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: 42


(a) $$(AB)^{-1}=\left[ \begin {array}{cc} {-\frac{4}{77}}& {\frac{9}{77}}\\ {-\frac{9}{77}} & {\frac{1}{77}} \end {array} \right] .$$ (b) $$(A^T)^{-1}= \left[ \begin {array}{cc} {-\frac{2}{7}}& {\frac{3}{7}}\\ {\frac{1}{7}} & {\frac{2}{7}} \end {array} \right].$$ (c) $$(2A)^{-1}= \left[ \begin {array}{cc} {-\frac{2}{14}}& {\frac{1}{14}}\\ {\frac{3}{14}} & {\frac{2}{14}} \end {array} \right].$$

Work Step by Step

(a) $$(AB)^{-1}=B^{-1}A^{-1}=\left[ \begin {array}{cc} {\frac{5}{11}}& {\frac{2}{11}}\\ {\frac{3}{11}} & {-\frac{1}{11}} \end {array} \right]\left[ \begin {array}{cc} {-\frac{2}{7}}& {\frac{1}{7}}\\ {\frac{3}{7}} & {\frac{2}{7}} \end {array} \right] =\left[ \begin {array}{cc} {-\frac{4}{77}}& {\frac{9}{77}}\\ {-\frac{9}{77}} & {\frac{1}{77}} \end {array} \right] .$$ (b) $$(A^T)^{-1}=(A^{-1})^T=\left[ \begin {array}{cc} {-\frac{2}{7}}& {\frac{3}{7}}\\ {\frac{1}{7}} & {\frac{2}{7}} \end {array} \right].$$ (c) $$(2A)^{-1}=\frac{1}{2}A^{-1}=\frac{1}{2}\left[ \begin {array}{cc} {-\frac{2}{7}}& {\frac{1}{7}}\\ {\frac{3}{7}} & {\frac{2}{7}} \end {array} \right]=\left[ \begin {array}{cc} {-\frac{2}{14}}& {\frac{1}{14}}\\ {\frac{3}{14}} & {\frac{2}{14}} \end {array} \right].$$
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