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 - Review Exercises - Page 138: 28

Answer

(a) $|A|=-31.$ (b) $|A^{-1}|= -\frac{1}{31}.$

Work Step by Step

Since $$A=\left[ \begin {array}{cccc}2&-1&4\\ 5&0&3\\1&-2&0 \end {array} \right]=\left[ \begin {array}{cccc}2&-1&4\\ 5&0&3\\-3&0&-8 \end {array} \right],$$ we have (a) $|A|=-31.$ (b) $|A^{-1}|=|A|^{-1}=-\frac{1}{31}.$
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