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 - 1.1 Introduction to Systems of Linear Equations - 1.1 Exercises - Page 12: 81


This system won't have any solutions for $$k=\frac{8}{3}.$$

Work Step by Step

We need to make these two equations inconsistent by choosing the right $k$. To do this follow the steps below: Step 1: Multiply the first equation by $3$: \begin{align*} 3x+6y+3kz = &18\\ 3x+6y+8z =& 4. \end{align*} If we choose $3k = 8$ i.e. $k=8/3$ then the left sides of both equations will be equal which would mean that their right sides must be equal. Since obviously $18\neq4$ this is impossible so for this particular value of $k$ the system is inconsistent i.e. it has no solutions.
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.