TITLE: Determinant of a sub-matrix of the classical adjoint
QUESTION [7 upvotes]: Let $A$ be a  square matrix of order $n$, say with complex coefficients, and let $M$ be the plain matrix of minors  of $A$ of order $n-1$ (no transpose, no sing changes). Let $I$ and $J$ be $r$-subsets of the index set $[n]$. Then, apparently
$$\det M_{I\times J}=\det A_{([n]\setminus I)\times ([n]\setminus J)}\det(A)^{r-1}.$$
(One can also write the identity for the classical adjoint of $A$ instead of $M$, or also assume $A$ invertible, and express the identity in terms of $A^{-1}$).
So for $r=1$ this is just the definition of $M_{ij}$. For small values of $n$, it is possible to check the identity, keeping track of the terms of the expansion of the determinant.  But is there a more synthetic proof, and an interpretation of it?

REPLY [8 votes]: For me, a standard book to look for such things is Prasolov's linear algebra. This is Theorem 1.2.6.1. Here is the Russian edition, but the proof is essentially a formula.

REPLY [4 votes]: You might try using the very interesting combinatorial approach of Doron Zeilberger in A Combinatorial Approach to Matrix Algebra, Discrete Mathematics 56 (1985), 61-72. There he gives short proofs of various matrix results such as the Cayley-Hamilton Theorem and the Matrix-Tree Theorem as combinatorial identities, which for me is the "right" way to understand these, rather than appealing to, for example, properties of complex matrices and eigenvalues.