Let \[ A = \begin{bmatrix} 2 & -1 & 1 \\ 1 & 2 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 3 \\ 1 \\ -1 \end{bmatrix}, \qquad C = \begin{bmatrix} 1 & -1 \end{bmatrix}. \] Compute \(ABC\) and \(CAB\).
Solution
First compute \[ AB= \begin{bmatrix}2&-1&1\\1&2&1\end{bmatrix} \begin{bmatrix}3\\1\\-1\end{bmatrix} =\begin{bmatrix}4\\4\end{bmatrix}. \] Hence \[ ABC=\begin{bmatrix}4\\4\end{bmatrix}\begin{bmatrix}1&-1\end{bmatrix} =\begin{bmatrix}4&-4\\4&-4\end{bmatrix}. \] Also \[ CA=\begin{bmatrix}1&-1\end{bmatrix} \begin{bmatrix}2&-1&1\\1&2&1\end{bmatrix} =\begin{bmatrix}1&-3&0\end{bmatrix}, \] so \[ CAB=\begin{bmatrix}1&-3&0\end{bmatrix} \begin{bmatrix}3\\1\\-1\end{bmatrix}=0. \] Therefore \[ABC=\begin{bmatrix}4&-4\\4&-4\end{bmatrix},\qquad CAB=\begin{bmatrix}0\end{bmatrix}.\]