## Precalculus (6th Edition) Blitzer

The required matrix is, $AB=\left[ \begin{array}{*{35}{r}} 0 & 0 & 4 \\ 0 & 2 & 2 \\ \end{array} \right]$, 90° counterclockwise rotation about the origin.
The matrix is: $B=\left[ \begin{array}{*{35}{r}} 0 & 2 & 2 \\ 0 & 0 & -4 \\ \end{array} \right]$ The multiplication matrix is $A=\left[ \begin{array}{*{35}{r}} 0 & -1 \\ 1 & 0 \\ \end{array} \right]$ Consider the given matrices, \begin{align} & AB=\left[ \begin{array}{*{35}{r}} 0 & -1 \\ 1 & 0 \\ \end{array} \right]\left[ \begin{array}{*{35}{r}} 0 & 2 & 2 \\ 0 & 0 & -4 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} 0 & 0 & 4 \\ 0 & 2 & 2 \\ \end{array} \right] \end{align} That is, the transformed graph is a triangle with vertices $\left( 0,0 \right),\left( 0,2 \right),\left( 4,2 \right)$ Therefore, the transformed triangle will be based on a 90° counterclockwise rotation about the origin. The matrix $AB=\left[ \begin{array}{*{35}{r}} 0 & 0 & 4 \\ 0 & 2 & 2 \\ \end{array} \right]$ and the transformed triangle will be based on a 90° counterclockwise rotation about the origin.