College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 6 - Matrices and Determinants - Exercise Set 6.4 - Page 641: 90

Answer

$(A^{-1})^{-1}=A$

Work Step by Step

If $A=\left[\begin{array}{ll} a & b\\ c & d \end{array}\right]$, then$ A$ is invertible if and only if $ad-bc\neq 0$, and, $A^{-1}=\displaystyle \frac{1}{ad-bc}\left[\begin{array}{ll} d & -b\\ -c & a \end{array}\right]$. --------- $\displaystyle \frac{1}{ad-bc}=\frac{1}{3(4)-5(2)}=\frac{1}{2}$ $A^{-1}$ exists and $B=A^{-1}=\displaystyle \frac{1}{2}\left[\begin{array}{ll} 4 & -5\\ -2 & 3 \end{array}\right]=\displaystyle \left[\begin{array}{ll} 2 & -5/2\\ -1 & 3/2 \end{array}\right]$ $B$ is the inverse of A. Therefore $BA=AB=I.$ We also read this as "A is the inverse of B" or $B^{-1}=A$ or $(A^{-1})^{-1}=A$
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