Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 2 - Matrices - 2.3 The Inverse of a Matrix - 2.3 Exercises - Page 71: 19


$$A^{-1}=\left[ \begin {array}{cccccc} \frac{1}{2}&0&0\\ 0&\frac{1}{3}&0\\ 0&0&\frac{1}{5}\end {array} \right].$$

Work Step by Step

To find $A^{-1}$, we have $$\left[ A \ \ I \right]= \left[ \begin {array}{cccccc} 2&0&0&1&0&0\\ 0&3&0&0 &1&0\\ 0&0&5&0&0&1\end {array} \right] . $$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[I \ \ A^{-1} \right]=\left[ \begin {array}{cccccc} 1&0&0&\frac{1}{2}&0&0\\ 0&1&0 &0&\frac{1}{3}&0\\0&0&1&0&0&\frac{1}{5}\end {array} \right] . $$ Then $A^{-1}$ is given by $$A^{-1}=\left[ \begin {array}{cccccc} \frac{1}{2}&0&0\\ 0&\frac{1}{3}&0\\ 0&0&\frac{1}{5}\end {array} \right].$$
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.