TITLE: Find the determinant of a matrix given the determinant of all $p\times p$ sub-matrices?
QUESTION [16 upvotes]: Is it possible to find the determinant of an $n\times n$- matrix, only given the determinant of all $p\times p$ sub-matrices in it? Here $p\leq n$ is fixed. This is obviously true if $p=1,n$. But what happens in other cases?

REPLY [16 votes]: Here is a way to see that, for all $p$ (even if $p \nmid n$), you can reconstruct the absolute value of the matrix's determinant, which was suggested by Will Sawin's comment. The condition $p \mid n$ is only needed to get the sign (or complex phase, if the ground field is $\mathbb{C}$).
If you know the determinants of all $p \times p$ submatrices of $A$, that means exactly that you know $C_p(A)$: the $p$-th compound matrix of $A$. Since $\det(C_p(A)) = (\det(A))^\binom{n-1}{p-1}$, you can compute
$$
\lvert\det(A)\rvert = \lvert\det(C_p(A))\rvert^{1/\binom{n-1}{p-1}}.
$$