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: 43


(a) $$(AB)^{-1}= =\left[ \begin {array}{ccc} {\frac {69}{8}}&\frac{7}{2}&-{\frac {21}{4}} \\ {\frac {37}{16}}&{\frac {13}{8}}&-{\frac {71}{16} }\\ \frac{3}{2}&{\frac {17}{8}}&\frac{3}{16}\end {array} \right] $$ (b) $$(A^T)^{-1}=\left[ \begin {array}{ccc} 1& \frac{3}{2}&\frac{1}{4}\\ -\frac{1}{2}&\frac{1}{2}&1 \\ \frac{3}{4}&-2&\frac{1}{2}\end {array} \right]$$ (c) $$(2A)^{-1}=\left[ \begin {array}{ccc} \frac{1}{2}&-\frac{1}{4}&\frac{3}{8}\\ \frac{3}{4}&\frac{1}{4}&-1 \\ \frac{1}{8}&\frac{1}{2}&\frac{1}{4}\end {array} \right]$$

Work Step by Step

(a) $$(AB)^{-1}=B^{-1}A^{-1}=\left[ \begin {array}{ccc} 2&4&\frac{5}{2}\\ -\frac{3}{4}&2&\frac{1}{4} \\ \frac{1}{4}&\frac{1}{2}&2\end {array} \right] \left[ \begin {array}{ccc} 1&-\frac{1}{2}&\frac{3}{4}\\ \frac{3}{2}&\frac{1}{2}&-2 \\ \frac{1}{4}&1&\frac{1}{2}\end {array} \right] =\left[ \begin {array}{ccc} {\frac {69}{8}}&\frac{7}{2}&-{\frac {21}{4}} \\ {\frac {37}{16}}&{\frac {13}{8}}&-{\frac {71}{16} }\\ \frac{3}{2}&{\frac {17}{8}}&\frac{3}{16}\end {array} \right] $$ (b) $$(A^T)^{-1}=(A^{-1})^T=\left[ \begin {array}{ccc} 1& \frac{3}{2}&\frac{1}{4}\\ -\frac{1}{2}&\frac{1}{2}&1 \\ \frac{3}{4}&-2&\frac{1}{2}\end {array} \right]$$ (c) $$(2A)^{-1}=\frac{1}{2}A^{-1}=\frac{1}{2}\left[ \begin {array}{ccc} 1&-\frac{1}{2}&\frac{3}{4}\\ \frac{3}{2}&\frac{1}{2}&-2 \\ \frac{1}{4}&1&\frac{1}{2}\end {array} \right]=\left[ \begin {array}{ccc} \frac{1}{2}&-\frac{1}{4}&\frac{3}{8}\\ \frac{3}{4}&\frac{1}{4}&-1 \\ \frac{1}{8}&\frac{1}{2}&\frac{1}{4}\end {array} \right]$$
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