we will define two quantities on the matrix $A_G$ which are related to 1-factors and Hamiltonian cycles respectively.\
\
**Definition 1.** The permanent of an $n\times n$ matrix $A=(A(i,j))_{n\times n}$ is $$\operatorname{per}(A)=
\sum\limits_{\sigma}\prod\limits_{i=1}^{n}A(i,\sigma(i)),$$ where $\sigma$ ranges over all the permutations of $\{1,\cdots,n\}$.\
\
**Definition 2.** The Hamilton of an $n\times n$ matrix $A=(A(i,j))_{n\times n}$ is defined as $$\operatorname{ham}(A)= \sum\limits_{\{k_1 \cdots,
k_{n-1}\}}A(k_1,1)A(k_2,k_1)\cdots A(k_{n-1},k_{n-2})A(1,k_{n-1}),$$ where $\{k_1 \cdots, k_{n-1}\}$ ranges over all the permutations of $\{2,\cdots,n\}$ when $n\geq 2$, and $\operatorname{ham}(A)=A(1,1)$, if $n=1$.\
By the definition of permanent and Hamilton, it is not difficult to see that $\operatorname{per}(A)\geq \operatorname{ham}(A)$ if the entries of $A$ are all nonnegative. Suppose $A=A_G$. For any permutation $(k_1 \cdots, k_{n-1})$ of $(2,\cdots,n)$, $A_G(1,k_{n-1})$, $A_G(k_{n-1},k_{n-2})$, $\cdots$, $A_G(k_2,k_1)$, $A_G(k_1,1)$ are the edge weight of the Hamiltonian cycle $(1,k_{n-1},\cdots, k_1,1)$ in $G$ if and only if they are all positive. Therefore, we have $$W_H(G)=\operatorname{ham}(A_G).$$ Note that the diagonal entries of $A_G$ are all zero, and for any permutation $\sigma$ over $\{1,2,\cdots,n\}$, $A(i,\sigma(i))> 0$, $i=1,\cdots,n$ if and only if their corresponding edges in $G$ form an 1-factor of $G$. Hence $$W_F(G)=\operatorname{per}(A_G).$$

Next we present the Laplacian expansion formulas for the permanent and the Hamilton.\
\
**Lemma 3.** *Let $A=(A(i,j))_{n\times n}$ be an 