Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Review Exercises - Page 950: 31

Answer

The matrix is, $\left[ \begin{array}{*{35}{r}} -2 & 0 & 0 \\ 1 & 1 & -3 \\ \end{array} \right]$

Work Step by Step

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 $\left[ \begin{array}{*{35}{r}} -2 & 0 & 0 \\ 1 & 1 & -3 \\ \end{array} \right]$.
Small 1570430103
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.