## 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: 25

#### Answer

a basis for the nullspace of $A$ consists of the vector \left[\begin{aligned} 1\\ 2 \end{aligned}\right].

#### Work Step by Step

Given the matrix $$A=\left[ \begin{array}{rr}{2} & {-1} \\ {-6} & {3}\end{array} \right]$$ The reduced row echelon form is given by $$\left[ \begin {array}{cc} 1&-\frac{1}{2}\\ 0&0\end {array} \right] .$$ The corresponding system is \begin{aligned} 2x_{1}- x_{2} &=0 \end{aligned}. The solution of the above system is $x_1=t$ and $x_2=2t$. This means that the solution space $Ax = 0$ of consists of all solution vectors of the following form x=\left[\begin{aligned} x_{1}\\ x_{2} \end{aligned}\right]=\left[\begin{aligned} t\\ 2t \end{aligned}\right]=t\left[\begin{aligned} 1\\ 2 \end{aligned}\right]. So, a basis for the nullspace of $A$ consists of the vector \left[\begin{aligned} 1\\ 2 \end{aligned}\right].

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