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: 27


$$A=\left[ \begin {array}{cc} \frac{3}{42}&\frac{1}{42}\\ -\frac{1}{21}&\frac{2}{21} \end {array} \right] .$$

Work Step by Step

Since $$((3A)^{-1})^{-1}=3A,$$ then by calculating the inverse of the matrix $ \left[ \begin {array}{cc}4&-1\\2&3 \end {array} \right] $ we have $$3A=\left[ \begin {array}{cc} \frac{3}{14}&\frac{1}{14}\\ -\frac{1}{7}&\frac{2}{7} \end {array} \right] .$$ Hence, $A$ is given by $$A=\frac{1}{3}\left[ \begin {array}{cc} \frac{3}{14}&\frac{1}{14}\\ -\frac{1}{7}&\frac{2}{7} \end {array} \right]=\left[ \begin {array}{cc} \frac{3}{42}&\frac{1}{42}\\ -\frac{1}{21}&\frac{2}{21} \end {array} \right] .$$
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