2x2 subdeterminants of a matrix

If N>2, it is well known that if two invertible NxN matrices A and B have the same determinants of any 2x2 corresponding submatrices, then A=B or A=-B.
Given then all these 2x2 determinants of an invertible matrix A, is there an "explicit" way to recover/write down A?
If N=3 it is easy, as you can get the determinant of A (up to the sign) and all its cofactors, so you can obtain the inverse matrix of A or -A, but when N>3?

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Something is wrong, as $A$ and $-A$ have the same $2\times2$ minors. – Denis Serre Jan 16 at 15:44
You're right, I should have written A=B or A=-B. The same for the last sentence, you can get the determinant of A up to the sign. Sorry for the inaccuracy. – Carlo Mantegazza Jan 16 at 23:24

By "corresponding submatrices" I presume you mean those $2\times2$ minors obtained by deleting $n-2$ colums and $n-2$ rows, where these columns and rows have the same $n-2$ indices. Once you've calculated the determinants of these submatrices you recover the action of $A$ on the exterior square $\Lambda^2 V$.

Now the following paper: "An algorithm for recognising the exterior square of a matrix" by Catherine Greenhill explains how to then obtain the original matrix $A$. Here's the relevant quote:

One computational problem which presents itself immediately is this: how can we determine whether a given matrix $Y$ is equal to the exterior square of another matrix $X$? In particular, if such an $X$ exists then we would like to construct one. A polynomial-time algorithm which solves this problem is described in Section 5.

One needs to be slightly careful here, because the exterior square does not quite determine the matrix $X$ uniquely. Here is another quote from the paper:

We prove in Section 4 that two matrices $X$, $X'$ with rank at least three have the same exterior square if and only $X'\in \{X, -X\}$.

So if the rank is at least three (which it is, since you are assuming invertibility), then we are pretty much done. I'm guessing that the situation where the rank is $\leq 2$ would be easy enough to resolve but in any case that's outside the scope of the question...

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Thanks, very interesting reference. Just a comment on my question: it comes from determining explicitely the unique (up to the sign) second fundamental form $A$ of a hypersurface of dimension $N>3$ in $R^{N+1}$. As $Riem=A*A$ (where $*$ is the Kulkarni-Nomizu product of two symmetric 2-forms) all the 2x2 determinants of the matrix associated to $A$ in an orthonormal basis are given by the sectional curvatures, hence they are all "intrinsic", so also $A$, up to the sign. – Carlo Mantegazza Jan 16 at 23:44
Sorry $N\geq3$ here above. – Carlo Mantegazza Jan 17 at 4:22