Let $M$ be **any** $n\times n$ matrix.


We define the usual cofactors: $C\_{i,j}$ is $(-1)^{i+j}$ times the determinant of the submatrix obtained by deleting row $i$ and column $j$ of $M$.


We can write the determinant of $M$ using Laplace expansion along column $p$ as:


$$\det M = \sum\_{q=1}^n M\_{q,p} C\_{q,p}$$


Now, for any $k, p\in \{1,...,n\}$ consider the sum:


$$W\_{n,p}(k) = \sum\_{q=1}^n {M\_{q,p} C\_{q,p}^2 \prod\_{j\ne q}{C\_{j,k}}}$$


Clearly $W\_{n,p}(p)$ can always be factored, with $\det M$ as one factor:


$$W\_{n,p}(p) = \sum\_{q=1}^n{M\_{q,p} C\_{q,p}^2 \prod\_{j\ne q}{C\_{j,p}}}
= \sum\_{q=1}^n{M\_{q,p} C\_{q,p} \prod\_{j=1}^n{C\_{j,p}}}
= \left(\det M\right) \prod\_{j=1}^n{C\_{j,p}}$$


But in all the specific cases I've examined, $\det M$ appears as a factor of $W\_{n,p}(k)$ even when $k\ne p$. For example, with $n=3$, $p=1$ and $k=2$:


$$W\_{3,1}(2)=\sum\_{q=1}^3{M\_{q,1} C\_{q,1}^2 \prod\_{j\ne q}{C\_{j,2}}}=\\
\left(\det M\right)\left(M\_{2,2} M\_{2,3} M\_{3,1}^2 M\_{1,3}^2+M\_{2,1} M\_{2,3} M\_{3,1} M\_{3,2} M\_{1,3}^2-M\_{2,1} M\_{2,2} M\_{3,1} M\_{3,3} M\_{1,3}^2-M\_{2,1}^2 M\_{3,2}
 M\_{3,3} M\_{1,3}^2-M\_{1,2} M\_{2,3}^2 M\_{3,1}^2 M\_{1,3}+M\_{1,2} M\_{2,1}^2 M\_{3,3}^2 M\_{1,3}+M\_{1,1} M\_{2,1} M\_{2,2} M\_{3,3}^2
 M\_{1,3}-M\_{1,1} M\_{2,3}^2 M\_{3,1} M\_{3,2} M\_{1,3}-M\_{1,1} M\_{1,2} M\_{2,1} M\_{2,3} M\_{3,3}^2-M\_{1,1}^2 M\_{2,2} M\_{2,3} M\_{3,3}^2+M\_{1,1}
 M\_{1,2} M\_{2,3}^2 M\_{3,1} M\_{3,3}+M\_{1,1}^2 M\_{2,3}^2 M\_{3,2} M\_{3,3}\right)
$$


It might be worth noting that the second factor here **cannot** be written as a linear combination of products of the cofactors, with purely numeric coefficients. This is in stark contrast to the case $k=p$, when the quotient is simply a product of cofactors.


I am seeking a proof that $\det M$ always divides $W\_{n,p}(k)$, and a general formula for the quotient in the non-trivial case $k\ne p$.



Factor a sum of products of cofactors
This is a proof that $\det M$ divides $W\_{n,p}(k)$.
I can assume that the entries of $M$ are variables of polynomial ring $\mathbb{C}[M\_{11},\ldots,M\_{nn}]$; then, it suffices to prove that points from the (irreducible) variety $\det M=0$ satisfy $W\_{n,p}(k)=0$.


Indeed, let $M'$ be a generic point on this variety; denote by $v^i$ the $i$th column vector of $M'$. Since $M'$ is generic, there is $l\notin\{p,k\}$ such that $v^l=\sum\_{t\neq l} \lambda\_t v^t$ with all $\lambda\_t$ nonzero. Then, the ratio of cofactors $C\_{qp}/C\_{qk}$ equals $\alpha=\lambda\_p/\lambda\_k$ and, in particular, it does not depend on $q$. We conclude $$W\_{n,p}(k)=\sum\limits\_{q=1}^{n}\,M'\_{qp}C\_{qp}^2\prod\limits\_{j\neq q}C\_{jk}=\sum\limits\_{q=1}^{n}\,M'\_{qp}C\_{qp}\alpha\prod\limits\_{j=1}^nC\_{jk}=\alpha\prod\limits\_{j=1}^nC\_{jk}\cdot\left(\sum\limits\_{q=1}^{n}\,M'\_{qp}C\_{qp}\right)=0.$$


