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 62: 25


$ \begin{bmatrix} \blacksquare&*\\ 0&\blacksquare\\ 0&0\\ 0&0 \end{bmatrix} $ $ \begin{bmatrix} 0&\blacksquare\\ 0&0\\ 0&0\\ 0&0 \end{bmatrix} $

Work Step by Step

Since the second column is not a multiple of the first column, it is possible that the columns are linearly independent, giving rise to a pivot position in each. Hence the first possibility. However, the information given does not exclude the possibility that the first column is the zero vector and that the second column is some nonzero vector. In this case, the second column could not be a multiple of the first (since $k\cdot\vec{0}=\vec{0}$ for all scalars $k$), but the columns would still be linearly dependent. Hence the second possibility.
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