In the theory of Smith Normal Form, there is a result:
If the diagonal elements in the Smith normal form are $d_1,d_2,d_3,\dots$, then $d_1d_2\dots d_k$ is the gcd of all the $k\times k$ minors of the original matrix.
Say the original matrix $A$ is $n\times n$ and has determinant 0.
Then $d_1d_2\dots d_n=\gcd(\det A)=\gcd(0)$.
That brings us to how do we define $\gcd(0)$ in this case? Do we take it to be 0?
The gcd of $\{0\}$ is $0$, as you expected. But your wording "we take it to be" strikes me as misleading; it sounds as though we made some arbitrary decision here, whereas in fact we just applied the definition of "gcd". Specifically, the gcd of any set $A$ of integers is defined to be an integer $d$ such that (1) $d$ divides all elements of $A$ and (2) if $x$ is any integer that divides all elements of $A$, then $x$ also divides $d$. (This leaves a sign ambiguity, because when $d$ satisfies this definition then so does $-d.$ One usually chooses the non-negative option, but that's irrelevant when $d=-d=0$.) Now if $A=\{0\}$ then all integers divide all elements of $A$ (because every integer divides $0$). So requirement (1) is satisfied by every integer. Requirement (2), on the other hand, says that every integer $x$ must divide $d$. The only $d$ with this property is $0$. So the definition of "gcd" tells us that the gcd of $\{0\}$ is $0$.