## Elementary Linear Algebra 7th Edition

(a) A basis for the column space is $$S=\left\{\left[\begin{array}{rrrrr}{2} \\ {2} \\ {4} \\ {2} \\ {0} \end{array}\right],\left[\begin{array}{rrrrr} {4} \\ {5} \\ {3} \\ {-4} \\ {1} \end{array}\right],\left[\begin{array}{rrrrr} {-2} \\ {4} \\ {1} \\ {2} \\ {4} \end{array}\right],\left[\begin{array}{rrrrr} {1} \\ {-2} \\ {1} \\ {-1} \\ {2} \end{array}\right],\left[\begin{array}{rrrrr} {1} \\ {2} \\ {2} \\ {1} \\ {-1}\end{array}\right]\right\}.$$ (b) The rank of the matrix is $5$.
Given the matrix $$\left[\begin{array}{rrrrr}{2} & {4} & {-2} & {1} & {1} \\ {2} & {5} & {4} & {-2} & {2} \\ {4} & {3} & {1} & {1} & {2} \\ {2} & {-4} & {2} & {-1} & {1} \\ {0} & {1} & {4} & {2} & {-1}\end{array}\right].$$ The reduced row echelon form is given by $$\left[ \begin {array}{ccccc} 1&0&0&0&0\\ 0&1&0&0&0 \\ 0&0&1&0&0\\ 0&0&0&1&0 \\ 0&0&0&0&1\end {array} \right] .$$ (a) A basis for the column space are the columns corresponding to the columns that have the leading 1's. That is, $$S=\left\{\left[\begin{array}{rrrrr}{2} \\ {2} \\ {4} \\ {2} \\ {0} \end{array}\right],\left[\begin{array}{rrrrr} {4} \\ {5} \\ {3} \\ {-4} \\ {1} \end{array}\right],\left[\begin{array}{rrrrr} {-2} \\ {4} \\ {1} \\ {2} \\ {4} \end{array}\right],\left[\begin{array}{rrrrr} {1} \\ {-2} \\ {1} \\ {-1} \\ {2} \end{array}\right],\left[\begin{array}{rrrrr} {1} \\ {2} \\ {2} \\ {1} \\ {-1}\end{array}\right]\right\}.$$ (b) The rank of the matrix is $5$.