College Algebra (6th Edition)

Published by Pearson

Chapter 6 - Matrices and Determinants - Exercise Set 6.4 - Page 639: 19

Answer

$A^{-1}=\left[\begin{array}{lll} 1/2 & 0 & 0\\ 0 & 1/4 & 0\\ 0 & 0 & 1/6 \end{array}\right]$

Work Step by Step

1. Form $[A|I]$ 2. Row operations: $\left[\begin{array}{lllllll} 2 & 0 & 0 & | & 1 & 0 & 0\\ 0 & 4 & 0 & | & 0 & 1 & 0\\ 0 & 0 & 6 & | & 0 & 0 & 1 \end{array}\right]\left\{\begin{array}{l} \div 2\\ \div 4\\ \div 6 \end{array}\right.$ $\left[\begin{array}{lllllll} 1 & 0 & 0 & | & 1/2 & 0 & 0\\ 0 & 1 & 0 & | & 0 & 1/4 & 0\\ 0 & 0 & 1 & | & 0 & 0 & 1/6 \end{array}\right]$ 3. We have obtained $[I|B], B=A^{-1}$ $A^{-1}=\left[\begin{array}{lll} 1/2 & 0 & 0\\ 0 & 1/4 & 0\\ 0 & 0 & 1/6 \end{array}\right]$ Checking: $AA^{-1}=$ $\left[\begin{array}{lll} 2(\frac{1}{2})+0+0 & 0+0+0 & 0+0+0\\ 0+0+0 & 0+4(\frac{1}{4})+0 & 0+0+0\\ 0+0+0 & 0+0+0 & 0+0+6(\frac{1}{6}) \end{array}\right]=I_{3}$ $A^{-1}A=$ $\left[\begin{array}{lll} (\frac{1}{2})(2)+0+0 & 0+0+0 & 0+0+0\\ 0+0+0 & 0+(\frac{1}{4})(4)+0 & 0+0+0\\ 0+0+0 & 0+0+0 & 0+0+(\frac{1}{6})(6) \end{array}\right]=I_3$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.