Determinant identity: $\det M \det N = \det M_{ii} \det M_{jj} - \det M_{ij}\det M_{ji}$ Let $M$ be a (real) $n \times n$ matrix. For $1 \leq i, j \leq n$ we denote by $M_{ij}$ the $(n-1) \times (n-1)$ matrix that we get when the $i$th row and $j$th column of $M$ are removed. Now, consider fixed $i$ and $j$ with $i\neq j$. Let $N$ be the $(n-2) \times (n-2)$ matrix that we get when removing both the $i$th and $j$th row and the $i$th and $j$th column from $M$. Then the following identity holds:
$$
\det M \det N = \det M_{ii}\det M_{jj} - \det M_{ij} \det M_{ji}.
$$
We were able to prove this looking at all different terms that can occur on both sides when evaluating the determinant as a sum of $n!$ terms (for instance, terms containing $a_{ji}a_{ij}$ are counted both $1$ time in the LHS and $1$ time in the RHS).
We are looking for a slick proof that does not involve writing out the determinant.
Any suggestions or approaches to this problem are appreciated!
 A: Amazingly, this is exactly Lemma 3 in my paper "Almost all integer matrices have no integer eigenvalues" with Erick B. Wong (other than restricting to $i=1,j=2$ which is trivially generalized). There's a short proof there which seems to fit your needs. There's also a reference to a book on curious determinant identities of this sort in the bibliography (which is where we got the proof in the first place).
A: I wrote below a variation of Greg Martin's argument.
First replace rows $i$ and $j$ of $M$ by the i-th vector and the j-th vector of the canonical basis of  $\mathbb{R}^n$, respectively, to form $N'$.
Notice that $\det(N')$ is equal to the determinant of your $N$ and has the same order of $M$ and its adjugate.
Now, multiply $N'$ by the adjugate of $M$ to get $N'.Adj(M)=$ $$=\begin{pmatrix}\det(M)&0&\ldots&\ldots & 0\\
0&\det(M)&0&\ldots&0\\
\vdots &\vdots&\ddots&\ldots&\vdots\\
(-1)^{i+1} \det(M_{1,i})& (-1)^{i+2} \det(M_{2,i}) & (-1)^{i+3} \det(M_{3,i})&\ldots & (-1)^{i+n} \det(M_{n,i})\\
\vdots &\vdots&\vdots&\ldots&\vdots\\
(-1)^{j+1} \det(M_{1,j})& (-1)^{j+2} \det(M_{2,j}) & (-1)^{j+3} \det(M_{3,j})&\ldots & (-1)^{j+n} \det(M_{n,j})\\
\vdots &\vdots&\vdots&\ldots&\vdots\\
0&0&0&\ldots&\det(M)\\
\end{pmatrix}$$
Notice that the only rows of $N'.Adj(M)$ that coincide with the rows of $Adj(M)$ are the rows $i$ and $j$. Therefore,
$$\det(N'.Adj(M))=\det(M)^{n-2}(\det(M_{ii})\det(M_{jj})-\det(M_{ij})\det(M_{ji})).$$
Finally, $\det(Adj(M))\det(M)=\det(Adj(M)M)=\det(M)^{n}$. Hence, if $\det(M)\neq 0$ then $\det(Adj(M))=\det(M)^{n-1}$, which completes the proof for the case which $M$ is invertible.
By continuity the equality holds for any matrix.
