## Precalculus (6th Edition) Blitzer

The matrix is, $\left[ \begin{array}{*{35}{r}} 5 & -2 \\ 1 & -1 \\ 4 & -1 \\ \end{array} \right]$
Matrix multiplication: consider two matrices ${{X}_{2\times 2}},{{Y}_{2\times 2}}$. A matrix multiplication is possible only if the rows of the first matrix are equal to the columns of the second matrix and the resulting matrix formed will have the order as: rows of the second matrix and columns of the first 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} Using the above method, we will solve the given expression as below: \begin{align} & AB=\left[ \begin{array}{*{35}{r}} 3 & 1 \\ 1 & 0 \\ 2 & 1 \\ \end{array} \right]\left[ \begin{array}{*{35}{r}} 1 & -1 \\ 2 & 1 \\ \end{array} \right] \\ & =\left[ \begin{matrix} 3+2 & -3+1 \\ 1+0 & -1+0 \\ 2+2 & -2+1 \\ \end{matrix} \right] \\ & =\left[ \begin{array}{*{35}{r}} 5 & -2 \\ 1 & -1 \\ 4 & -1 \\ \end{array} \right] \end{align} Therefore, $AB=\left[ \begin{array}{*{35}{r}} 5 & -2 \\ 1 & -1 \\ 4 & -1 \\ \end{array} \right]$.