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: 23

Answer

$$A^{-1}=\left[ \begin {array}{cccccc} 1&0&0\\ - \frac{3}{4}&\frac{1}{4}&0\\ {\frac {7}{20}}&-\frac{1}{4}&\frac{1}{5} \end {array} \right] .$$

Work Step by Step

To find $A^{-1}$, we have $$\left[ A \ \ I \right]= \left[ \begin {array}{cccccc} 1&0&0&1&0&0\\ 3&4&0&0 &1&0\\ 2&5&5&0&0&1\end {array} \right] . $$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[I \ \ A^{-1} \right]= \left[ \begin {array}{cccccc} 1&0&0&1&0&0\\ 0&1&0&- \frac{3}{4}&\frac{1}{4}&0\\ 0&0&1&{\frac {7}{20}}&-\frac{1}{4}&\frac{1}{5} \end {array} \right] . $$ Then $A^{-1}$ is given by $$A^{-1}=\left[ \begin {array}{cccccc} 1&0&0\\ - \frac{3}{4}&\frac{1}{4}&0\\ {\frac {7}{20}}&-\frac{1}{4}&\frac{1}{5} \end {array} \right] .$$
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