Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.4 Exercises - Page 42: 42

Answer

See solution

Work Step by Step

Row reducing, we get $\begin{bmatrix} 11&7&-7&-9&-6\\ 0&65&-10&-5&191\\ 0&0&0&1&-2\\ 0&0&0&0&1 \end{bmatrix}$. From here, we can see that we can either remove column 2 or 3 and get a 4 by 4 matrix that has a pivot in each row. There cannot be a pivot in each row if there are only 3 columns, so we can't delete more than 1 column.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.