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: 25

Answer

$$|A|=5, \quad \left|A^{-1}\right|= \frac{1}{5}.$$

Work Step by Step

Since $$A=\left[\begin{array}{rrr} {2} & {3} \\ {1} & {4}\end{array}\right],$$ then $$A^{-1}=\frac{1}{5}\left[\begin{array}{rrr} {4} & {-3} \\ {-1} & {2}\end{array}\right].$$ Now, we have $$|A|=5, \quad \left|A^{-1}\right|=\frac{1}{5^2}(8-3)=\frac{1}{5}.$$ Hence, we verify that $$\left|A^{-1}\right|=\frac{1}{|A|}.$$
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