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 - 4.6 Rank of a Matrix and Systems of Linear Equations - 4.6 Exercises - Page 200: 43

Answer

(a) A basis for the solution space is $$ \left[\begin{aligned}-4 \\-1 \\1\\0 \end{aligned}\right],\left[\begin{aligned} -3 \\-\frac{2}{3} \\0\\1 \end{aligned}\right] .$$ (b) The dimension of the solution space is $2$.

Work Step by Step

The coefficient matrix is given by $$ \left[ \begin {array}{cccc} 3&3&15&11\\ 1&-3&1&1 \\ 2&3&11&8\end {array} \right] . $$ The reduced row echelon form is $$ \left[ \begin {array}{cccc} 1&0&4&3\\ 0&1&1&\frac{2}{3} \\ 0&0&0&0\end {array} \right] . $$ The corresponding system is $$ \begin{aligned} x_1 +4x_3+3x_4&=0\\ x_2+ x_3+\frac{2}{3}x_4&=0\\ \end{aligned}. $$ The solution of the above system is $x_1=-4s-3t$,$x_2=- s-\frac{2}{3}t$,$x_3=s$, $x_4=t$. This means that the solution space of $Ax = 0 $ consists of the vectors on the form $$x= \left[\begin{aligned} x_1\\ x_2\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned}-4s-3t\\-s-\frac{2}{3}t \\s\\t \end{aligned}\right] =s\left[\begin{aligned}-4 \\-1 \\1\\0 \end{aligned}\right]+t\left[\begin{aligned} -3 \\-\frac{2}{3} \\0\\1 \end{aligned}\right] .$$ (a) A basis for the solution space is $$ \left[\begin{aligned}-4 \\-1 \\1\\0 \end{aligned}\right],\left[\begin{aligned} -3 \\-\frac{2}{3} \\0\\1 \end{aligned}\right] .$$ (b) The dimension of the solution space is $2$.
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