In another post, I'm trying to understand the author's logic in a certain line of the accepted answer.
"Now one uses that $nI-J=L(K_n)$, i.e. the Laplacian of the complete graph, hence $\text{adj}(nI-J)$ equals $n^{n-2}J$." Also, $J$ is a matrix entirely filled with ones.
From observation, it seems that the determinant of $L(K_n)$ is $0$, so one could use the previously established fact that $JL=LJ=0$, where $L$ is any Laplacian matrix, and show that $\text{adj}(L(K_n))=J$. The factor of $n^{n-2}$ is arbitrary and is included out of convenience for the rest of the proof.
However, if this is the case how would one show that $\text{det}(L(K_n))=0$?
The result comes from the matrix tree theorem which says that the number of spanning trees of the graph is given by any cofactor of the Laplacian, so that the adjugate is $tJ$ where $t$ is the number of spanning trees. The fact that $t=n^{n-2}$ in this case follows from a theorem of Caley, which is proved in the linked Wikipedia article.
The Laplacian of any graph is singular, since the row sums are all $0$.