## Elementary Linear Algebra 7th Edition

(a) A basis for the column space is $$S=\left\{\left[\begin{array}{ccc}{4} \\ {6} \\ {2} \end{array}\right],\left[\begin{array}{ccc} {20} \\ {-5} \\ {-11} \end{array}\right]\right\}.$$ (b) The rank of the matrix is $2$.
Given the matrix $$\left[\begin{array}{ccc}{4} & {20} & {31} \\ {6} & {-5} & {-6} \\ {2} & {-11} & {-16}\end{array}\right].$$ The reduced row echelon form is given by $$\left[ \begin {array}{ccc} 1&0&\frac{1}{4}\\ 0&1&\frac{3}{2} \\ 0&0&0\end {array} \right] ] .$$ (a) A basis for the column space are the columns corresponding to the columns that have the leading 1's. That is, $$S=\left\{\left[\begin{array}{ccc}{4} \\ {6} \\ {2} \end{array}\right],\left[\begin{array}{ccc} {20} \\ {-5} \\ {-11} \end{array}\right]\right\}.$$ (b) The rank of the matrix is $2$.