Below is the excerpt from this link to prove the following proposition.
Proposition $\ $Let $A$ be a square matrix over an arbitrary field with all row and column sums zero. Then all cofactors of $A$ are equal.
Proof $\ $ Let $A$ be $n × n$, and let $J$ be the all-1 matrix of size $n × n.$ We evaluate $\det(A + J)$ by performing row and column operations. We distinguish the first row and column for simplicity, but the same applies to any row and column. We do the following:
• Add all other rows of $A + J$ to the first. Then every entry in the first row is n, while the other rows are unaffected.
• Add all other columns to the first. In the result, the $(1, 1)$ entry is $n$ 2 ; all other entries of the first row and column are $n$; and the remaining entries are unaffected.
• Take out a factor n from the first row.
• Subtract the first row from each of the other rows. The $(1, 1)$ entry of the result is $n;$ the remaining entries of the first column are zero; and the entries not in the first row or column are precisely those of $A,$ since we subtract $1$ from each of them.
Thus, $\det(A + J) = n$ 2A11, where A11 is the $(1, 1)$ cofactor of $A.$ As remarked, the same applies to any cofactor; this shows that all cofactors are equal.
I am trying to change this proof to prove more general case, where just row sum is zero.
I am following the above proof lines but can't understand the bold-faced step, which says "Take out a factor n".
How's factor n defined?
Means that you should divide each element in the row in question by $n.$