Let $C = (c_{ij})_{n \times n}$ be a real and symmetric matrix and $C_{ij}$ be this matrix with $i$-th row and $j$-th column deleted. Is it true, that
$$ \sum_{ij} (-1)^{i+j} c_{ij} \det C_{ik} \det C_{jl} = (-1)^{l+k} \det C \det C_{kl}$$
for any fixed $k,l$? Can the formula be generalized to non-symmetric matrices?
Suppose for the moment that $C$ is invertible. Let $B$ be its inverse. Then $B=BCB$. Therefore $$ \begin{align} (-1)^{k+l}\frac{\det C_{lk}}{\det C}=b_{kl} &=\sum_{i,j}b_{ki}c_{ij}b_{jl}\\ &=\sum_{i,j}\frac{(-1)^{i+k}\det C_{ik}}{\det C}c_{ij}\frac{(-1)^{l+j}\det C_{lj}}{\det C}\\ &=(-1)^{k+l}\sum_{i,j}\frac{(-1)^{i+j}c_{ij}\det C_{ik}\det C_{lj}}{\det C^2}.\tag{1}\\ \end{align} $$ Hence $$ \sum_{i,j}(-1)^{i+j}c_{ij}\det C_{ik}\det C_{lj} =\det C\det C_{lk}\tag{2} $$ and by a continuity argument, the above identity holds for every square matrix $C$, even when $C$ is singular. Note that there isn't a factor $(-1)^{k+l}$ in $(2)$ because we have cancelled it out on both sides of $(1)$.
When $C$ is symmetric, since $\det C_{ik}=\det C_{ki},\,\det C_{lj}=\det C_{jl}$ and $\det C_{lk}=\det C_{kl}$, you may interchange the order of the indices under each symbol $C$.