Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 1 - Systems of Linear Equations - 1.1 Introduction to Systems of Linear Equations - 1.1 Exercises: 29

Answer

$x_{1} = -t$ $x_{2} = 2t$ $x_{3} = t$ $t$ is any real number.

Work Step by Step

Start by writing the equations in reduced row echelon form. We can accomplish this by interchanging Equation 1 and 2 and then row reducing. $2x_{1} + x_{2} = 0$ $5x_{1} + 2x_{2} + x_{3} = 0$ Dividing Equation 1 by 2 gives... $x_{1} + \frac{1}{2}x_{2} = 0$ $5x_{1} + 2x_{2} + x_{3} = 0$ Adding -5 times Equation 1 to Equation 2 will eliminate the $x_1$ variable from Equation 2. $x_{1} + \frac{1}{2}x_{2} = 0$ $0x_{1} + -\frac{1}{2}x_{2} + x_{3} = 0$ Multiply Equation 2 by -2. $x_{1} + \frac{1}{2}x_{2} = 0$ $0x_{1} + x_{2} -2 x_{3} = 0$ Because there 2 pivots, $x_{3}$ is a free variable. Solve both equations for $x_{1}$ and $x_{2}$ respectively. $x_{1} = -\frac{1}{2}x_{2}$ $x_{2} = 2x_{3}$ Since $x_3$ is free, it can be any real number $t$. $x_3 = t$ Through back substitution, we can express $x_{1}$ and $x_{2}$ in terms of $t$. $x_{2} = 2t$ $x_{1} = -\frac{1}{2}*2t = -t$
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