Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 1 - Systems of Linear Equations - Review Exercises: 30

Answer

$x=5$ $y=2$ $z=-6$

Work Step by Step

Write the augmented matrix of the system of linear equations. $ \begin{bmatrix} 2 & 3 & 1 & 10\\ 2 &-3 & -3 & 22\\ 4 & -2 & 3 & -2 \end{bmatrix} $ Divide the first row by 2. $ \begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 5\\ 2 &-3 & -3 & 22\\ 4 & -2 & 3 & -2 \end{bmatrix} $ Add -2 times the 1st row to the 2nd row to produce a new 2nd row. Add -4 times the 1st row to the 3rd row to produce a new 3rd row. $ \begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 5\\ 0 &-6 & -4 & 12\\ 0 & -8 & 1 & -22 \end{bmatrix} $ Divide the second row by -6. $ \begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 5\\ 0 &1 & \frac{2}{3} & -2\\ 0 & -8 & 1 & -22 \end{bmatrix} $ Add 8 times the 2nd row to the 3rd row to produce a new 3rd row. $ \begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 5\\ 0 &1 & \frac{2}{3} & -2\\ 0 & 0 & \frac{19}{3} & -38 \end{bmatrix} $ Multiply the third row by $\frac{3}{19}$ $ \begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 5\\ 0 &1 & \frac{2}{3} & -2\\ 0 & 0 & 1 & -6 \end{bmatrix} $ Use back-substitution to find the solution. $z=-6$ $y=2$ $x=5$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.