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 72: 58

Answer

$A=(1/256)*\left[\begin{array}{cc} 8&-16\\ 12&8 \end{array} \right]$

Work Step by Step

Given $(4A)^{-1}=\left[\begin{array}{cc} 2&4\\ -3&2 \end{array} \right]$ thus $ (1/4)A^{-1}=\left[\begin{array}{cc} 2&4\\ -3&2 \end{array} \right]$ $A^{-1}=\left[\begin{array}{cc} 8&16\\ -12&8 \end{array} \right]$ Since $A=(A^{-1})^{-1}=(1/|A^{-1}|)*\left[\begin{array}{cc} 8&-16\\ 12&8 \end{array} \right]=(1/256)*\left[\begin{array}{cc} 8&-16\\ 12&8 \end{array} \right]$
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