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: 14

Answer

$x_3$ and $x_4$ free $$x_1=-7x_3-9$$ $$x_2=6x_3+3x_4+2$$ $$x_4=0$$

Work Step by Step

We're given the augmented matrix: $$ \begin{bmatrix} 1 & 2 & -5 & -6 & 0 & -5 \\ 0 & 1 & -6 & -3 & 0 & 2 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix} $$ To put it into reduced row echelon form, add $-2$ times the second row to the first row: $$ \begin{bmatrix} 1 & 0 & 7 & 0 & 0 & -9 \\ 0 & 1 & -6 & -3 & 0 & 2 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix} $$ Now, we can see that $x_3$ and $x_4$ are free. From the matrix, we have: $$x_1=-7x_3-9$$ $$x_2=6x_3+3x_4+2$$ $$x_4=0$$
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