Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 0-32198-238-X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.2 Exercises: 11

Answer

The general solution of the matrix is: 3 -2 4 0 0 0 0 0 0 0 0 0 In parametric form, this is expressed as: $x_{1}$ = $\frac{2}{3}x_{2}$ - $\frac{4}{3}x_{3}$ $x_{2}$ is free $x_{3}$ is free

Work Step by Step

We begin with the original matrix. 3 -2 4 0 9 -6 12 0 6 -4 8 0 We divide the second row by 3, and the third row by 2. 3 -2 4 0 3 -2 4 0 3 -2 4 0 Rows two and three can be reduced to rows of zeroes, meaning that they are redundant and that the system has no unique solution and two free variables. 3 -2 4 0 0 0 0 0 0 0 0 0
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