Products of adjugate matrices Let $S$ and $A$ be a symmetric and a skew-symmetric $n \times n$ matrix over $\mathbb{R}$, respectively. When calculating (numerically) the product $S^{-1} A S^{-1}$ I keep getting the factor $\det S$ in the denominator, while I would expect to get the square $$S^{-1} A S^{-1} = \frac{(\text{adj }S) A (\text{adj }S)}{(\det S)^2},$$ where $\text{adj }S$ is the adjugate of $S$.
Is there a way to prove that the combination $(\text{adj }S) A (\text{adj }S)$ already contains a factor of $\det S$?
 A: Assume that $\det(S)=0$. Then $adj(S)$ has rank $1$ and is symmetric; then $adj(S)=avv^T$ where $a\in \mathbb{R}$ and $v$ is a vector. Thus $adj(S)Aadj(S)=a^2v(v^TAv)v^T$. Since $A$ is skew-symmetric, $v^TAv=0$ and $adj(S)Aadj(S)=0$. We use the Darij's method; here, the condition is that $\det(S)$ is an irreducible polynomial when $S$ is a generic symmetric matrix; if it is true, then $\det(S)$ is a factor of every entry of $adj(S)Aadj(S)$.
EDIT 1. For the proof  that $\det(S)$ is an irreducible polynomial when $S$ is a generic symmetric matrix, cf. https://mathoverflow.net/questions/50362/irreducibility-of-determinant-of-symmetric-matrix
and we are done !
EDIT 2. @ darij grinberg , hi Darij, I read quickly your Theorem 1 (for $K$, a commutative ring with unity) and I think that your proof works; yet it is complicated! I think (as you wrote in your comment above) that it suffices to prove the result when $K$ is a field; yet I do not kwow how to write it rigorously...
STEP 1. $K$ is a field. If $\det(S)=0$, then $adj(S)=vw^T$ and $adj(S).A.adj(S)=v(w^TAw)v^T=0$ (even if $char(K)=2$). Since $\det(.)$ is irreducible over $M_n(K)$, we conclude as above.
STEP 2. Let $S=[s_{ij}],A=[a_{i,j}]$. We work in the ring of polynomials $\mathbb{Z}[(s_{i,j})_{i,j},(a_{i,j})_{i<j}]$ in the indeterminates $(s_{i,j}),(a_{i,j})$. This ring has no zero-divisors, is factorial and its characteristic is $0$ and even is integrally closed. Clearly the entries of $adj(S).A.adj(S)$ are in $\mathbb{Z}[(s_{i,j})_{i,j},(a_{i,j})_{i<j}]$; moreover they formally have $\det(S)$ as a factor.
Now, if $K$ is a commutative ring with unity, we must use an argument using a variant of Gauss lemma showing that the factor $\det(S)$ is preserved over $K$. What form of the lemma can be used and how to write it correctly ?  
I just see that the OP takes for himself the green chevron; we are our own best advocates
A: Here is a proof.
In the following, we fix a commutative ring $\mathbb{K}$. All matrices are
over $\mathbb{K}$.

Theorem 1. Let $n\in\mathbb{N}$. Let $S$ be an $n\times n$-matrix. Let $A$
  be an alternating $n\times n$-matrix. (This means that $A^{T}=-A$ and that the
  diagonal entries of $A$ are $0$.) Then, each entry of the matrix $\left(
\operatorname*{adj}S\right)  \cdot A\cdot\left(  \operatorname*{adj}\left(
S^{T}\right)  \right)  $ is divisible by $\det S$ (in $\mathbb{K}$).

[UPDATE: A slight modification of the below proof of Theorem 1 can be
found in the solution to Exercise 6.42 in
my Notes on the combinatorial
fundamentals of algebra, version of 10 January 2019. More precisely,
said Exercise 6.42 claims that each entry of the matrix
$\left(\operatorname{adj} S\right)^T \cdot A \cdot
\left(\operatorname{adj} S\right)$ is divisible by $\det S$; now it remains
to substitute $S^T$ for $S$ and recall that
$\left(\operatorname{adj} S\right)^T = \operatorname{adj} \left(S^T\right)$,
and this immediately yields Theorem 1 above. Still, the following (shorter)
version of this proof might be useful as well.]
The main workhorse of the proof of Theorem 1 is the following result, which
is essentially (up to some annoying switching of rows and columns) the
Desnanot-Jacobi identity used in Dodgson condensation:

Theorem 2. Let $n\in\mathbb{N}$. Let $S$ be an $n\times n$-matrix. For
  every $u\in\left\{  1,2,\ldots,n\right\}  $ and $v\in\left\{  1,2,\ldots
,n\right\}  $, we let $S_{\sim u,\sim v}$ be the $\left(  n-1\right)
\times\left(  n-1\right)  $-matrix obtained by crossing out the $u$-th row and
  the $v$-th column in $S$. (Thus, $\operatorname*{adj}S=\left(  \left(
-1\right)  ^{i+j}S_{\sim j,\sim i}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$.)
  For every four elements $u$, $u^{\prime}$, $v$ and $v^{\prime}$ of $\left\{
1,2,\ldots,n\right\}  $ with $u\neq u^{\prime}$ and $v\neq v^{\prime}$, we let
  $S_{\left(  \sim u,\sim u^{\prime}\right)  ,\left(  \sim v,\sim v^{\prime
}\right)  }$ be the $\left(  n-2\right)  \times\left(  n-2\right)  $-matrix
  obtained by crossing out the $u$-th and $u^{\prime}$-th rows and the $v$-th
  and $v^{\prime}$-th columns in $S$. Let $u$, $i$, $v$ and $j$ be four elements
  of $\left\{  1,2,\ldots,n\right\}  $ with $u\neq v$ and $i\neq j$. Then,
  \begin{align}
& \det\left(  S_{\sim i,\sim u}\right)  \cdot\det\left(  S_{\sim j,\sim
v}\right)  -\det\left(  S_{\sim i,\sim v}\right)  \cdot\det\left(  S_{\sim
j,\sim u}\right) \\
& = \left(  -1\right)  ^{\left[  i<j\right]  +\left[  u<v\right]  }\det
S\cdot\det\left(  S_{\left(  \sim i,\sim j\right)  ,\left(  \sim u,\sim
v\right)  }\right) .
\end{align}
  Here, we use the Iverson bracket notation (that is, we write
  $\left[  \mathcal{A}\right]  $ for the truth value of a statement
  $\mathcal{A}$; this is defined by $\left[  \mathcal{A}\right]  =
\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases}
$).

There are several ways to prove Theorem 2: I am aware of one argument
that derives it from the Plücker
relations (the simplest ones, where just one column is being shuffled around).
There is at least one combinatorial argument that proves Theorem 2 in the
case when $i = 1$, $j = n$, $u = 1$ and $v = n$ (see Zeilberger's paper);
the general case can be reduced to this case by permuting rows and columns
(although it is quite painful to track how the signs change under these
permutations). (See also a paper by Berliner and Brualdi for a
generalization of Theorem 2, with a combinatorial proof too.) There is
at least one short algebraic proof of Theorem 2 (again in the case when
$i = 1$, $j = n$, $u = 1$ and $v = n$ only) which relies on "formal" division
by $\det S$ (that is, it proves that
\begin{align}
& \det S \cdot \left(\det\left(  S_{\sim 1,\sim 1}\right)  \cdot\det\left(  S_{\sim 2,\sim
2}\right)  -\det\left(  S_{\sim 1,\sim 2}\right)  \cdot\det\left(  S_{\sim
2,\sim 1}\right) \right) \\
& = \left(\det S\right)^2
\cdot\det\left(  S_{\left(  \sim 1,\sim 2\right)  ,\left(  \sim 1,\sim
2\right)  }\right)  ,
\end{align}
and then argues that $\det S$ can be cancelled because the determinant of a
"generic" square matrix is invertible). (This proof appears in Bressoud's
Proofs and Confirmations; a French version can also be found in
lecture notes by Yoann Gelineau.) Unfortunately, none of these proofs
seems to release the reader from the annoyance of dealing with the signs.
Maybe exterior powers are the best thing to use here, but I do not see how.
I have written up a division-free (but laborious and annoying) proof of
Theorem 2 in my determinant notes; more precisely, I have written up
the proof of the $i < j$ and $u < v$ case, but the general case can easily
be obtained from it as follows:
Proof of Theorem 2. We need to prove the equality
\begin{align}
& \det\left(  S_{\sim i,\sim u}\right)  \cdot\det\left(  S_{\sim j,\sim
v}\right)  -\det\left(  S_{\sim i,\sim v}\right)  \cdot\det\left(  S_{\sim
j,\sim u}\right)  \\
& = \left(  -1\right)  ^{\left[  i<j\right]  +\left[  u<v\right]  }\det
S\cdot\det\left(  S_{\left(  \sim i,\sim j\right)  ,\left(  \sim u,\sim
v\right)  }\right)  .
\label{darij.eq.1}
\tag{1}
\end{align}
If we interchange $u$ with $v$, then the left hand side of this equality
gets multiplied by $-1$ (because its subtrahend and its minuend switch places),
whereas the right hand side also gets multiplied by $-1$ (since
$S_{\left(  \sim i,\sim j\right)  ,\left(  \sim u,\sim
v\right)}$ does not change, but $\left[ u<v\right]$ either changes from $0$
to $1$ or changes from $1$ to $0$). Hence, if we interchange $u$ with $v$,
then the equality \eqref{darij.eq.1} does not change its truth value.
Thus, we can WLOG assume that $u \leq v$ (since otherwise we can just
interchange $u$ with $v$). Assume this. For similar reasons, we can WLOG
assume that $i \leq j$; assume this too. From $u \leq v$ and $u \neq v$,
we obtain $u < v$. From $i \leq j$ and $i \neq j$, we obtain $i < j$.
Thus, Theorem 6.126 in my Notes on the combinatorial
fundamentals of algebra, version of 10 January 2019 (applied to $A=S$,
$p=i$ and $q=j$) shows that
\begin{align}
& \det
S\cdot\det\left(  S_{\left(  \sim i,\sim j\right)  ,\left(  \sim u,\sim
v\right)  }\right)  \\
& = \det\left(  S_{\sim i,\sim u}\right)  \cdot\det\left(  S_{\sim j,\sim
v}\right)  -\det\left(  S_{\sim i,\sim v}\right)  \cdot\det\left(  S_{\sim
j,\sim u}\right)
\label{darij.eq.2}
\tag{2}
\end{align}
(indeed, what I am calling $S_{\left(  \sim i,\sim j\right)  ,\left(  \sim u,\sim
v\right)  }$ here is what I am calling
$\operatorname{sub}^{1,2,\ldots,\widehat{u},\ldots,\widehat{v},\ldots,n}_{1,2,\ldots,\widehat{i},\ldots,\widehat{j},\ldots,n} A$ in my notes).
But both $\left[i < j\right]$ and $\left[u < v\right]$ equal $1$ (since
$i < j$ and $u < v$). Thus,
$\left(  -1\right)  ^{\left[  i<j\right]  +\left[  u<v\right]  }
= \left(-1\right)^{1+1} = 1$. Therefore
\begin{align}
& \underbrace{\left(  -1\right)  ^{\left[  i<j\right]  +\left[  u<v\right]  }}_{=1}\det
S\cdot\det\left(  S_{\left(  \sim i,\sim j\right)  ,\left(  \sim u,\sim
v\right)  }\right)  \\
& = \det
S\cdot\det\left(  S_{\left(  \sim i,\sim j\right)  ,\left(  \sim u,\sim
v\right)  }\right)  \\
& = \det\left(  S_{\sim i,\sim u}\right)  \cdot\det\left(  S_{\sim j,\sim
v}\right)  -\det\left(  S_{\sim i,\sim v}\right)  \cdot\det\left(  S_{\sim
j,\sim u}\right)
\end{align}
(by \eqref{darij.eq.2}). This proves Theorem 2. $\blacksquare$
Finally, here is an obvious lemma:

Lemma 3. Let $n\in\mathbb{N}$. For every $i\in\left\{  1,2,\ldots
,n\right\}  $ and $j\in\left\{  1,2,\ldots,n\right\}  $, let $E_{i,j}$ be the
  $n\times n$-matrix whose $\left(  i,j\right)  $-th entry is $1$ and whose all
  other entries are $0$. (This is called an elementary matrix.) Then, every
  alternating $n\times n$-matrix is a $\mathbb{K}$-linear combination of the
  matrices $E_{i,j}-E_{j,i}$ for pairs $\left(  i,j\right)  $ of integers
  satisfying $1\leq i<j\leq n$.

Proof of Theorem 1. We shall use the notation $E_{i,j}$ defined in Lemma 3.
We need to prove that every entry of the matrix $\left(  \operatorname*{adj}
S\right)  \cdot A\cdot\left(  \operatorname*{adj}\left(  S^{T}\right)
\right)  $ is divisible by $\det S$. In other words, we need to prove that,
for every $\left(  u,v\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$, the
$\left(  u,v\right)  $-th entry of the matrix $\left(  \operatorname*{adj}
S\right)  \cdot A\cdot\left(  \operatorname*{adj}\left(  S^{T}\right)
\right)  $ is divisible by $\det S$. So, fix $\left(  u,v\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}$.
We need to show that the $\left(  u,v\right)  $-th entry of the matrix
$\left(  \operatorname*{adj}S\right)  \cdot A\cdot\left(  \operatorname*{adj}
\left(  S^{T}\right)  \right)  $ is divisible by $\det S$. This statement is
clearly $\mathbb{K}$-linear in $A$ (in the sense that if $A_{1}$ and $A_{2}$
are two alternating $n\times n$-matrices such that this statement holds both
for $A=A_{1}$ and for $A=A_{2}$, and if $\lambda_{1}$ and $\lambda_{2}$ are
two elements of $\mathbb{K}$, then this statement also holds for
$A=\lambda_{1}A_{1}+\lambda_{2}A_{2}$). Thus, we can WLOG assume that $A$ has
the form $E_{i,j}-E_{j,i}$ for a pair $\left(  i,j\right)  $ of integers
satisfying $1\leq i<j\leq n$ (according to Lemma 3). Assume this, and consider
this pair $\left(  i,j\right)  $.
We have $\operatorname*{adj}S=\left(  \left(  -1\right)  ^{x+y}\det\left(
S_{\sim y,\sim x}\right)  \right)  _{1\leq x\leq n,\ 1\leq y\leq n}$ and
\begin{align}
\operatorname*{adj}\left(  S^{T}\right)
& = \left(  \left(  -1\right)
^{x+y}\det\left(  \underbrace{\left(  S^{T}\right)  _{\sim y,\sim x}
}_{=\left(  S_{\sim x,\sim y}\right)  ^{T}}\right)  \right)  _{1\leq x\leq
n,\ 1\leq y\leq n} \\
& = \left(  \left(  -1\right)  ^{x+y}\underbrace{\det\left(  \left(  S_{\sim
x,\sim y}\right)  ^{T}\right)  }_{=\det\left(  S_{\sim x,\sim y}\right)
}\right)  _{1\leq x\leq n,\ 1\leq y\leq n} \\
& = \left(  \left(  -1\right)  ^{x+y}\det\left(  S_{\sim x,\sim y}\right)
\right)  _{1\leq x\leq n,\ 1\leq y\leq n} .
\end{align}
Hence,
\begin{align}
& \underbrace{\left(  \operatorname*{adj}S\right)  }_{=\left(  \left(
-1\right)  ^{x+y}\det\left(  S_{\sim y,\sim x}\right)  \right)  _{1\leq x\leq
n,\ 1\leq y\leq n}}\cdot\underbrace{A}_{=E_{i,j}-E_{j,i}}\cdot
\underbrace{\left(  \operatorname*{adj}\left(  S^{T}\right)  \right)
}_{=\left(  \left(  -1\right)  ^{x+y}\det\left(  S_{\sim x,\sim y}\right)
\right)  _{1\leq x\leq n,\ 1\leq y\leq n}} \\
& =\left(  \left(  -1\right)  ^{x+y}\det\left(  S_{\sim y,\sim x}\right)
\right)  _{1\leq x\leq n,\ 1\leq y\leq n}\cdot\left(  E_{i,j}-E_{j,i}\right) \\
& \qquad \qquad
\cdot\left(  \left(  -1\right)  ^{x+y}\det\left(  S_{\sim x,\sim y}\right)
\right)  _{1\leq x\leq n,\ 1\leq y\leq n} \\
& = \left(  \left(  -1\right)  ^{x+i}\det\left(  S_{\sim i,\sim x}\right)
\cdot\left(  -1\right)  ^{j+y}\det\left(  S_{\sim j,\sim y}\right) \right. \\
& \qquad \qquad
\left. -\left(
-1\right)  ^{x+j}\det\left(  S_{\sim j,\sim x}\right)  \cdot\left(  -1\right)
^{i+y}\det\left(  S_{\sim i,\sim y}\right)  \right)  _{1\leq x\leq n,\ 1\leq
y\leq n} .
\end{align}
Hence, the $\left(  u,v\right)  $-th entry of the matrix $\left(
\operatorname*{adj}S\right)  \cdot A\cdot\left(  \operatorname*{adj}\left(
S^{T}\right)  \right)  $ is
\begin{align}
& \left(  -1\right)  ^{u+i}\det\left(  S_{\sim i,\sim u}\right)  \cdot\left(
-1\right)  ^{j+v}\det\left(  S_{\sim j,\sim v}\right)  -\left(  -1\right)
^{u+j}\det\left(  S_{\sim j,\sim u}\right)  \cdot\left(  -1\right)  ^{i+v}
\det\left(  S_{\sim i,\sim v}\right)  \\
& = \left(  -1\right)  ^{i+j+u+v}\left(  \det\left(  S_{\sim i,\sim
u}\right)  \cdot\det\left(  S_{\sim j,\sim v}\right)  -\det\left(  S_{\sim
i,\sim v}\right)  \cdot\det\left(  S_{\sim j,\sim u}\right)  \right)  .
\label{darij.eq.3}
\tag{3}
\end{align}
We need to prove that this is divisible by $\det S$. If $u=v$, then this is
obvious (because if $u=v$, then the right hand side of \eqref{darij.eq.3} is $0$). Hence,
we WLOG assume that $u\neq v$. Thus, \eqref{darij.eq.3} shows that the $\left(
u,v\right)  $-th entry of the matrix $\left(  \operatorname*{adj}S\right)
\cdot A\cdot\left(  \operatorname*{adj}\left(  S^{T}\right)  \right)  $ is
\begin{align}
& \left(  -1\right)  ^{i+j+u+v}\underbrace{\left(  \det\left(  S_{\sim i,\sim
u}\right)  \cdot\det\left(  S_{\sim j,\sim v}\right)  -\det\left(  S_{\sim
i,\sim v}\right)  \cdot\det\left(  S_{\sim j,\sim u}\right)  \right)
}_{\substack{=\left(  -1\right)  ^{\left[  i<j\right]  +\left[
u<v\right]  }\det S\cdot\det\left(  S_{\left(  \sim i,\sim j\right)  ,\left(
\sim u,\sim v\right)  }\right)  \\\text{(by Theorem 2)}}} \\
& = \left(  -1\right)  ^{i+j+u+v}\left(  -1\right)  ^{\left[
i<j\right]  +\left[  u<v\right]  }\det S\cdot\det\left(  S_{\left(  \sim
i,\sim j\right)  ,\left(  \sim u,\sim v\right)  }\right) ,
\end{align}
which is clearly divisible by $\det S$. Theorem 1 is thus proven. $\blacksquare$
A: I have managed to prove the claim in a pedestrian way. I'll just present the final result because the proof is quite involved and probably of interest only to me.
In the following, I will use the (abstract) index notation and Einstein summation convention.
First of all, the determinant of a square $n \times n$ matrix $S$ is given by
$$\det S = \frac{1}{n!} \epsilon_{a_1 \ldots a_n} \epsilon_{b_1 \ldots b_n} S_{b_1 a_1} \ldots S_{b_n a_n}.$$
Second, the adjugate of $S$ is given by a similar expression
$$(\operatorname{ajd} S)_{a_1 b_1} = \frac{1}{(n-1)!} \epsilon_{a_1 a_2 \ldots a_n} \epsilon_{b_1 b_2 \ldots b_n} S_{b_2 a_2} \ldots S_{b_n a_n}.$$
Finally, we will need the another tensor of order $4$, defined as
$$(\operatorname{ajd}_2 S)_{a_1 a_2,b_1 b_2} = \frac{1}{(n-2)!} \epsilon_{a_1 a_2 a_3 \ldots a_n} \epsilon_{b_1 b_2 b_3 \ldots b_n} S_{b_3 a_3} \ldots S_{b_n a_n}.$$
Then the following identity holds
$$((S^{-1}) A (S^{-1})^{T})_{ab} = \frac{1}{2} \frac{(\operatorname{ajd}_2 S)_{ab,cd} A_{cd}}{\det S}$$
for $A$ skew-symmetric.
A: Because $S\mbox{adj}(S)=\mbox{adj}(S)S=\mbox{det}(S)I$, then $S$ is invertible iff $\mbox{det}(S)\ne 0$, and, in that case
$$
                         S^{-1} = \frac{1}{\mbox{det}(S)}\mbox{adj}(S).
$$
Therefore,
$$
                         S^{-1}AS^{-1} = \frac{1}{\mbox{det}(S)^{2}}\mbox{adj}(S)\,A\,\mbox{adj}(S).
$$
