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: 41

Answer

Column 4 or 3

Work Step by Step

For $\begin{bmatrix} 12&-7&11&-9&5\\ -9&4&-8&7&-3\\ -6&11&-7&3&-9\\ 4&-6&10&-5&12 \end{bmatrix}$ to span $R^4$, each column must be a pivot column. By finding the row reduced echelon form, we can see that column 4 is not a pivot column, therefore it can be removed. $\begin{bmatrix} 1&0&0&-\frac{10}{21}&0\\ 0&1&0&-\frac{25}{84}&0\\ 0&0&1&-\frac{41}{84}&0\\ 0&0&0&0&1 \end{bmatrix}$ However, it is also possible to delete column 3
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