Is it true that, to determine the rank of a matrix in trapezoidal form (row echelon form but pivots can be 0), you need to count the number of rows (or columns) with the same number of zeroes, except the null rows (columns)?
For example, take the matrix $$\begin{pmatrix}1&2&3&4&5\\0&1&2&3&4\\0&0&0&1&2\\0&0&0&3&4\\0&0&0&0&1\end{pmatrix}$$ which is in upper trapezoidal form. Row 1 has no zeroes, row 2 has 1 zero, rows 3 and 4 have each 3 zeroes and row 4 has 4 zeroes, thus the rank is 4.
Sometimes when doing Gaussian elimination it is more convenient to leave the pivot zero (probably because the elements below are already zero), so does this rule always work?
No, that is not accurate. To determine the rank of a matrix in trapezoidal form, you do not count the number of rows/columns with the same number of zeroes. Instead, you count the number of non-zero rows/columns with the leading entry (pivot) in each row/column.
Here's how you determine the rank of a matrix in trapezoidal form:
Count the number of non-zero rows. This count is the rank of the matrix. If you are working with a matrix in column echelon form (where pivots are the leftmost non-zero entry in each column), you can also count the number of non-zero columns with the leading entry (pivot) in each column. This will give you the same rank as counting non-zero rows.
The number of zeroes in the rows or columns with non-zero entries does not affect the rank of the matrix. The rank is solely determined by the number of linearly independent rows or columns in the matrix.