Precalculus (6th Edition) Blitzer

$\left( A+B \right)\left( C-D \right)=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]$
First we will find the sum of $A\text{ and }B$ as below: \begin{align} & A+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 \\ 0+0 & 1-1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 2 & 0 \\ 0 & 0 \\ \end{matrix} \right] \end{align} Next we will find the difference between $C$ and $D$ as below: \begin{align} & C-D=\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 \\ 0-0 & 1+1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0 & 0 \\ 0 & 2 \\ \end{matrix} \right] \end{align} Now consider, \begin{align} & \left( A+B \right)\left( C-D \right)=\left[ \begin{matrix} 2 & 0 \\ 0 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 0 \\ 0 & 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 2\left( 0 \right)+0\left( 0 \right) & 2\left( 0 \right)+0\left( 2 \right) \\ 0\left( 0 \right)+0\left( 0 \right) & 0\left( 0 \right)+0\left( 2 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right] \end{align}