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: 13

Answer

$x_3$ and $x_5$ are free $$x_1=3x_5+5$$ $$x_2=4x_5+1$$ $$x_4=-9x_5+4$$

Work Step by Step

We're given the augmented matrix $$ \begin{bmatrix} 1 & -3 & 0 & -1 & 0 & -2 \\ 0 & 1 & 0 & 0 & -4 & 1 \\ 0 & 0 & 0 & 1 & 9 & 4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$ First, add $3$ times the second row to the first $$ \begin{bmatrix} 1 & 0 & 0 & -1 & -12 & 1 \\ 0 & 1 & 0 & 0 & -4 & 1 \\ 0 & 0 & 0 & 1 & 9 & 4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$ And add the third row to the first. $$ \begin{bmatrix} 1 & 0 & 0 & 0 & -3 & 5 \\ 0 & 1 & 0 & 0 & -4 & 1 \\ 0 & 0 & 0 & 1 & 9 & 4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$ The matrix is now in reduced row echelon form, so we can see the solution to the system: $x_3$ and $x_5$ are free because their corresponding columns have no pivot entries. From the first three rows, we can clearly see that $$x_1=3x_5+5$$ $$x_2=4x_5+1$$ $$x_4=-9x_5+4$$
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