## Precalculus (6th Edition) Blitzer

The matrix $BD$ is, $BD=\left[ \begin{matrix} -6 & 4 & -8 \\ 0 & 5 & 11 \\ -17 & 13 & -19 \\ \end{matrix} \right]$.
Here we need to find $BD$. Therefore consider, \begin{align} & BD=\left[ \begin{array}{*{35}{l}} 0 & -2 \\ 3 & 2 \\ 1 & -5 \\ \end{array} \right]\left[ \begin{array}{*{35}{l}} -2 & 3 & 1 \\ 3 & -2 & 4 \\ \end{array} \right] \\ & =\left[ \begin{matrix} 0\left( -2 \right)-2\left( 3 \right) & 0\left( 3 \right)-2\left( -2 \right) & 0\left( 1 \right)-2\left( 4 \right) \\ 3\left( -2 \right)+2\left( 3 \right) & 3\left( 3 \right)+2\left( -2 \right) & 3\left( 1 \right)+2\left( 4 \right) \\ 1\left( -2 \right)-5\left( 3 \right) & 1\left( 3 \right)-5\left( -2 \right) & 1\left( 1 \right)-5\left( 4 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -6 & 4 & -8 \\ -6+6 & 9-4 & 3+8 \\ -2-15 & 3+10 & 1-20 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -6 & 4 & -8 \\ 0 & 5 & 11 \\ -17 & 13 & -19 \\ \end{matrix} \right] \end{align} Thus, $BD=\left[ \begin{matrix} -6 & 4 & -8 \\ 0 & 5 & 11 \\ -17 & 13 & -19 \\ \end{matrix} \right]$