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 - Review Exercises - Page 222: 45

Answer

(a) A basis for the solution space is $$ \left[\begin{aligned} \frac{2}{7} \\\frac{3}{7} \\1\\ 0 \end{aligned}\right], \left[\begin{aligned}-1\\0\\0\\ 1 \end{aligned}\right] .$$ (b) The dimension of the solution space is $2$.

Work Step by Step

The coefficient matrix is given by $$ \left[ \begin {array}{cccc} 1&-3&1&1\\ 2&1&-1&2 \\ 1&4&-2&1\\ 5&-8&2&5\end {array} \right] . $$ The reduced row echelon form is $$ \left[ \begin {array}{cccc} 1&0&-\frac{2}{7}&1\\ 0&1&-\frac{3}{7}&0 \\ 0&0&0&0\\ 0&0&0&0\end {array} \right] . $$ The corresponding system is $$ \begin{aligned} x_1 -\frac{2}{7}x_3 +x_4&=0\\ x_2-\frac{3}{7}x_3 &=0\\ \end{aligned}. $$ The solution of the above system is $x_1= \frac{2}{7}s-t$,$x_2=\frac{3}{7}s$,$x_3=s$, $x_4=t$. This means that the solution space of $Ax = 0 $ consists of the vectors on the form $$x= \left[\begin{aligned} x_1\\ x_2\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned} \frac{2}{7}s-t\\\frac{3}{7}s\\s\\ t \end{aligned}\right] = s \left[\begin{aligned} \frac{2}{7} \\\frac{3}{7} \\1\\ 0 \end{aligned}\right] +t \left[\begin{aligned}-1\\0\\0\\ 1 \end{aligned}\right] .$$ (a) A basis for the solution space is $$ \left[\begin{aligned} \frac{2}{7} \\\frac{3}{7} \\1\\ 0 \end{aligned}\right], \left[\begin{aligned}-1\\0\\0\\ 1 \end{aligned}\right] .$$ (b) The dimension of the solution space is $2$.
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