \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix} \]
Solution:
Step 1: Compute B + C
\[ B + C = \begin{bmatrix} -1 & 0 \\ 2 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 3 & 0 \end{bmatrix} \]Step 2: Compute A(B + C)
\[ A(B + C) = \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 3 & 0 \end{bmatrix} \] \[ A(B + C) = \begin{bmatrix} -4 & 1 \\ 6 & 0 \end{bmatrix} \]Step 3: Compute AB
\[ AB = \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} -3 & -1 \\ 4 & 2 \end{bmatrix} \]Step 4: Compute AC
\[ AC = \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 2 \\ 2 & -2 \end{bmatrix} \]Step 5: Compute AB + AC
\[ AB + AC = \begin{bmatrix} -3 & -1 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} -1 & 2 \\ 2 & -2 \end{bmatrix} = \begin{bmatrix} -4 & 1 \\ 6 & 0 \end{bmatrix} \]Final Result:
\[ A(B + C) = AB + AC = \begin{bmatrix} -4 & 1 \\ 6 & 0 \end{bmatrix} \]Hence, the distributive property is verified.
Conclusion:
Thus, matrix multiplication is distributive over matrix addition.