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: 32

Answer

$|A|=12$ and $|A^{-1}|=\frac{1}{12}$.

Work Step by Step

Let $A$ be given by $$A=\left[ \begin {array}{ccc} -1&1&2\\ 2&4&8 \\ -1&1&0\end {array} \right] ,$$ then we have $A^{-1}=\left[ \begin {array}{ccc} -\frac{2}{3}&\frac{1}{6}&0\\ -\frac{2}{3}&\frac{1}{6}&1 \\ \frac{1}{2}&0&-\frac{1}{2}\end {array} \right] $. Now, $|A|=12$ and $|A^{-1}|=\frac{1}{12}$. Hence, we verify that $$|A^{-1}=\frac{1}{|A|}.$$
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