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: 44

Answer

(a) $$(AB)^{-1} =\left[ \begin {array}{ccc} -6&-25&24\\ -6&10&7 \\ 17&7&15\end {array} \right] $$ (b) $$(A^T)^{-1}=\left[ \begin {array}{ccc} 1&0&4\\ -4&1&2 \\ 2&3&1\end {array} \right] $$ (c) $$(2A)^{-1} =\left[ \begin {array}{ccc} \frac{1}{2}&-2&1\\ 0&\frac{1}{2}&\frac{3}{2} \\ 2&1&\frac{1}{2}\end {array} \right].$$

Work Step by Step

(a) $$(AB)^{-1}=B^{-1}A^{-1}=\left[ \begin {array}{ccc} 6&5&-3\\ -2&4&-1 \\ 1&3&4\end {array} \right] \left[ \begin {array}{ccc} 1&-4&2\\ 0&1&3 \\ 4&2&1\end {array} \right] =\left[ \begin {array}{ccc} -6&-25&24\\ -6&10&7 \\ 17&7&15\end {array} \right] $$ (b) $$(A^T)^{-1}=(A^{-1})^T=\left[ \begin {array}{ccc} 1&0&4\\ -4&1&2 \\ 2&3&1\end {array} \right] $$ (c) $$(2A)^{-1}=\frac{1}{2}A^{-1}=\frac{1}{2}\left[ \begin {array}{ccc} 1&-4&2\\ 0&1&3 \\ 4&2&1\end {array} \right] =\left[ \begin {array}{ccc} \frac{1}{2}&-2&1\\ 0&\frac{1}{2}&\frac{3}{2} \\ 2&1&\frac{1}{2}\end {array} \right].$$
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