Linear Algebra and Its Applications, 4th Edition

Published by Brooks Cole
ISBN 10: 0030105676
ISBN 13: 978-0-03010-567-8

Chapter 1 - Section 1.2 - The Geometry of Linear Equations - Problem Set: 8

Answer

The right hand side of the third equation must be -1 instead of 0 for the system to have solution. In that case any point on the plane $u+v+w = 2$ is a solution. For example, $u = 0$, $v = 0$, $w = 2$ is point on that plane and it would satisfy the modified set of equations.

Work Step by Step

Observe that if you do, Equation 1 - Equation 2 + Equation 3 , we would get $0 = 1$! This implies that the given system cannot have a solution. Now if the constant on the Right hand side of the last equation were -1, then performing the same operation as above on the three equations would result in $0 = 0$ Now, for the modified system if you do Equation 2 - Equation 3 one would get back Equation 1. This implies that the entire plane represented by Equation 1 is the solution set.
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