## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Section 8.3 - Matrix Operations and Their Applications - Exercise Set - Page 917: 11

#### Answer

a) The resultanat sum of matrices is $A+B=\left[ \begin{array}{*{35}{l}} 3 & 2 \\ 6 & 2 \\ 5 & 7 \\ \end{array} \right]$ B) The resultanat difference of matrices is $A-B=\left[ \begin{array}{*{35}{l}} -1 & 4 \\ 0 & 6 \\ 5 & 5 \\ \end{array} \right]$ c) The resultant matrix is $\left( -4 \right)A=\left[ \begin{array}{*{35}{l}} -4 & -12 \\ -12 & -16 \\ -20 & -24 \\ \end{array} \right]$ d)The resultant matrix is $3A+2B=\left[ \begin{array}{*{35}{l}} 7 & 7 \\ 15 & 8 \\ 15 & 20 \\ \end{array} \right]$

#### Work Step by Step

(a) Perform the addition of the matrices $A$ and $B$ as below: \begin{align} & A+B=\left[ \begin{array}{*{35}{l}} 1 & 3 \\ 3 & 4 \\ 5 & 6 \\ \end{array} \right]+\left[ \begin{array}{*{35}{l}} 2 & -1 \\ 3 & -2 \\ 0 & 1 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 1+2 & 3-1 \\ 3+3 & 4-2 \\ 5+0 & 6+1 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 3 & 2 \\ 6 & 2 \\ 5 & 7 \\ \end{array} \right] \end{align} Hence, $A+B=\left[ \begin{array}{*{35}{l}} 3 & 2 \\ 6 & 2 \\ 5 & 7 \\ \end{array} \right]$ (b) Perform the subtraction of the matrices $A$ and $B$ as below: \begin{align} & A-B=\left[ \begin{array}{*{35}{l}} 1 & 3 \\ 3 & 4 \\ 5 & 6 \\ \end{array} \right]-\left[ \begin{array}{*{35}{l}} 2 & -1 \\ 3 & -2 \\ 0 & 1 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 1-2 & 3+1 \\ 3-3 & 4+2 \\ 5-0 & 6-1 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} -1 & 4 \\ 0 & 6 \\ 5 & 5 \\ \end{array} \right] \end{align} Hence, $A-B=\left[ \begin{array}{*{35}{l}} -1 & 4 \\ 0 & 6 \\ 5 & 5 \\ \end{array} \right]$ (c) Perform the multiplication of the matrices $A$ with constant number as below: \begin{align} & \left( -4 \right)A=\left( -4 \right)\left[ \begin{array}{*{35}{l}} 1 & 3 \\ 3 & 4 \\ 5 & 6 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} -4 & -12 \\ -12 & -16 \\ -20 & -24 \\ \end{array} \right] \end{align} Hence, $\left( -4 \right)A=\left[ \begin{array}{*{35}{l}} -4 & -12 \\ -12 & -16 \\ -20 & -24 \\ \end{array} \right]$ (d) Perform the multiplication with a content number and addition of the matrices $A$ and $B$ as below: \begin{align} & 3A+2B=3\left[ \begin{array}{*{35}{l}} 1 & 3 \\ 3 & 4 \\ 5 & 6 \\ \end{array} \right]+2\left[ \begin{array}{*{35}{l}} 2 & -1 \\ 3 & -2 \\ 0 & 1 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 3 & 9 \\ 9 & 12 \\ 15 & 18 \\ \end{array} \right]+\left[ \begin{array}{*{35}{l}} 4 & -2 \\ 6 & -4 \\ 0 & 2 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{l}} 7 & 7 \\ 15 & 8 \\ 15 & 20 \\ \end{array} \right] \end{align} Hence, $3A+2B=\left[ \begin{array}{*{35}{l}} 7 & 7 \\ 15 & 8 \\ 15 & 20 \\ \end{array} \right]$

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