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 $k\ne p$.