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: 3

Answer

Nul A= Span $\{ \left[\begin{array}{l} 7\\ -4\\ 1\\ 0 \end{array}\right],\ \left[\begin{array}{l} -6\\ 2\\ 0\\ 1 \end{array}\right] \}$

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 $0]$ = $\left[\begin{array}{lllll} 1 & 3 & 5 & 0 & 0\\ 0 & 1 & 4 & -2 & 0 \end{array}\right] \left(\begin{array}{l} R_{1}=R_{1}-3R_{2}\\ . \end{array}\right)$ $\sim\left[\begin{array}{lllll} 1 & 0 & -7 & 6 & 0\\ 0 & 1 & 4 & -2 & 0 \end{array}\right]$ With $x_{3}$ and $x_{4}$ free parameters (any real numbers), $x_{1}=7x_{3}-6x_{4}$ $x_{2}=-4x_{2}+2x_{4}$ $x=\left[\begin{array}{l} 7x_{3}-6x_{4}\\ -4x_{3}+2x_{4}\\ x_{3}\\ x_{4} \end{array}\right]=x_{3}\left[\begin{array}{l} 7\\ -4\\ 1\\ 0 \end{array}\right]+x_{4}\left[\begin{array}{l} -6\\ 2\\ 0\\ 1 \end{array}\right]$ Nul A= Span $\{ \left[\begin{array}{l} 7\\ -4\\ 1\\ 0 \end{array}\right],\ \left[\begin{array}{l} -6\\ 2\\ 0\\ 1 \end{array}\right] \}$ (the answer in the back of the book is different, but this is correct)
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