## Precalculus (6th Edition) Blitzer

First we will consider the product $CB$, \begin{align} & CB=\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1\left( 1 \right)-0\left( 0 \right) & -1\left( 0 \right)-0\left( -1 \right) \\ 0\left( 1 \right)-1\left( 0 \right) & 0\left( 0 \right)-1\left( -1 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \end{align} Now we will find the product $A\left( CB \right)$ as follows: \begin{align} & A\left( CB \right)=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1(-1)-0(0) & 1(0)-1(1) \\ 0(-1)-1(0) & 0(0)-1(1) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1 & -1 \\ 0 & -1 \\ \end{matrix} \right] \end{align} Now we will find the product $AC$, \begin{align} & AC=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\left( -1 \right)-0\left( 0 \right) & 1\left( 0 \right)-0\left( 1 \right) \\ 0\left( -1 \right)-1\left( 0 \right) & 0\left( 0 \right)-1\left( 1 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right] \end{align} Now we will find the product $(AC)B$ as follows: \begin{align} & (AC)B=\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1(1)-0(0) & -1(0)+1(-1) \\ 0(1)+1(0) & 0(0)+1(-1) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1 & -1 \\ 0 & -1 \\ \end{matrix} \right] \end{align} So, $A\left( CB \right)=(AC)B$ Hence the associative property is verified.