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 41: 12

Answer

Yes because $Ax=u$ is consistent.

Work Step by Step

If $u$ is in the plane spanned by the columns of $A$, then there exists an $x$ such that $Ax=u$. This can be determined by solving the equation through row reduction. This is the augmented matrix of the system: $$ \begin{bmatrix} 3&-5&0\\ -2&6&4\\ 1&1&4 \end{bmatrix} $$ For convenience, switch the third and first rows: $$ \begin{bmatrix} 1&1&4\\ 3&-5&0\\ -2&6&4\\ \end{bmatrix} $$ Now add multiples of the first row to the second and third rows to make a pivot entry for row 1: $$ \begin{bmatrix} 1&1&4\\ 0&-8&-12\\ 0&8&12\\ \end{bmatrix} $$ Add the second row to the third: $$ \begin{bmatrix} 1&1&4\\ 0&-8&-12\\ 0&0&0\\ \end{bmatrix} $$ And divide the second row by 4: $$ \begin{bmatrix} 1&1&4\\ 0&1&1.5\\ 0&0&0\\ \end{bmatrix} $$ The matrix is in row echelon form and there is no row that reduces to $0=1$. Thus, the equation is consistent, and $u$ is in the span of the columns of $A$.
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