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: 39

Answer

First method: $$A^{-2}=(A^2)^{-1}=\left[ \begin {array}{ccc} \frac{1}{4}&0&0\\ 0&1&0 \\ 0&0&\frac{1}{9}\end {array} \right] . $$ Second method; $$A^{-2}=(A^{-1})^{2}=\left[ \begin {array}{ccc} \frac{1}{4}&0&0\\ 0&1&0 \\ 0&0&\frac{1}{9}\end {array} \right] .$$ One can see that the two methods are equivalent.

Work Step by Step

First method; we find $A^2$ which is given by $$A^{2}=\left[ \begin {array}{ccc} 4&0&0\\ 0&1&0 \\ 0&0&9\end {array} \right] .$$ By calculating the inverse of $A^2$, we have $$A^{-2}=(A^2)^{-1}=\left[ \begin {array}{ccc} \frac{1}{4}&0&0\\ 0&1&0 \\ 0&0&\frac{1}{9}\end {array} \right] . $$ Second method; we find $A^{-1}$ as follows $$A^{-1}= \left[ \begin {array}{ccc} -\frac{1}{2}&0&0\\ 0&1&0 \\ 0&0&\frac{1}{3}\end {array} \right] . $$ Now, we have $$A^{-2}=(A^{-1})^{2}=\left[ \begin {array}{ccc} \frac{1}{4}&0&0\\ 0&1&0 \\ 0&0&\frac{1}{9}\end {array} \right] .$$ One can see that the two methods are equivalent.
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