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 $\det S$?
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 $\mathbb{Z}[(s_{i,j})_{i,j},(a_{i,j})_{i&lt;j}]$ 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 $\mathbb{Z}[(s_{i,j})_{i,j},(a_{i,j})_{i&lt;j}]$; 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
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,
$$
                         S^{-1}AS^{-1} = \frac{1}{\mbox{det}(S)^{2}}\mbox{adj}(S)\,A\,\mbox{adj}(S).
$$