## Precalculus (6th Edition) Blitzer

The matrix is, $\left[ \begin{array}{*{35}{r}} -2 & 0 & 0 \\ 1 & 1 & -3 \\ \end{array} \right]$
The matrix representing the triangle is $B=\left[ \begin{array}{*{35}{r}} 0 & 2 & 2 \\ 0 & 0 & -4 \\ \end{array} \right]$ The translation matrix to move the triangle 2 units to the left will be, ${{T}_{1}}=\left[ \begin{array}{*{35}{r}} -2 & -2 & -2 \\ 0 & 0 & 0 \\ \end{array} \right]$ The translation matrix to move the triangle 1 unit up will be, ${{T}_{2}}=\left[ \begin{array}{*{35}{r}} 0 & 0 & 0 \\ 1 & 1 & 1 \\ \end{array} \right]$ Therefore, the required translation matrix is given by: \begin{align} & T={{T}_{1}}+{{T}_{2}} \\ & =\left[ \begin{array}{*{35}{r}} -2 & -2 & -2 \\ 0 & 0 & 0 \\ \end{array} \right]+\left[ \begin{array}{*{35}{r}} 0 & 0 & 0 \\ 1 & 1 & 1 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} -2 & -2 & -2 \\ 1 & 1 & 1 \\ \end{array} \right] \end{align} Hence the resultant matrix representing the newly transformed triangle will be given by, \begin{align} & B+T=\left[ \begin{array}{*{35}{r}} 0 & 2 & 2 \\ 0 & 0 & -4 \\ \end{array} \right]+\left[ \begin{array}{*{35}{r}} -2 & -2 & -2 \\ 1 & 1 & 1 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} -2 & 0 & 0 \\ 1 & 1 & -3 \\ \end{array} \right] \end{align} That is, the transformed graph is a triangle with vertices $\left( -2,0 \right),\left( 0,1 \right),\left( 0,-3 \right)$ The matrix representing the transformed triangle is $\left[ \begin{array}{*{35}{r}} -2 & 0 & 0 \\ 1 & 1 & -3 \\ \end{array} \right]$.