Chapter 8 - Section 8.3 - Matrix Operations and Their Applications - Exercise Set - Page 918: 40

$A\left( B+C \right)=\left[ \begin{array}{*{35}{l}} 24 & 0 \\ -33 & -5 \\ -3 & -1 \\ \end{array} \right]$

Consider, $A\left( B+C \right)$. By the distributive law, $A\left( B+C \right)=AB+AC$. So, consider the matrix product, $\begin{align} & AB=\left[ \begin{array}{*{35}{l}} 4 & 0 \\ -3 & 5 \\ 0 & 1 \\ \end{array} \right]\left[ \begin{matrix} 5 & 1 \\ -2 & -2 \\ \end{matrix} \right] \\ & =\left[ \begin{array}{*{35}{l}} 4\left( 5 \right)+0\left( -2 \right) & 4\left( 1 \right)+0\left( -2 \right) \\ -3\left( 5 \right)+5\left( -2 \right) & -3\left( 1 \right)+5\left( -2 \right) \\ 0\left( 5 \right)+1\left( -2 \right) & 0\left( 1 \right)+1\left( -2 \right) \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 20-0 & 4-0 \\ -15-10 & -3-10 \\ 0-2 & 0-2 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 20 & 4 \\ -25 & -13 \\ -2 & -2 \\ \end{array} \right] \end{align}$ Next we will consider the matrix product $AC$ as below: $\begin{align} & AC=\left[ \begin{array}{*{35}{l}} 4 & 0 \\ -3 & 5 \\ 0 & 1 \\ \end{array} \right]\left[ \begin{matrix} 1 & -1 \\ -1 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{array}{*{35}{l}} 4\left( 1 \right)+0\left( -1 \right) & 4\left( -1 \right)+0\left( 1 \right) \\ -3\left( 1 \right)+5\left( -1 \right) & -3\left( -1 \right)+5\left( 1 \right) \\ 0\left( 1 \right)+1\left( -1 \right) & 0\left( -1 \right)+1\left( 1 \right) \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 4+0 & -4+0 \\ -3-5 & 3+5 \\ 0-1 & 0+1 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 4 & -4 \\ -8 & 8 \\ -1 & 1 \\ \end{array} \right] \end{align}$ Now add these matrices as below: $\begin{align} & A\left( B+C \right)=AB+AC \\ & =\left[ \begin{array}{*{35}{l}} 20 & 4 \\ -25 & -13 \\ -2 & -2 \\ \end{array} \right]+\left[ \begin{array}{*{35}{l}} 4 & -4 \\ -8 & 8 \\ -1 & 1 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 20+4 & 4-4 \\ -25-8 & -13+8 \\ -2-1 & -2+1 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 24 & 0 \\ -33 & -5 \\ -3 & -1 \\ \end{array} \right] \end{align}$