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: 31

Answer

$|A|=-10$ and $|A^{-1}|=-\frac{1}{10}$.

Work Step by Step

Let $A$ be given by $$A=\left[ \begin {array}{ccc} 1&0&1\\ 2&-1&4 \\ 2&6&0\end {array} \right] ,$$ then we have $A^{-1}=\left[ \begin {array}{ccc} {\frac {12}{5}}&-\frac{3}{5}&-\frac{1}{10} \\ -\frac{4}{5}&\frac{1}{5}&\frac{1}{5}\\ -\frac{7}{5}&\frac{3}{5}&\frac{1}{10} \end {array} \right] $. Now, $|A|=-10$ and $|A^{-1}|=-\frac{1}{10}$. Hence, we verify that $$|A^{-1}=\frac{1}{|A|}.$$
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