Let \[ A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 0 & 1 \\ 3 & 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 2 & -2 \\ 1 & 3 \\ 4 & 4 \end{bmatrix}. \] Verify directly that \(A(AB) = A^2B\).
Solution
Compute \[ AB=\begin{bmatrix}5&-1\\8&0\\10&-2\end{bmatrix}, \qquad A^2=\begin{bmatrix}2&-1&1\\5&-2&3\\6&-3&4\end{bmatrix}. \] Then \[ A(AB)= \begin{bmatrix}7&-3\\20&-4\\25&-5\end{bmatrix} \] and \[ A^2B= \begin{bmatrix}7&-3\\20&-4\\25&-5\end{bmatrix}. \] Thus \(A(AB)=A^2B\), as required.