Let us compute the rank and nullity of \begin{equation*} A=\begin{bmatrix} 2 & -4 & -1 & 3 & 2 & 1 & -4\\ 1 & -2 & 0 & 0 & 4 & 0 & 1\\ -2 & 4 & 1 & 0 & -5 & -4 & -8\\ 1 & -2 & 1 & 1 & 6 & 1 & -3\\ 2 & -4 & -1 & 1 & 4 & -2 & -1\\ -1 & 2 & 3 & -1 & 6 & 3 & -1 \end{bmatrix} \end{equation*}
To do this, we will first row-reduce the matrix since that will help us determine bases for the null space and column space. \begin{equation*} \begin{bmatrix} \leading{1} & -2 & 0 & 0 & 4 & 0 & 1\\ 0 & 0 & \leading{1} & 0 & 3 & 0 & -2\\ 0 & 0 & 0 & \leading{1} & -1 & 0 & -3\\ 0 & 0 & 0 & 0 & 0 & \leading{1} & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
From this row-equivalent matrix in reduced row-echelon form we record $D=\set{1,\,3,\,4,\,6}$ and $F=\set{2,\,5,\,7}$.
For each index in $D$, Theorem BCS creates a single basis vector. In total the basis will have $4$ vectors, so the column space of $A$ will have dimension $4$ and we write $\rank{A}=4$.
For each index in $F$, Theorem BNS creates a single basis vector. In total the basis will have $3$ vectors, so the null space of $A$ will have dimension $3$ and we write $\nullity{A}=3$.