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: 6

Answer

(a) $$S=\left\{(1,-3,2),(0,1,-\frac{1}{2})\right\}.$$ (b) The rank of the matrix is $2$.

Work Step by Step

(a)The reduced form of the matrix is given as follows $$\left[\begin{array}{rrr}{1} & {-3} & {2} \\ {4} & {2} & {1}\end{array}\right] \rightarrow\left[\begin{array}{rrr}{1} & {-3} & {2} \\ {0} & {1} & {-\frac{1}{2}}\end{array}\right].$$ Hence, the basis for the row space is $$S=\left\{(1,-3,2),(0,1,-\frac{1}{2})\right\}.$$ (b) Since the dimension of the row space is $2$, then 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.