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 71: 20


$A$ has no inverse, or is non-invertible (or singular).

Work Step by Step

To find $A^{-1}$, we have $$\left[ A \ \ I \right]= \left[ \begin {array}{cccccc} -\frac{5}{6}&\frac{1}{3}&{\frac {11}{6}}&1&0&0 \\ 0&\frac{2}{3}&2&0&1&0\\ 1&-\frac{1}{2}&-\frac{5}{2}&0&0 &1\end {array} \right] . $$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$ \left[ \begin {array}{cccccc} 1&0&-1&0&\frac{3}{4}&1\\ 0&1& 3&0&\frac{3}{2}&0\\ 0&0&0&1&\frac{1}{8}&\frac{5}{6}\end {array} \right] . $$ Note that the portion of the matrix $A$ has a row of zeros. So it is not possible to rewrite the matrix $\left[ A \ \ I \right] $ in the form $\left[I \ \ A^{-1} \right]$. This means that has no inverse or is non-invertible (or singular).
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