## Precalculus (6th Edition) Blitzer

The product is, $\left[ \begin{array}{*{35}{r}} 2 & -1 \\ 13 & 1 \\ \end{array} \right]$.
Consider the given matrices \begin{align} & A=\left[ \begin{array}{*{35}{r}} 1 & -1 & 0 \\ 2 & 1 & 3 \\ \end{array} \right] \\ & B=\left[ \begin{array}{*{35}{r}} 4 & -1 \\ 2 & 0 \\ 1 & 1 \\ \end{array} \right] \end{align} Therefore, the order of matrix $A$ is $2\times 3$ and $B$ is $3\times 2$; therefore, the product of the matrices is defined and the order of the matrix will be $2\times 2$ \begin{align} & AB=\left[ \begin{array}{*{35}{r}} 1 & -1 & 0 \\ 2 & 1 & 3 \\ \end{array} \right]\left[ \begin{array}{*{35}{r}} 4 & -1 \\ 2 & 0 \\ 1 & 1 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} \left( 1 \right)\left( 4 \right)+\left( -1 \right)\left( 2 \right)+\left( 0 \right)\left( 1 \right) & \left( 1 \right)\left( -1 \right)+\left( -1 \right)\left( 0 \right)+\left( 0 \right)\left( 1 \right) \\ \left( 2 \right)\left( 4 \right)+\left( 1 \right)\left( 2 \right)+\left( 3 \right)\left( 1 \right) & \left( 2 \right)\left( -1 \right)+\left( 1 \right)\left( 0 \right)+\left( 3 \right)\left( 1 \right) \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} 4-2+0 & -1+0+0 \\ 8+2+3 & -2+0+3 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} 2 & -1 \\ 13 & 1 \\ \end{array} \right] \end{align} The product of the matrices is $\left[ \begin{array}{*{35}{r}} 2 & -1 \\ 13 & 1 \\ \end{array} \right]$.