Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 4 - Section 4.3 - Matrix Inversion - Exercises - Page 263: 10

Answer

$A^{-1}=\left[\begin{array}{rr} 1/4 & 0\\ 0 & 1/2 \end{array}\right]$

Work Step by Step

$1.\quad $Start with $[A|I]$, $2.\quad $ Row reduce, $3.\quad $ If the reduced form is $[I|B], $then $B=A^{-1}.$ If not, then A is singular (has no inverse) ---- $\left[\begin{array}{lllll} 4 & 0 & | & 1 & 0\\ 0 & 2 & | & 0 & 1 \end{array}\right]\left\{\begin{array}{l} \div 4.\\ \div 2. \end{array}\right.$ $\rightarrow\left[\begin{array}{lllll} 1 & 0 & | & 1/4 & 0\\ 0 & 1 & | & 0 & 1/2 \end{array}\right]$ $A^{-1}=\left[\begin{array}{rr} 1/4 & 0\\ 0 & 1/2 \end{array}\right]$ Check: $\left[\begin{array}{ll} 4 & 0\\ 0 & 2 \end{array}\right]\left[\begin{array}{ll} 1/4 & 0\\ 0 & 1/2 \end{array}\right]=\left[\begin{array}{ll} 1+0 & 0+0\\ 0+0 & 0+1 \end{array}\right]=I_{2}$
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