Linear Algebra and Its Applications (5th Edition)

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

Chapter 1 - Linear Equations in Linear Algebra - 1.7 Exercises - Page 61: 6

Answer

Yes, the columns of this matrix form a linearly independent set of vectors.

Work Step by Step

To see if the columns are linearly independent, we reduce the matrix to row echelon form: $ \begin{bmatrix} -4&-3&0\\0&-1&4\\1&0&3\\5&4&6 \end{bmatrix}\sim \begin{bmatrix} 1&0&3\\0&-1&4\\-4&-3&0\\5&4&6 \end{bmatrix}\sim \begin{bmatrix} 1&0&3\\0&-1&4\\0&-3&12\\0&4&-9 \end{bmatrix} \sim \begin{bmatrix} 1&0&3\\0&-1&4\\0&0&7\\0&0&0 \end{bmatrix} $ Since every column of the row-equivalent echelon matrix has a pivot, there are no free variables, so the columns are linearly independent.
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