Kirchhoff's matrix-tree theorem

Delete row $i$ from $B$ to obtain an $(n-1)\times \mathrm{\mid }E(G)\mathrm{\mid }$ matrix $B^{(i)}$. By [L1], the matrix $L^{(i)}$ obtained by deleting row $i$ and column $i$ from $L(G)$ is $L^{(i)}=B^{(i)}(B^{(i)}{)}^{\mathsf{T}}$. Applying [L3] gives $\det L^{(i)}={\sum }_{\mathrm{\mid }S\mathrm{\mid }=n-1}\det (B^{(i)}[S]{)}^2$, where $S$ ranges over all $(n-1)$-edge subsets of $E(G)$.

Fix such a set $S$, and let $G_S$ be the spanning subgraph with edge set $S$. If $G_S$ is disconnected, then the vertex-indicator vector of the component of $v_i$ gives a nonzero linear relation among the rows of $B^{(i)}[S]$, so $\det (B^{(i)}[S])=0$. If $G_S$ is connected, then $G_S$ has $n-1$ edges and is therefore a tree by [L2]. In that case choose the orientation of every edge of $G_S$ away from the root $v_i$ and order the nonroot vertices so that every parent precedes its children. In this order the matrix $B^{(i)}[S]$ is triangular with diagonal entries all $\pm 1$, hence $\det (B^{(i)}[S])=\pm 1$. Therefore $\det (B^{(i)}[S]{)}^2$ is $1$ when $S$ is the edge set of a spanning tree and $0$ otherwise.

Substituting step 1.2 into the sum of step 1.1 shows that $\det L^{(i)}$ counts exactly the spanning trees of $G$, namely $\tau (G)$ by [F1]. Since the right-hand side does not depend on $i$, every principal cofactor of the Laplacian is equal.