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: 42

Answer

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

Work Step by Step

The coefficient matrix is given by $$ \left[ \begin {array}{ccc} 1&2&-4\\ -3&-6&-12 \end {array} \right] . $$ The reduced row echelon form is $$ \left[ \begin {array}{ccc} 1&2&0\\ 0&0&1 \end {array} \right] . $$ The corresponding system is $$ \begin{aligned} x +2y&=0\\ z&=0\\ \end{aligned}. $$ The solution of the above system is $x=-2t$,$y=t$, $z=0$. This means that the solution space of $Ax = 0 $ consists of the vectors on the form $$x= \left[\begin{aligned} x\\ y\\z \end{aligned}\right]= \left[\begin{aligned}-2t\\t \\0 \end{aligned}\right] =t\left[\begin{aligned}-2\\1 \\0 \end{aligned}\right] .$$ (a) A basis for the solution space is $$ \left[\begin{aligned}-2\\1 \\0 \end{aligned}\right] .$$ (b) The dimension of the solution space is $1$.
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