## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 10 - 10.2 - Operations with Matrices - 10.2 Exercises - Page 724: 34

#### Answer

X = $\begin{bmatrix} \frac{10}{3} & \frac{7}{3} & -13\\ \frac{23}{3} & 0 & -\frac{14}{3}\\ \end{bmatrix}$

#### Work Step by Step

5A = 6B - 3X 3X = 6B - 5A X = $\frac{1}{3}$(6B - 5A) First multiply matrix A by the scalar multiple 5: $\begin{bmatrix} -2(5) & 1(5) & 3(5)\\ -1(5) & 0(5) & 4(5)\\ \end{bmatrix}$ = $\begin{bmatrix} -10 & 5 & 15\\ -5 & 0 & 20\\ \end{bmatrix}$ Then multiply matrix B by the scalar multiple 6: $\begin{bmatrix} 0(6) & 2(6) & -4(6)\\ 3(6) & 0(6) & 1(6)\\ \end{bmatrix}$ = $\begin{bmatrix} 0 & 12 & -24\\ 18 & 0 & 6\\ \end{bmatrix}$ Then perform 6B - 5A to get matrix Y: Y = $\begin{bmatrix} 0+10 & 12-5 & -24-15\\ 18+5 & 0-0 & 6-20\\ \end{bmatrix}$ = $\begin{bmatrix} 10 & 7 & -39\\ 23 & 0 & -14\\ \end{bmatrix}$ Then multiply Y by the scalar $\frac{1}{3}$ to get X: X = $\begin{bmatrix} 10(\frac{1}{3}) & 7(\frac{1}{3}) & -39(\frac{1}{3})\\ 23(\frac{1}{3}) & 0(\frac{1}{3}) & -14(\frac{1}{3})\\ \end{bmatrix}$ = $\begin{bmatrix} \frac{10}{3} & \frac{7}{3} & -13\\ \frac{23}{3} & 0 & -\frac{14}{3}\\ \end{bmatrix}$

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