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: 7

Answer

The final solution to the augmented matrix is: 1 3 0 -5 0 0 1 3 In parametric form, the solution is: $x_{1}$ = -5 - $3x_{2}$ $x_{2}$ is free $x_{3}$ = 3

Work Step by Step

We begin with the original augmented matrix: 1 3 4 7 3 9 7 6 We replace the second row by the second row plus the first row multiplied by -3. 1 3 4 7 0 0 -5 -15 We divide the second row by -5. 1 3 4 7 0 0 1 3 We then eliminate all non-zero entries except for the leading entries from every column. We replace the first row by the first row plus the second row multiplied by -4. This gives us the solution of the matrix. 1 3 0 -5 0 0 1 3
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