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 - Review Exercises - Page 98: 17


$$A^{-1}= \left[ \begin {array}{cccccc} {\frac {3}{20}}&{\frac {3}{20}}&\frac{1}{10} \\ \frac{3}{10}&-\frac{1}{30}&-\frac{2}{15}\\ -\frac{1}{5}&-\frac{1}{5}&\frac{1}{5}\end {array} \right] .$$

Work Step by Step

Let $A$ be a matrix given by $$A=\left[ \begin {array}{cccccc} 2&3&1 \\ 2&-3&-3 \\ 4&0&3 \end {array} \right].$$ To find $A^{-1}$, we have $$\left[ A \ \ I \right]=\left[ \begin {array}{cccccc} 2&3&1&1&0&0\\ 2&-3&-3 &0&1&0\\ 4&0&3&0&0&1\end {array} \right] . $$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[I \ \ A^{-1} \right]= \left[ \begin {array}{cccccc} 1&0&0&{\frac {3}{20}}&{\frac {3}{20}}&\frac{1}{10} \\ 0&1&0&\frac{3}{10}&-\frac{1}{30}&-\frac{2}{15}\\ 0&0 &1&-\frac{1}{5}&-\frac{1}{5}&\frac{1}{5}\end {array} \right] . $$ Then $A^{-1}$ is given by $$A^{-1}= \left[ \begin {array}{cccccc} {\frac {3}{20}}&{\frac {3}{20}}&\frac{1}{10} \\ \frac{3}{10}&-\frac{1}{30}&-\frac{2}{15}\\ -\frac{1}{5}&-\frac{1}{5}&\frac{1}{5}\end {array} \right] .$$
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