## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Review Exercises - Page 950: 19

#### Answer

The matrix $-2A+4D$ is, $-2A+4D=\left[ \begin{array}{*{35}{l}} -12 & 14 & 0 \\ 2 & -14 & 18 \\ \end{array} \right]$.

#### Work Step by Step

Here we need to find $-2A+4D$. Therefore consider, \begin{align} & -2A+4D=-2\left[ \begin{array}{*{35}{l}} 2 & -1 & 2 \\ 5 & 3 & -1 \\ \end{array} \right]+4\left[ \begin{array}{*{35}{l}} -2 & 3 & 1 \\ 3 & -2 & 4 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} -2\times 2 & -2\times -1 & -2\times 2 \\ -2\times 5 & -2\times 3 & -2\times -1 \\ \end{array} \right]+\left[ \begin{array}{*{35}{l}} 4\times -2 & 4\times 3 & 4\times 1 \\ 4\times 3 & 4\times -2 & 4\times 4 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} -4 & 2 & -4 \\ -10 & -6 & 2 \\ \end{array} \right]+\left[ \begin{array}{*{35}{l}} -8 & 12 & 4 \\ 12 & -8 & 16 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} -12 & 14 & 0 \\ 2 & -14 & 18 \\ \end{array} \right] \end{align} Thus, $-2A+4D=\left[ \begin{array}{*{35}{l}} -12 & 14 & 0 \\ 2 & -14 & 18 \\ \end{array} \right]$.

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