Jacobi's equality between complementary minors of inverse matrices 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.
 A: The key word under which you will find this result in modern books is "Schur complement". Here is a self-contained proof. Assume $I$ and $J$ are $(1,2,\dots,k)$ for some $k$ without loss of generality (you may reorder rows/columns). Let the matrix be
$$
M=\begin{bmatrix}A & B\\\\ C & D\end{bmatrix}, 
$$
where the blocks $A$ and $D$ are square. Assume for now that $A$ is invertible --- you may treat the general case with a continuity argument. Let $S=D-CA^{-1}B$ be the so-called Schur complement of $A$ in $M$.
You may verify the following identity ("magic wand Schur complement formula")
$$
\begin{bmatrix}A & B\\\\ C & D\end{bmatrix} =
\begin{bmatrix}I & 0\\\\ CA^{-1} & I\end{bmatrix}
\begin{bmatrix}A & 0\\\\ 0 & S\end{bmatrix}
\begin{bmatrix}I & A^{-1}B\\\\ 0 & I\end{bmatrix}. \tag{1}
$$
By taking determinants, $$\det M=\det A \det S. \tag{2}$$ Moreover, if you invert term-by-term the above formula you can see that the (2,2) block of $M^{-1}$ is $S^{-1}$. So your thesis is now (2).
Note that the "magic formula" (1) can be derived via block Gaussian elimination and is much less magic than it looks at first sight.
A: This is nothing but the Schur complement formula. See my book Matrices; Theory and Applications, 2nd ed., Springer-Verlag GTM 216, page 41. 
Up to some permutation of rows and columns, we may assume that $I=J=[1,p]$. Let us write blockwise
$$A=\begin{pmatrix} W & X \\\\ Y & Z \end{pmatrix}.$$
Assume WLOG that $W$ is invertible.
On the one hand, we have (Schur C.F)
$$\det A=\det W\cdot\det(Z-YW^{-1}X).$$
Finally, we have
$$A^{-1}=\begin{pmatrix} \cdot & \cdot \\\\ \cdot & (Z-YW^{-1}X)^{-1} \end{pmatrix},$$which gives the desired result.
These formulas are obtained by factorizing $A$ into $LU$ (namely, $L= \begin{pmatrix} I_* & 0 \\ YW^{-1} & I_* \end{pmatrix}$ and $U = \begin{pmatrix} W & X \\ 0 & Z-YW^{-1}X \end{pmatrix}$, with the $I_*$ being identity matrices of appropriate size).
A: 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
$$ (\det A) (\det A^{-1}[J^c,I^c]) = \det A[I,J] $$.
