Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 4 - Vector Spaces - 4.6 Rank of a Matrix and Systems of Linear Equations - 4.6 Exercises - Page 199: 29

Answer

a basis for the nullspace of $A$ consists of the vectors $$ \left[\begin{aligned} -3\\ 0\\1 \end{aligned}\right] .$$

Work Step by Step

Given the matrix $$ A=\left[\begin{array}{lll}{1} & {2} & {3} \\ {0} & {1} & {0}\end{array}\right]. $$ The reduced row echelon form is given by $$ \left[ \begin {array}{ccc} 1&0&3\\ 0&1&0 \end {array} \right] . $$ The corresponding system is $$ \begin{aligned} x_{1} +3x_3&=0 \\ x_2&=0.\\ \end{aligned}. $$ The solution of the above system is $x_1=-3t$, $x_2=0$ and $x_3=t$. This means that the solution space of $Ax = 0 $ consists of the vectors on the following form $$x= \left[\begin{aligned} x_{1}\\ x_{2} \\x_3 \end{aligned}\right]= \left[\begin{aligned} -3t\\ 0\\t \end{aligned}\right]=t\left[\begin{aligned} -3\\ 0\\1 \end{aligned}\right] .$$ So, a basis for the nullspace of $A$ consists of the vectors $$ \left[\begin{aligned} -3\\ 0\\1 \end{aligned}\right] .$$
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