finding bases for row space and null space of matrix. My problem is:
For the matrix 
$$A = \begin{bmatrix}
      1&  4&  5&  6&  9\\
      3& −2&  1&  4& −1\\
     −1&  0& −1& −2& −1\\
      2&  3&  5&  7&  8\end{bmatrix}$$
(a) Find a basis for the row space of A.
(b) Find a basis for the null space of A.
(c) Find the rank and nullity of A.
I tried searching online and I became more confused, take the example here. 
http://www2.kenyon.edu/Depts/Math/Paquin/PracticeExam1Solns.pdf
As you can see for the column space he takes the columns of the original matrix instead of the rref of A, which I don't understand. 

The row space is the span of the rows of $A$. (or the column space of $A^{T}$). 
$$
A^{T} = \begin{bmatrix} 1 & 3 & -1 & 2\\
4 & -2 & 0 & 3 \\
5 & 1 & -1 & 5 \\
6 & 4 & -2 & 7\\
9 & -1 & -1 & 8 \end{bmatrix} \vec{x} = \begin{bmatrix} 1\\ 4 \\ 5 \\ 6 \\ 9\end{bmatrix}x_1 + \begin{bmatrix} 3\\ -2 \\ 1 \\ 4 \\ -1\end{bmatrix}x_2  + \begin{bmatrix} -1\\ 0 \\ -1 \\ -2 \\ -1\end{bmatrix}x_3 + \begin{bmatrix} 2\\ 3 \\ 5 \\ 7 \\ 8\end{bmatrix}x_4
$$
So find a maximally linear independent subset $R$ of:
$$
\left\{\begin{bmatrix} 1\\ 4 \\ 5 \\ 6 \\ 9\end{bmatrix} , \begin{bmatrix} 3\\ -2 \\ 1 \\ 4 \\ -1\end{bmatrix}  , \begin{bmatrix} -1\\ 0 \\ -1 \\ -2 \\ -1\end{bmatrix} , \begin{bmatrix} 2\\ 3 \\ 5 \\ 7 \\ 8\end{bmatrix} \right\}
$$
Now the null space is the set: $$\{\vec{x}: A\vec{x} = \vec{0} \}$$ 
$$
A = \begin{bmatrix}
      1&  4&  5&  6&  9\\
      3& −2&  1&  4& −1\\
     −1&  0& −1& −2& −1\\
      2&  3&  5&  7&  8\end{bmatrix}\vec{x} = \vec{0} \longrightarrow \; ?
$$
So you must find a maximally linear independent subset $S$ of
$$\{\vec{x}: A\vec{x} = \vec{0} \}$$ Now once you find these two things,
$$
\text{rank}(A) = |R|,  \; \; \text{nullity}(A)= |S|
$$
