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: 39

Answer

$\sim \left[\begin{array}{ r r r r r }10&\ -7&1&\ \ \ \ 4&6\\0&\ -8/5&\ \ \ -26/5&-34/5&9/5\\0&\ \ \ \ \ 0&-193/8&-193/8&49/16\\0&\ \ \ \ \ 0&0&\ \ \ \ \ 0&4715/386\end{array}\right]$ The original matrix has a pivot in every row, so its columns span $R^4$, by Theorem 4.

Work Step by Step

$\left[\begin{array}{ r r r r r }10&-7&1&4&6\\-8&4&-6&-10&-3\\-7&11&-5&-1&-8\\3&-1&10&12&12\end{array}\right]\sim \left[\begin{array}{ r r r r r }10&-7&1&4&6\\0&-8/5&-26/5&-34/5&9/5\\0&61/10&-43/10&9/5&-19/5\\0&11/10&97/10&54/5&51/5\end{array}\right]$ $\sim \left[\begin{array}{ r r r r r }10&-7&1&4&6\\0&-8/5&-26/5&-34/5&9/5\\0&0&-193/8&-193/8&49/16\\0&0&49/8&49/8&183/16\end{array}\right]\sim \left[\begin{array}{ r r r r r }10&\ -7&1&\ \ \ \ 4&6\\0&\ -8/5&\ \ \ -26/5&-34/5&9/5\\0&\ \ \ \ \ 0&-193/8&-193/8&49/16\\0&\ \ \ \ \ 0&0&\ \ \ \ \ 0&4715/386\end{array}\right]$ The original matrix has a pivot in every row, so its columns span $R^4$, by Theorem 4.
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.