Linear Algebra and Its Applications (5th Edition)

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

Chapter 4 - Vector Spaces - 4.2 Exercises - Page 208: 5

Answer

Nul A= Span $\{ \left[\begin{array}{l} 2\\ 1\\ 0\\ 0\\ 0 \end{array}\right],\ \left[\begin{array}{l} -4\\ 0\\ 9\\ 1\\ 0 \end{array}\right] \}$

Work Step by Step

By definition, Nul A =$\{$x$: $ x$\in \mathbb{R}^{n}$ and Ax=0 $\}$. --------- We find the general solution to Ax=0: $[$A $0]$ = $\left[\begin{array}{llllll} 1 & -2 & 0 & 4 & 0 & 0\\ 0 & 0 & 1 & -9 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right]$ With $x_{2}$ and $x_{4}$ free parameters (any real numbers), $x_{1}=2x_{2}-4x_{3}$ $x_{3}=9x_{4}$ $x_{5}=0$ $x=\left[\begin{array}{l} 2x_{2}-4x_{4}\\ x_{2}\\ 9x_{4}\\ x_{4}\\ 0 \end{array}\right]=x_{2}\left[\begin{array}{l} 2\\ 1\\ 0\\ 0\\ 0 \end{array}\right]+x_{4}\left[\begin{array}{l} -4\\ 0\\ 9\\ 1\\ 0 \end{array}\right]$ Nul A= Span $\{ \left[\begin{array}{l} 2\\ 1\\ 0\\ 0\\ 0 \end{array}\right],\ \left[\begin{array}{l} -4\\ 0\\ 9\\ 1\\ 0 \end{array}\right] \}$
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