Answer
(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 $5$.
Work Step by Step
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 $5$.
