TITLE: Jacobi's equality between complementary minors of inverse matrices
QUESTION [24 upvotes]: What's a quick way to prove the following fact about minors of an invertible matrix $A$ and its inverse?
Let $A[I,J]$ denote the submatrix of an $n \times n$ matrix $A$ obtained by keeping only the rows indexed by $I$ and columns indexed by $J$. Then 
$$ |\det A[I,J]| = | (\det A) \det A^{-1}[J^c,I^c]|,$$
where $I^c$ stands for $[n] \setminus I$, for $|I| = |J|$. It is trivial when $|I| = |J| = 1$ or $n-1$. This is apparently proved by Jacobi, but I couldn't find a proof anywhere in books or online. Horn and Johnson listed this as one of the advanced formulas in their preliminary chapter, but didn't give a proof. In general what's a reliable source to find proofs of all these little facts? I ran into this question while reading Macdonald's book on symmetric functions and Hall polynomials, in particular page 22 where he is explaining the determinantal relation between the elementary symmetric functions $e_\lambda$ and the complete symmetric functions $h_\lambda$. 
I also spent 3 hours trying to crack this nut, but can only show it for diagonal matrices :(
Edit: It looks like Ferrar's book on Algebra subtitled determinant, matrices and algebraic forms, might carry a proof of this in chapter 5. Though the book seems to have a sexist bias.

REPLY [10 votes]: Here is a short proof inspired by darij grinberg's answer and not using Schur complements (in particular, no invertibility of sub-matrices or continuity assumptions are needed).
WLOG let $I=J=[k]$. Let $A_i$ denote the $i$th column of a matrix $A$. Consider the product
$$
A
\left[
\begin{array}{c|c|c|c|c|c}
e_1 & \cdots & e_k & A^{-1}_{k+1} & \cdots & A^{-1}_n
\end{array}
\right]
=
\left[
\begin{array}{c|c|c|c|c|c}
A_1 & \cdots & A_k & e_{k+1} & \cdots & e_n
\end{array}
\right]
$$
which can also be written
$$
A
\begin{pmatrix}
I_k & A^{-1}[J,I^c] \\ 0 & A^{-1}[J^c,I^c]
\end{pmatrix}
=
\begin{pmatrix}
A[I,J] & 0 \\ A[I^c,J] & I_{n-k}
\end{pmatrix}
$$
Taking the determinant of both sides yields
$$ (\det A) (\det A^{-1}[J^c,I^c]) = \det A[I,J] $$.