## Precalculus (6th Edition) Blitzer

The matrix $3A+2D$ is, $3A+2D=\left[ \begin{array}{*{35}{l}} 2 & 3 & 8 \\ 21 & 5 & 5 \\ \end{array} \right]$
Here we need to find $3A+2D$. Therefore consider, \begin{align} & 3A+2D=3\left[ \begin{array}{*{35}{l}} 2 & -1 & 2 \\ 5 & 3 & -1 \\ \end{array} \right]+2\left[ \begin{array}{*{35}{l}} -2 & 3 & 1 \\ 3 & -2 & 4 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 3\times 2 & 3\times \left( -1 \right) & 3\times 2 \\ 3\times 5 & 3\times 3 & 3\times \left( -1 \right) \\ \end{array} \right]+\left[ \begin{array}{*{35}{l}} 2\times \left( -2 \right) & 2\times 3 & 2\times 1 \\ 2\times 3 & 2\times \left( -2 \right) & 2\times 4 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 6 & -3 & 6 \\ 15 & 9 & -3 \\ \end{array} \right]+\left[ \begin{array}{*{35}{l}} -4 & 6 & 2 \\ 6 & -4 & 8 \\ \end{array} \right] \end{align} Now we will add these matrices as shown below: \begin{align} & \left[ \begin{array}{*{35}{l}} 6 & -3 & 6 \\ 15 & 9 & -3 \\ \end{array} \right]+\left[ \begin{array}{*{35}{l}} -4 & 6 & 2 \\ 6 & -4 & 8 \\ \end{array} \right]=\left[ \begin{array}{*{35}{l}} 6-4 & -3+6 & 6+2 \\ 15+6 & 9-4 & -3+8 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 2 & 3 & 8 \\ 21 & 5 & 5 \\ \end{array} \right] \end{align} Thus, $3A+2D=\left[ \begin{array}{*{35}{l}} 2 & 3 & 8 \\ 21 & 5 & 5 \\ \end{array} \right]$