Consider $A \in M_n(\mathbb{C})$ such that row $l$ is permutable. Denote by $\Gamma_{li}$ the algebraic complement of $a_{li}$, $i = 1, \dots, n$. Suppose $i, j, k, p \in \{1, 2, \dots, n\}$ are such that $a_{li} \neq a_{lj}$ and $\Gamma_{lk} \neq \Gamma_{lp}$. Consider matrices $B$ and $C$ obtained from $A$ by permuting elements in row $l$, such that $B$ contains $a_{li}$ in place $(l, k)$ and element $a_{lj}$ in place $(l, p)$ and, matrix $C$ contains $a_{lj}$ in place $(l, k)$ and $a_{li}$ in place $(l, p)$; the two matrices having in the other positions the initial elements.

Developing determinants along row $l$, we get $\det(B) - \det(C) = (a_{li} - a_{lj})(\Gamma_{lk} - \Gamma_{lp}) \neq 0$, in contradiction with the hypothesis.

We deduce that row $l$ of $A$ is permutable if and only if all its elements are equal or, if all algebraic complements of elements in line $l$ are equal.

Consider now a matrix $A$ that admits two permutable rows. Then:

* if both rows are constant $\det(A) = 0$;
* if both rows have constant algebraic complements then $A^*$ has two constant rows; that is $\det(A^*) = 0$, which implies $\det(A) = 0$;
* if one row has all elements equal to $a$, and another row has all algebraic complements equal to $b$, then from $AA^* = \det(A)I_n$ we deduce $ab = 0$, so $a = 0$ or $b = 0$.

Thus, in every case, $\det(A) = 0$.