Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.4 Exercises - Page 42: 30

Answer

~$\begin{bmatrix} 2 & 4 & 6\\ 5 & 1 & 7\\ 6 & 3 & 10 \end{bmatrix} $

Work Step by Step

~$\begin{bmatrix} 2 & 4 & 6\\ 5 & 1 & 7\\ 6 & 3 & 10 \end{bmatrix} $ Divide elements of row 1 by 2 ~$\begin{bmatrix} 1 & 2 & 3\\ 5 & 1 & 7\\ 6 & 3 & 10 \end{bmatrix} $ Add row 2 to row 1 ~$\begin{bmatrix} 6 & 3 & 10\\ 5 & 1 & 7\\ 6 & 3 & 10 \end{bmatrix} $ Subtract row 1 from row 3 ~$\begin{bmatrix} 6 & 3 & 10\\ 5 & 1 & 7\\ 0 & 0 & 0 \end{bmatrix} $ Multiply elements of row 2 by 6/5 ~$\begin{bmatrix} 6 & 3 & 10\\ 6 & 6/5 & 42/5\\ 0 & 0 & 0 \end{bmatrix} $ Multiply elements of row 2 by 6/5 ~$\begin{bmatrix} 6 & 3 & 10\\ 6 & 6/5 & 42/5\\ 0 & 0 & 0 \end{bmatrix} $ Subtract row 1 from row 2 ~$\begin{bmatrix} 6 & 3 & 10\\ 0 & 9/5 & 8/5\\ 0 & 0 & 0 \end{bmatrix} $ Subtract row 1 from row 2 Because all elements of the last row is 0, it columns don't span $R^3$
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.