## Precalculus (6th Edition) Blitzer

$A(BC)=\left[ \begin{matrix} 12 & -4 \\ 22 & -7 \\ -4 & 1 \\ \end{matrix} \right]$
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} Use the above method to solve the provided expression as below: \begin{align} & (BC)=\left[ \begin{matrix} 4 & 1 \\ -6 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 4\cdot -1+1\cdot 0 & 4\cdot 0+1\cdot 1 \\ -6\cdot -1+-2\cdot 0 & -6\cdot 0+-2\cdot -2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -4 & 1 \\ 6 & -2 \\ \end{matrix} \right] \end{align} Now, \begin{align} & A(BC)=\left[ \begin{matrix} 0 & 2 \\ -1 & 3 \\ 1 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} -4 & 1 \\ 6 & -2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 12 & -4 \\ 22 & -7 \\ -4 & 1 \\ \end{matrix} \right] \end{align} Hence, $A(BC)=\left[ \begin{matrix} 12 & -4 \\ 22 & -7 \\ -4 & 1 \\ \end{matrix} \right]$