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