## Elementary Linear Algebra 7th Edition

Published by Cengage Learning

# Chapter 4 - Vector Spaces - Review Exercises - Page 222: 44

#### Answer

(a) A basis for the solution space is \left[\begin{aligned} 3\\-\frac{1}{2}\\-4\\ 1 \end{aligned}\right] . (b) The dimension of the solution space is $1$.

#### Work Step by Step

The coefficient matrix is given by $$\left[ \begin {array}{cccc} 3&8&2&3\\ 4&6&2&-1 \\ 3&4&1&-3\end {array} \right] .$$ The reduced row echelon form is $$\left[ \begin {array}{cccc} 1&0&0&-3\\ 0&1&0&\frac{1}{2} \\ 0&0&1&4\end {array} \right] .$$ The corresponding system is \begin{aligned} x_1 - 3x_4 &=0\\ x_2+\frac{1}{2}x_4 &=0\\ x_3+4x_4 &=0\\ \end{aligned}. The solution of the above system is $x_1= 3t$,$x_2=-\frac{1}{2}t$,$x_3=-4t$, $x_4=t$. This means that the solution space of $Ax = 0$ consists of the vectors on the form x= \left[\begin{aligned} x_1\\ x_2\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned} 3t\\-\frac{1}{2}t\\-4t\\ t \end{aligned}\right] = t\left[\begin{aligned} 3\\-\frac{1}{2}\\-4\\ 1 \end{aligned}\right] . (a) A basis for the solution space is \left[\begin{aligned} 3\\-\frac{1}{2}\\-4\\ 1 \end{aligned}\right] . (b) The dimension of the solution space is $1$.

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