## 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 200: 38

#### Answer

(a) A basis for the solution space is \left[\begin{aligned}1\\1 \end{aligned}\right]. (b) The dimension of the solution space is $1$.

#### Work Step by Step

The coefficient matrix is given by $$\left[ \begin {array}{cc} 1&-1\\ -1&1\end {array} \right] .$$ The reduced row echelon form is $$\left[ \begin {array}{cc} 1&-1\\ 0&0\end {array} \right] .$$ The corresponding system is \begin{aligned} x_{1}-x_{2} &=0\\ \end{aligned}. The solution of the above system is $x_1=t$,$x_2=t$. 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} \end{aligned}\right]= \left[\begin{aligned}t\\t \end{aligned}\right]=t \left[\begin{aligned}1\\1 \end{aligned}\right] . (a) A basis for the solution space is \left[\begin{aligned}1\\1 \end{aligned}\right]. (b) The dimension of the solution space is $1$.

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