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 - 2.4 Elementary Matrices - 2.4 Exercises - Page 82: 21


$$A^{-1}= \left[ \begin {array}{cccccc} \frac{1}{k}&0&0\\ 0&1&0\\ 0&0&1\end {array} \right] .$$

Work Step by Step

Let $A$ be a matrix given by $$A=\left[ \begin {array}{cccccc} k&0&0 \\ 0&1&0 \\ 0&0&1 \end {array} \right]$$ To find $A^{-1}$, we have $$\left[ A \ \ I \right]=\left[ \begin {array}{cccccc} k&0&0&1&0&0\\ 0&1&0&0 &1&0\\ 0&0&1&0&0&1\end {array} \right] . $$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[I \ \ A^{-1} \right]= \left[ \begin {array}{cccccc} 1&0&0&\frac{1}{k}&0&0\\ 0 &1&0&0&1&0\\ 0&0&1&0&0&1\end {array} \right] . $$ Then $A^{-1}$ is given by $$A^{-1}= \left[ \begin {array}{cccccc} \frac{1}{k}&0&0\\ 0&1&0\\ 0&0&1\end {array} \right] .$$
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