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 639: 15

Answer

$A^{-1}=\left[\begin{array}{ll} 1 & 1/2\\ 2 & 3/2 \end{array}\right]$

Work Step by Step

$ad-bc=6-4= 2$ $A^{-1}=\displaystyle \frac{1}{2}\left[\begin{array}{ll} 2 & 1\\ 4 & 3 \end{array}\right]=\displaystyle \left[\begin{array}{ll} 1 & 1/2\\ 2 & 3/2 \end{array}\right]$ Checking, $AA^{-1}=\left[\begin{array}{ll} 3(1)-1(2) & 3(\frac{1}{2})-1(\frac{3}{2})\\ -4(1)+2(2) & -4(\frac{1}{2})+2(\frac{3}{2}) \end{array}\right]=\left[\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right]=I_{2}$ $A^{-1}A=\left[\begin{array}{ll} 1(3)+\frac{1}{2}(-4) & 1(-1)+\frac{1}{2}(2)\\ 2(3)+\frac{3}{2}(-4) & 2(-1)+\frac{3}{2}(2) \end{array}\right]=\left[\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right]=I_{2}$
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