## Precalculus (6th Edition) Blitzer

We prove that, $\left( AB \right)C=A\left( BC \right)$ holds for any matrix $A,B$ and $C$. Take the three matrices $A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \\ \end{matrix} \right]\text{ and }C=\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$. To prove that, $\left( AB \right)C=A\left( BC \right)$ : First we will find the product $AB$. That is, \begin{align} & AB=\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 find the product $\left( AB \right)C$ as below: \begin{align} & \left( AB \right)C=\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)+\left( -1 \right)\left( 1 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right] \end{align} Therefore, $\left( AB \right)C=\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right]$ Next we will find the product $BC$. That is, \begin{align} & BC=\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)+\left( -1 \right)\left( 0 \right) & 0\left( 0 \right)+\left( -1 \right)\left( 1 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right] \end{align} And then we will find the product, \begin{align} & A\left( BC \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\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} Thus, clearly $\left( AB \right)C=A\left( BC \right)$ and hence the associative property is verified.