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: 29

Answer

$$|A|=24, \quad \left|A^{-1}\right|=\frac{1}{24}.$$

Work Step by Step

Since $$A=\left[\begin{array}{rrr} 1&0&-1&3\\ 1&0&3&-2 \\ 2&0&2&-1\\ 1&-3&1&-2 \end{array}\right],$$ then one can calculate its inverse , that is, $$A^{-1}= \left[\begin{array}{rrr} -\frac{1}{8}&-\frac{5}{8}&{\frac {7}{8}}&0 \\ -\frac{1}{4}&-\frac{1}{4}&{\frac {5}{12}}&-\frac{1}{3} \\ \frac{3}{8}&{\frac {7}{8}}&-\frac{5}{8}&0\\ \frac{1}{2} &\frac{1}{2}&- \frac{1}{2}&0 \end{array}\right].$$ Now, we have $$|A|=24, \quad \left|A^{-1}\right|=\frac{1}{24}.$$ Hence, we verify that $$\left|A^{-1}\right|=\frac{1}{|A|}.$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.