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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 $X'\in {X, \{X, -X}$.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... 3 edited last paragraph 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. The paper can be downloaded here. One slight problem is that 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. When I'm guessing that the situation where the rank is less than three I'm not sure what happens - you will need to make use of other$2\times \leq 2$minors than would be easy enough to resolve but in any case that's outside the ones described above..scope of the question... 2 added caveat! 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. The paper can be downloaded here. One slight problem is that 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, then we are pretty much done. When the rank is less than three I'm not sure what happens - you will need to make use of other$2\times 2\$ minors than the ones described above...