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: 17


$$A^{-1}=\left[ \begin {array}{cccccc} -\frac{3}{2}&\frac{3}{2}&1\\ \frac{9}{2}&-\frac{7}{2}&-3\\ -1&1&1\end {array} \right].$$

Work Step by Step

To find $A^{-1}$, we have $$\left[ A \ \ I \right]= \left[ \begin {array}{cccccc} 1&1&2&1&0&0\\ 3&1&0&0 &1&0\\ -2&0&3&0&0&1\end {array} \right] . $$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[I \ \ A^{-1} \right]=\left[ \begin {array}{cccccc} 1&0&0&-\frac{3}{2}&\frac{3}{2}&1\\ 0& 1&0&\frac{9}{2}&-\frac{7}{2}&-3\\ 0&0&1&-1&1&1\end {array} \right] . $$ Then $A^{-1}$ is given by $$A^{-1}=\left[ \begin {array}{cccccc} -\frac{3}{2}&\frac{3}{2}&1\\ \frac{9}{2}&-\frac{7}{2}&-3\\ -1&1&1\end {array} \right].$$
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