Chapter 8 - Section 8.3 - Matrix Operations and Their Applications - Exercise Set - Page 918: 43

$A\left( BC \right)=\left[ \begin{array}{*{35}{l}} 16 & -16 \\ -12 & 12 \\ 0 & 0 \\ \end{array} \right]$

First we will consider the product $BC$ , $\begin{align} & BC=\left[ \begin{matrix} 5 & 1 \\ -2 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -1 \\ -1 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 5\left( 1 \right)+1\left( -1 \right) & 5\left( -1 \right)+1\left( 1 \right) \\ -2\left( 1 \right)-2\left( -1 \right) & -2\left( -1 \right)-2\left( 1 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 5-1 & -5+1 \\ -2+2 & 2-2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 4 & -4 \\ 0 & 0 \\ \end{matrix} \right] \end{align}$ Now we will find the product $A\left( BC \right)$ as follows: $\begin{align} & A\left( BC \right)=\left[ \begin{array}{*{35}{l}} 4 & 0 \\ -3 & 5 \\ 0 & 1 \\ \end{array} \right]\left[ \begin{matrix} 4 & -4 \\ 0 & 0 \\ \end{matrix} \right] \\ & =\left[ \begin{array}{*{35}{l}} 4\left( 4 \right)+0\left( 0 \right) & 4\left( -4 \right)+0\left( 0 \right) \\ -3\left( 4 \right)+5\left( 0 \right) & -3\left( -4 \right)+5\left( 0 \right) \\ 0\left( 4 \right)+1\left( 0 \right) & 0\left( -4 \right)+1\left( 0 \right) \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 16+0 & -16+0 \\ -12+0 & 12+0 \\ 0+0 & 0+0 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 16 & -16 \\ -12 & 12 \\ 0 & 0 \\ \end{array} \right] \end{align}$