## Elementary Linear Algebra 7th Edition

Published by Cengage Learning

# Chapter 4 - Vector Spaces - 4.6 Rank of a Matrix and Systems of Linear Equations - 4.6 Exercises - Page 199: 23

#### Answer

(a) A basis for the column is $$S=\left\{\left[\begin{array}{rrrr}{2} \\ {7} \\ {-2} \\ {2} \end{array}\right],\left[\begin{array}{rrrr} {-3} \\ {-6} \\ {1} \\ {-2} \end{array}\right]\right\}.$$ (b) The rank of the matrix is $2$.

#### Work Step by Step

Given the matrix $$\left[\begin{array}{rrrr}{2} & {4} & {-3} & {-6} \\ {7} & {14} & {-6} & {-3} \\ {-2} & {-4} & {1} & {-2} \\ {2} & {4} & {-2} & {-2}\end{array}\right].$$ The reduced row echelon form is given by $$\left[ \begin {array}{cccc} 1&2&0&3\\ 0&0&1&4 \\ 0&0&0&0\\ 0&0&0&0\end {array} \right] .$$ (a) A basis for the column space are the columns corresponding to the columns that have the leading 1's. That is, $$S=\left\{\left[\begin{array}{rrrr}{2} \\ {7} \\ {-2} \\ {2} \end{array}\right],\left[\begin{array}{rrrr} {-3} \\ {-6} \\ {1} \\ {-2} \end{array}\right]\right\}.$$ (b) The rank of the matrix is $2$.

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