Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 3 - Determinants - 3.3 Properties of Determinants - 3.3 Exercises - Page 125: 9

Answer

$54.$

Work Step by Step

Since we have $$\left[\begin{array}{rrr} {-3} & {6}&{9} \\ {6} & {9}&{12}\\{9}&{12}&{15}\end{array}\right]=3\left[\begin{array}{rrr} {-1} & {2}&{3} \\ {2} & {3}&{4}\\{3}&{4}&{5}\end{array}\right]=3\left[\begin{array}{rrr} {-1} & {2}&{3} \\ {0} & {7}&{10}\\{0}&{10}&{14}\end{array}\right]$$ then, we get $$\left|\begin{array}{rrr} {-3} & {6}&{9} \\ {6} & {9}&{12}\\{9}&{12}&{15}\end{array}\right|=3^3\left|\begin{array}{rrr}{-1} & {2}&{3} \\ {0} & {7}&{10}\\{0}&{10}&{14}\end{array}\right|=27(-1)(98-100)=54.$$
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