Elementary Linear Algebra 7th Edition

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

Chapter 4 - Vector Spaces - 4.6 Rank of a Matrix and Systems of Linear Equations - 4.6 Exercises - Page 199: 21

Answer

(a) A basis for the column space is $$S=\left\{\left[\begin{array}{rrr}{1} \\ {-1} \end{array}\right],\left[\begin{array}{rrr} {2} \\ {2} \end{array}\right]\right\}.$$ (b) The rank of the matrix is $2$.

Work Step by Step

Given the matrix $$ \left[\begin{array}{rrr}{1} & {2} & {4} \\ {-1} & {2} & {1}\end{array}\right] $$ The reduced row echelon form is given by $$ \left[ \begin {array}{ccc} 1&0&0\\ 0&1&1/2 \end {array} \right] . $$ (a) A basis for the column space are the columns corresponding to the columns that have the leading 1's. That is, $$S=\left\{\left[\begin{array}{rrr}{1} \\ {-1} \end{array}\right],\left[\begin{array}{rrr} {2} \\ {2} \end{array}\right]\right\}.$$ (b) The rank of the matrix is $2$.
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.