Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 12 - Matrices - 12-1 Adding and Subtracting Matrices - Practice and Problem-Solving Exercises - Page 769: 28

Answer

$a=2,\text{ }b=\dfrac{9}{4},\text{ }c=-1,\text{ }d=0,\text{ }f=\dfrac{1}{2},\text{ }g=-4 $

Work Step by Step

Equating each corresponding elements in the given matrix equation, $ \begin{bmatrix} 4b+2 & -3 & 4d \\ -4a & 2 & 3 \\ 2f-1 & -14 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 2c-1 & 0 \\ -8 & 2 & 3 \\ 0 & 3g-2 & 1 \end{bmatrix} ,$ and then solving the resulting equation, then \begin{align*} 4b+2&=11 \\ 4b+2-2&=11-2 \\ 4b&=9 \\ \dfrac{4b}{4}&=\dfrac{9}{4} \\ b&=\dfrac{9}{4} ,\\\\\\ -3&=2c-1 \\ -3+1&=2c-1+1 \\ -2&=2c \\ -\dfrac{2}{2}&=\dfrac{2c}{2} \\ -1&=c ,\\\\\\ 4d&=0 \\ \dfrac{4d}{4}&=\dfrac{0}{4} \\ d&=0 ,\\\\\\ -4a&=-8 \\ \dfrac{-4a}{-4}&=\dfrac{-8}{-4} \\ a&=2 ,\\\\\\ 2f-1&=0 \\ 2f-1+1&=0+1 \\ 2f&=1 \\ \dfrac{2f}{2}&=\dfrac{1}{2} \\ f&=\dfrac{1}{2} ,\\\\\\ -14&=3g-2 \\ -14+2&=3g-2+2 \\ -12&=3g \\ -\dfrac{12}{3}&=\dfrac{3g}{3} \\ -4&=g .\end{align*} Hence, $ a=2,\text{ }b=\dfrac{9}{4},\text{ }c=-1,\text{ }d=0,\text{ }f=\dfrac{1}{2},\text{ }g=-4 .$
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