## Precalculus (6th Edition) Blitzer

The matrix $BA$ is, $BA=\left[ \begin{matrix} -10 & -6 & 2 \\ 16 & 3 & 4 \\ -23 & -16 & 7 \\ \end{matrix} \right]$
Here we need to find $BA$. Therefore consider, \begin{align} & BA=\left[ \begin{array}{*{35}{l}} 0 & -2 \\ 3 & 2 \\ 1 & -5 \\ \end{array} \right]\left[ \begin{array}{*{35}{l}} 2 & -1 & 2 \\ 5 & 3 & -1 \\ \end{array} \right] \\ & =\left[ \begin{matrix} 0\left( 2 \right)-2\left( 5 \right) & 0\left( -1 \right)-2\left( 3 \right) & 0\left( 2 \right)-2\left( -1 \right) \\ 3\left( 2 \right)+2\left( 5 \right) & 3\left( -1 \right)+2\left( 3 \right) & 3\left( 2 \right)+2\left( -1 \right) \\ 1\left( 2 \right)-5\left( 5 \right) & 1\left( -1 \right)-5\left( 3 \right) & 1\left( 2 \right)-5\left( -1 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -10 & -6 & 2 \\ 6+10 & -3+6 & 6-2 \\ 2-25 & -1-15 & 2+5 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -10 & -6 & 2 \\ 16 & 3 & 4 \\ -23 & -16 & 7 \\ \end{matrix} \right] \end{align} Thus, $BA=\left[ \begin{matrix} -10 & -6 & 2 \\ 16 & 3 & 4 \\ -23 & -16 & 7 \\ \end{matrix} \right]$