Answer
(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 $2$.
Work Step by Step
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 $2$.
