Chapter 8 - Review Exercises - Page 950: 20

The matrix $-5\left( A+D \right)$ is, $-5\left( A+D \right)=\left[ \begin{array}{*{35}{l}} 0 & -10 & -15 \\ -40 & -5 & -15 \\ \end{array} \right]$.

Here we need to find $-5\left( A+D \right)$. Therefore consider, $\begin{align} & -5\left( A+D \right)=-5A-5D \\ & =-5\left[ \begin{array}{*{35}{l}} 2 & -1 & 2 \\ 5 & 3 & -1 \\ \end{array} \right]-5\left[ \begin{array}{*{35}{l}} -2 & 3 & 1 \\ 3 & -2 & 4 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} -5\times 2 & -5\times -1 & -5\times 2 \\ -5\times 5 & -5\times 3 & -5\times -1 \\ \end{array} \right]+\left[ \begin{array}{*{35}{l}} -5\times -2 & -5\times 3 & -5\times 1 \\ -5\times 3 & -5\times -2 & -5\times 4 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} -10 & 5 & -10 \\ -25 & -15 & 5 \\ \end{array} \right]+\left[ \begin{array}{*{35}{l}} 10 & -15 & -5 \\ -15 & 10 & -20 \\ \end{array} \right] \end{align}$ Now by adding the matrices as below, we get: $\begin{align} & -5\left( A+D \right)=\left[ \begin{array}{*{35}{l}} -10 & 5 & -10 \\ -25 & -15 & 5 \\ \end{array} \right]+\left[ \begin{array}{*{35}{l}} 10 & -15 & -5 \\ -15 & 10 & -20 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} -10+10 & 5-15 & -10-5 \\ -25-15 & -15+10 & 5-20 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 0 & -10 & -15 \\ -40 & -5 & -15 \\ \end{array} \right] \end{align}$ Thus, $-5\left( A+D \right)=\left[ \begin{array}{*{35}{l}} 0 & -10 & -15 \\ -40 & -5 & -15 \\ \end{array} \right]$