## Precalculus (6th Edition) Blitzer

The matrix $\left( A-D \right)C$ is, $\left( A-D \right)C=\left[ \begin{array}{*{35}{l}} 7 & 6 & 5 \\ 2 & -1 & 11 \\ \end{array} \right]$.
Here we need to find $\left( A-D \right)C$. Therefore consider, \begin{align} & \left( A-D \right)C=\left( \left[ \begin{array}{*{35}{l}} 2 & -1 & 2 \\ 5 & 3 & -1 \\ \end{array} \right]-\left[ \begin{array}{*{35}{l}} -2 & 3 & 1 \\ 3 & -2 & 4 \\ \end{array} \right] \right)\left[ \begin{matrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ -1 & 2 & 1 \\ \end{matrix} \right] \\ & =\left( \left[ \begin{array}{*{35}{l}} 2+2 & -1-3 & 2-1 \\ 5-3 & 3+2 & -1-4 \\ \end{array} \right] \right)\left[ \begin{matrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ -1 & 2 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{array}{*{35}{l}} 4 & -4 & 1 \\ 2 & 5 & -5 \\ \end{array} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ -1 & 2 & 1 \\ \end{matrix} \right] \end{align} It can be further simplified as: \begin{align} & \left[ \begin{array}{*{35}{l}} 4 & -4 & 1 \\ 2 & 5 & -5 \\ \end{array} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ -1 & 2 & 1 \\ \end{matrix} \right]=\left[ \begin{array}{*{35}{l}} 4\left( 1 \right)-4\left( -1 \right)+1\left( -1 \right) & 4\left( 2 \right)-4\left( 1 \right)+1\left( 2 \right) & 4\left( 3 \right)-4\left( 2 \right)+1\left( 1 \right) \\ 2\left( 1 \right)+5\left( -1 \right)-5\left( -1 \right) & 2\left( 2 \right)+5\left( 1 \right)-5\left( 2 \right) & 2\left( 3 \right)+5\left( 2 \right)-5\left( 1 \right) \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 4+4-1 & 8-4+2 & 12-8+1 \\ 2-5+5 & 4+5-10 & 6+10-5 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 7 & 6 & 5 \\ 2 & -1 & 11 \\ \end{array} \right] \end{align} Thus, $\left( A-D \right)C=\left[ \begin{array}{*{35}{l}} 7 & 6 & 5 \\ 2 & -1 & 11 \\ \end{array} \right]$