## Precalculus (6th Edition) Blitzer

a) $AB=\left[ \begin{matrix} 0 & 0 & -1 & -1 & -5 & -5 \\ 0 & 3 & 3 & 1 & 1 & 0 \\ \end{matrix} \right]$ b) See the graph below:
(a) \begin{align} & AB=\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 3 & 3 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 5 & 5 \\ \end{matrix} \right] \\ & AB=\left[ \begin{matrix} 0\left( 0 \right)+\left( -1 \right)\left( 0 \right) & 0\left( 3 \right)+\left( -1 \right)\left( 0 \right) & 0\left( 3 \right)+\left( -1 \right)\left( 1 \right) & 0\left( 1 \right)+\left( -1 \right)\left( 1 \right) & 0\left( 1 \right)+\left( -1 \right)\left( 5 \right) & 0\left( 0 \right)+\left( -1 \right)\left( 5 \right) \\ 1\left( 0 \right)+0\left( 0 \right) & 1\left( 3 \right)+0\left( 0 \right) & 1\left( 3 \right)+0\left( 1 \right) & 1\left( 1 \right)+0\left( 1 \right) & 1\left( 1 \right)+0\left( 5 \right) & 1\left( 0 \right)+0\left( 5 \right) \\ \end{matrix} \right] \\ & AB=\left[ \begin{matrix} 0-0 & 0-0 & 0-1 & 0-1 & 0-5 & 0-5 \\ 0+0 & 3+0 & 3+0 & 1+0 & 1+0 & 0+0 \\ \end{matrix} \right] \\ & AB=\left[ \begin{matrix} 0 & 0 & -1 & -1 & -5 & -5 \\ 0 & 3 & 3 & 1 & 1 & 0 \\ \end{matrix} \right] \\ \end{align} (b) The graph for matrix A and AB is drawn as shown above. From the graph, it can be inferred that the shape of letter “L” has been changed due to matrix multiplication. It has shifted from the positive quadrant to the negative quadrant and it has become a horizontal “L” from the vertical.