If $A$ is a invertible integer matrix, then $\det A$ is coprime with some cofactor

Question

Let $A\in \mathbb{Z}^{n\times n}$ with $\det A\neq 0$. I was wondering whether $A^{-1}$ must have an entry of the form $p/q$ with $q=\det A$ and $(p,q)=1$ (that is, an irreducible fraction).

By the adjugate form of the inverse, this is equivalent to some minor of $A$ being coprime with $\det A$.

Playing around with $2\times 2$ matrices, it seems to hold for $n=2$, and I tried using Laplace's expansion to generalize to higher dimensions, but got nowhere. Does it hold in general?

Answer 1

If $A = k I$ then $\det A = k^n$ while $A^{-1} = k^{-1} I$ which for $n > 1$ doesn't have $\det A$ as the denominator of any entry.