## Precalculus (6th Edition) Blitzer

$A(B+C)=\left[ \begin{array}{*{35}{l}} -12 & -2 \\ -21 & -4 \\ 3 & 1 \\ \end{array} \right]$
To solve the provided expression, follow the below methods: Matrix addition: Consider two given matrices $X,Y$ $X=\left[ \begin{matrix} {{x}_{1}} & {{x}_{2}} \\ {{x}_{3}} & {{x}_{4}} \\ \end{matrix} \right]$ and $Y=\left[ \begin{matrix} {{y}_{1}} & {{y}_{2}} \\ {{y}_{3}} & {{y}_{4}} \\ \end{matrix} \right]$ Now, \begin{align} & X+Y=\left[ \begin{matrix} {{x}_{1}} & {{x}_{2}} \\ {{x}_{3}} & {{x}_{4}} \\ \end{matrix} \right]+\left[ \begin{matrix} {{y}_{1}} & {{y}_{2}} \\ {{y}_{3}} & {{y}_{4}} \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} {{x}_{1}}+{{y}_{1}} & {{x}_{2}}+{{y}_{2}} \\ {{x}_{3}}+{{y}_{3}} & {{x}_{4}}+{{y}_{4}} \\ \end{matrix} \right] \end{align} Matrix multiplication: Consider two matrices ${{X}_{2\times 2}},{{Y}_{2\times 2}}$. The multiplication of two matrices is possible only if the number of rows of the first matrix equals the number of columns of the second matrix. $X=\left[ \begin{matrix} {{x}_{1}} & {{x}_{2}} \\ {{x}_{3}} & {{x}_{4}} \\ \end{matrix} \right]$ and $Y=\left[ \begin{matrix} {{y}_{1}} & {{y}_{2}} \\ {{y}_{3}} & {{y}_{4}} \\ \end{matrix} \right]$ Now, \begin{align} & X\times Y=\left[ \begin{matrix} {{x}_{1}} & {{x}_{2}} \\ {{x}_{3}} & {{x}_{4}} \\ \end{matrix} \right]\times \left[ \begin{matrix} {{y}_{1}} & {{y}_{2}} \\ {{y}_{3}} & {{y}_{4}} \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} {{x}_{1}}{{y}_{1}}+{{x}_{2}}{{y}_{3}} & {{x}_{1}}{{y}_{2}}+{{x}_{2}}{{y}_{4}} \\ {{x}_{3}}{{y}_{1}}+{{x}_{4}}{{y}_{3}} & {{x}_{3}}{{y}_{2}}+{{x}_{4}}{{y}_{4}} \\ \end{matrix} \right] \end{align} Substitute the value of $A=\left[ \begin{array}{*{35}{l}} 0 & 2 \\ -1 & 3 \\ 1 & 0 \\ \end{array} \right]$, $B=\left[ \begin{matrix} 4 & 1 \\ -6 & -2 \\ \end{matrix} \right]$ and $C=\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ in $A(B+C)$ \begin{align} & A(B+C)=\left[ \begin{array}{*{35}{l}} 0 & 2 \\ -1 & 3 \\ 1 & 0 \\ \end{array} \right]\left( \left[ \begin{matrix} 4 & 1 \\ -6 & -2 \\ \end{matrix} \right]+\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \right) \\ & =\left[ \begin{array}{*{35}{l}} 0 & 2 \\ -1 & 3 \\ 1 & 0 \\ \end{array} \right]\left( \left[ \begin{matrix} 4-1 & 1+0 \\ -6+0 & -2+1 \\ \end{matrix} \right] \right) \\ & =\left[ \begin{array}{*{35}{l}} 0 & 2 \\ -1 & 3 \\ 1 & 0 \\ \end{array} \right]\left( \left[ \begin{matrix} 3 & 1 \\ -6 & -1 \\ \end{matrix} \right] \right) \\ & =\left[ \begin{array}{*{35}{l}} -12 & -2 \\ -21 & -4 \\ 3 & 1 \\ \end{array} \right] \end{align} Hence $A(B+C)$ $=\left[ \begin{array}{*{35}{l}} -12 & -2 \\ -21 & -4 \\ 3 & 1 \\ \end{array} \right]$