For the $5\times 9$ matrix \begin{align*} B&= \begin{bmatrix} \leading{1}&5&0&0&2&8&0&5&-1\\ 0&0&\leading{1}&0&4&7&0&2&0\\ 0&0&0&\leading{1}&3&9&0&3&-6\\ 0&0&0&0&0&0&\leading{1}&4&2\\ 0&0&0&0&0&0&0&0&0 \end{bmatrix} \end{align*} in reduced row-echelon form we have \begin{align*} r&=4\\ d_1&=1 & d_2&=3 & d_3&=4 & d_4&=7\\ f_1&=2 & f_2&=5 & f_3&=6 & f_4&=8 & f_5&=9 \end{align*}
Notice that the sets \begin{align*} D&=\set{d_1,\,d_2,\,d_3,\,d_4}=\set{1,\,3,\,4,\,7} & F=\set{f_1,\,f_2,\,f_3,\,f_4,\,f_5}=\set{2,\,5,\,6,\,8,\,9} \end{align*} have nothing in common and together account for all of the columns of $B$ (we say it is a partition of the set of column indices).