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: 19

Answer

(a) A basis for the column space are the columns corresponding to the columns that have the leading 1's. That is, $$S=\left\{\left[\begin{array}{ll}{2} \\ {1} \end{array}\right],\left[\begin{array}{ll} {4} \\ {6}\end{array}\right]\right\}.$$ (b) The rank of the matrix is $2$.

Work Step by Step

Given the matrix $$ \left[\begin{array}{ll}{2} & {4} \\ {1} & {6}\end{array}\right]. $$ The reduced row echelon form is given by $$ \left[ \begin {array}{cc} 1&0\\0&1\end {array} \right].$$ (a) A basis for the column space are the columns corresponding to the columns that have the leading 1's. That is, $$S=\left\{\left[\begin{array}{ll}{2} \\ {1} \end{array}\right],\left[\begin{array}{ll} {4} \\ {6}\end{array}\right]\right\}.$$ (b) The rank of the matrix is $2$.
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