## Precalculus (6th Edition) Blitzer

The matrix $B\left( AC \right)$ is, $B\left( AC \right)=\left[ \begin{matrix} -6 & -22 & -40 \\ 9 & 43 & 58 \\ -14 & -48 & -94 \\ \end{matrix} \right]$
Here we need to find $B\left( AC \right)$. Therefore consider, \begin{align} & B\left( AC \right)=\left[ \begin{array}{*{35}{l}} 0 & -2 \\ 3 & 2 \\ 1 & -5 \\ \end{array} \right]\left( \left[ \begin{array}{*{35}{l}} 2 & -1 & 2 \\ 5 & 3 & -1 \\ \end{array} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ -1 & 2 & 1 \\ \end{matrix} \right] \right) \\ & =\left[ \begin{array}{*{35}{l}} 0 & -2 \\ 3 & 2 \\ 1 & -5 \\ \end{array} \right]\left[ \begin{array}{*{35}{l}} 2\left( 1 \right)-1\left( -1 \right)+2\left( -1 \right) & 2\left( 2 \right)-1\left( 1 \right)+2\left( 2 \right) & 2\left( 3 \right)-1\left( 2 \right)+2\left( 1 \right) \\ 5\left( 1 \right)+3\left( -1 \right)-1\left( -1 \right) & 5\left( 2 \right)+3\left( 1 \right)-1\left( 2 \right) & 5\left( 3 \right)+3\left( 2 \right)-1\left( 1 \right) \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 0 & -2 \\ 3 & 2 \\ 1 & -5 \\ \end{array} \right]\left[ \begin{array}{*{35}{l}} 2+1-2 & 4-1+4 & 6-2+2 \\ 5-3+1 & 10+3-2 & 15+6-1 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 0 & -2 \\ 3 & 2 \\ 1 & -5 \\ \end{array} \right]\left[ \begin{array}{*{35}{l}} 1 & 7 & 6 \\ 3 & 11 & 20 \\ \end{array} \right] \end{align} It can be further simplified as: \begin{align} & \left[ \begin{array}{*{35}{l}} 0 & -2 \\ 3 & 2 \\ 1 & -5 \\ \end{array} \right]\left[ \begin{array}{*{35}{l}} 1 & 7 & 6 \\ 3 & 11 & 20 \\ \end{array} \right]=\left[ \begin{matrix} 0\left( 1 \right)-2\left( 3 \right) & 0\left( 7 \right)-2\left( 11 \right) & 0\left( 6 \right)-2\left( 20 \right) \\ 3\left( 1 \right)+2\left( 3 \right) & 3\left( 7 \right)+2\left( 11 \right) & 3\left( 6 \right)+2\left( 20 \right) \\ 1\left( 1 \right)-5\left( 3 \right) & 1\left( 7 \right)-5\left( 11 \right) & 1\left( 6 \right)-5\left( 20 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -6 & -22 & -40 \\ 3+6 & 21+22 & 18+40 \\ 1-15 & 7-55 & 6-100 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -6 & -22 & -40 \\ 9 & 43 & 58 \\ -14 & -48 & -94 \\ \end{matrix} \right] \end{align} Thus, $B\left( AC \right)=\left[ \begin{matrix} -6 & -22 & -40 \\ 9 & 43 & 58 \\ -14 & -48 & -94 \\ \end{matrix} \right]$