Find the determinant of a matrix given the determinant of all $p\times p$ sub-matrices? 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?
 A: As mentioned by Will Sawin, a necessary condition is that $p$ divides $n$. Thus let us assume that $n=pk$. Denoting $e_1,\dotsc,e_n$ the canonical basis, the knowledge of the $p\times p$ minors is the knowledge of the $p$-vectors $$(Ae_{i_1})\wedge\cdots\wedge(Ae_{i_p})\in\Lambda^p(K^n),$$
where $K$ is the field of scalars (e.g. $\mathbb C$).
Splitting
$$(Ae_1)\wedge\cdots\wedge (Ae_n) = [(Ae_1)\wedge\cdots\wedge (Ae_p)]\wedge\cdots\wedge[(Ae_{n-p+1})\wedge\cdots\wedge (Ae_n)],
$$
we see that $(Ae_1)\wedge\cdots\wedge (Ae_n)$ is a polynomial function in the $p\times p$ minors. Since $(Ae_1)\wedge\dotsb\wedge (Ae_n)=(\det A)e_1\wedge\dotsb\wedge e_n$, we deduce the value of $\det A$.
Let me describe how it works when $n=4$ and $p=2$. The minors are denoted
$$A\binom{i\alpha}{j\beta}=a_{i\alpha}a_{j\beta}-a_{i\beta}a_{j\alpha}.$$
Then
$$(Ae_\alpha)\wedge(Ae_\beta)=\sum_{i<j}A\binom{i\alpha}{j\beta}e_i\wedge e_j.$$
We thus obtain
$$\det A=\sum_{\substack{\rho\in{\frak S}_4 \\ \rho(1)<\rho(2),\rho(3)<\rho(4)}}A\binom{\rho(1)1}{\rho(2)2}A\binom{\rho(3)3}{\rho(4)4}.$$
A: 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
$$
\lvert\det(A)\rvert = \lvert\det(C_p(A))\rvert^{1/\binom{n-1}{p-1}}.
$$
