Let $B$ denote the $n\times n$ matrix $I_x-A$. Let $[n]=\{1,2,\ldots,n\}$. For $I,J\subset [n]$ with the same number of elements, let $D_{I,J}$ denote the determinant of the matrix resulting from $B$ by deleting the rows indexed by the elements of $I$ and the columns indexed by the elements of $J$.

Assuming $i\neq j$, we clearly have $P=D_{\varnothing,\varnothing}$, $P_i=D_{\{i\},\{i\}}$, $P_j=D_{\{j\},\{j\}}$, and $P_{ij}=D_{\{i,j\},\{i,j\}}$. By the Dodgson condensation identity, $$ D_{\{i\},\{i\}}D_{\{j\},\{j\}}-D_{\{i\},\{j\}}D_{\{j\},\{i\}}= D_{\{i,j\},\{i,j\}}D_{\varnothing,\varnothing}\ . $$ Since $B$ is symmetric $D_{\{i\},\{j\}}=D_{\{j\},\{i\}}$. As a result $$ \Delta_{ij}=(D_{\{i\},\{j\}})^2\ . $$

In the tree case, the relation between $\Phi_{T-[v_i,v_j]}$ and $D_{\{i\},\{j\}}$ can be worked out using Theorem 1 of my article "The Grassmann–Berezin calculus and theorems of the matrix-tree type" in Adv. Appl. Math. 2004. For those without access to the journal, the preprint versions is here.

Let $B$ denote the $n\times n$ matrix $I_x-A$. Let $[n]=\{1,2,\ldots,n\}$. For $I,J\subset [n]$ with the same number of elements, let $D_{I,J}$ denote the determinant of the matrix resulting from $B$ when deleting the rows indexed by the elements of $I$ and the columns indexed by the elements of $J$.

Assuming $i\neq j$, we clearly have $P=D_{\varnothing,\varnothing}$, $P_i=D_{\{i\},\{i\}}$, $P_j=D_{\{j\},\{j\}}$, and $P_{ij}=D_{\{i,j\},\{i,j\}}$. By the Dodgson condensation identity, $$ D_{\{i\},\{i\}}D_{\{j\},\{j\}}-D_{\{i\},\{j\}}D_{\{j\},\{i\}}= D_{\{i,j\},\{i,j\}}D_{\varnothing,\varnothing}\ . $$ Since $B$ is symmetric $D_{\{i\},\{j\}}=D_{\{j\},\{i\}}$. As a result $$ \Delta_{ij}=(D_{\{i\},\{j\}})^2\ . $$

In the tree case, the relation between $\Phi_{T-[v_i,v_j]}$ and $D_{\{i\},\{j\}}$ can be worked out using Theorem 1 of my article "The Grassmann–Berezin calculus and theorems of the matrix-tree type" in Adv. Appl. Math. 2004. For those without access to the journal, the preprint version is here.

Let $B$ denote the $n\times n$ matrix $I_x-A$. Let $[n]=\{1,2,\ldots,n\}$. For $I,J\subset [n]$ with the same number of elements, let $D_{I,J}$ denote the determinant of the matrix resulting from $B$ by deleting the rows indexed by the elements of $I$ and the columns indexed by the elements of $J$.

Assuming $i\neq j$, we clearly have $P=D_{\varnothing,\varnothing}$, $P_i=D_{\{i\},\{i\}}$, $P_j=D_{\{j\},\{j\}}$, and $P_{ij}=D_{\{i,j\},\{i,j\}}$. By the Dodgson condensation identity, $$ D_{\{i\},\{i\}}D_{\{j\},\{j\}}-D_{\{i\},\{j\}}D_{\{j\},\{i\}}= D_{\{i,j\},\{i,j\}}D_{\varnothing,\varnothing}\ . $$ Since $B$ is symmetric $D_{\{i\},\{j\}}=D_{\{j\},\{i\}}$. As a result $$ \Delta_{ij}=(D_{\{i\},\{j\}})^2\ . $$

Let $B$ denote the $n\times n$ matrix $I_x-A$. Let $[n]=\{1,2,\ldots,n\}$. For $I,J\subset [n]$ with the same number of elements, let $D_{I,J}$ denote the determinant of the matrix resulting from $B$ by deleting the rows indexed by the elements of $I$ and the columns indexed by the elements of $J$.

Assuming $i\neq j$, we clearly have $P=D_{\varnothing,\varnothing}$, $P_i=D_{\{i\},\{i\}}$, $P_j=D_{\{j\},\{j\}}$, and $P_{ij}=D_{\{i,j\},\{i,j\}}$. By the Dodgson condensation identity, $$ D_{\{i\},\{i\}}D_{\{j\},\{j\}}-D_{\{i\},\{j\}}D_{\{j\},\{i\}}= D_{\{i,j\},\{i,j\}}D_{\varnothing,\varnothing}\ . $$ Since $B$ is symmetric $D_{\{i\},\{j\}}=D_{\{j\},\{i\}}$. As a result $$ \Delta_{ij}=(D_{\{i\},\{j\}})^2\ . $$

In the tree case, the relation between $\Phi_{T-[v_i,v_j]}$ and $D_{\{i\},\{j\}}$ can be worked out using Theorem 1 of my article "The Grassmann–Berezin calculus and theorems of the matrix-tree type" in Adv. Appl. Math. 2004. For those without access to the journal, the preprint versions is here.