## Elementary Linear Algebra 7th Edition

Published by Cengage Learning

# Chapter 4 - Vector Spaces - 4.6 Rank of a Matrix and Systems of Linear Equations - 4.6 Exercises - Page 199: 30

#### Answer

a basis for the nullspace of $A$ consists of the vectors \left[\begin{aligned} -4\\ 1\\0 \end{aligned}\right] .

#### Work Step by Step

Given the matrix $$A=\left[\begin{array}{lll}{1} & {4} & {2} \\ {0} & {0} & {1}\end{array}\right].$$ The reduced row echelon form is given by $$\left[ \begin {array}{ccc} 1&4&0\\ 0&0&1 \end {array} \right] .$$ The corresponding system is \begin{aligned} x_{1} +4x_2&=0\\ x_3&=0.\\ \end{aligned}. The solution of the above system is $x_1=-4t$, $x_2=t$ and $x_3=0$. 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} -4t\\ t\\0 \end{aligned}\right]=t\left[\begin{aligned} -4\\ 1\\0 \end{aligned}\right] . So, a basis for the nullspace of $A$ consists of the vectors \left[\begin{aligned} -4\\ 1\\0 \end{aligned}\right] .

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