Reading about graph theory they said that, for the Laplacian matrix of a graph, let's call it $A$, its adjoint $Adj(A)$ since $A$ is symmetric, ,
Here is an example of Laplacian matrix $A$ and its adjoint matrix.
$$A=\begin{pmatrix}2&-1&0&-1\\-1&3&-1&-1\\0&-1&1&0\\-1&-1&0&2 \end{pmatrix} \to Adj(A)=\begin{pmatrix} 3&3&3&3\\3&3&3&3\\3&3&3&3\\3&3&3&3 \end{pmatrix}$$
I can understand that when looking for the determinant of, for example, the entry $(3,1)$ It will be composed of the same entries as those of the determinant of its symmetric element $(1,3)$ since in both cases the entries of the main diagonal are eliminated in the same order and the remaining symmetric elements are taken
$$A_{(3,1)}=\begin{pmatrix}-1&0&-1\\3&-1&-1\\-1&0&2 \end{pmatrix} \to A_{(1,3)}=\begin{pmatrix} -1&3&-1\\0&-1&0\\-1&-1&2 \end{pmatrix}$$
But I do not understand how it is that The determinants of the main diagonal are equal to each other since there is no relationship that leads to that result.
$$A_{(1,1)}=\begin{pmatrix}3&-1&-1\\-1&1&0\\-1&0&2 \end{pmatrix} \to A_{(3,3)}=\begin{pmatrix} 2&-1&-1\\-1&3&-1\\-1&-1&2 \end{pmatrix}$$
As seen in $Adj(A)$ the cofactors of $A$ are all equal to $3$, therefore each submatrix presented has $det(A_{(i, i)})=3$.
I imagine that it will have to do with the type of entries in the matrix and the way in which they are related in each permutation of the order to obtain the cofactors, but I don't know what that relationship is and hence my doubt.
How do I prove that, if a matrix is symmetric, its adjoint is also symmetric?
The Graph Laplacian $A$ is always singular, because $Ae=0$ where $e$ denotes the vector of ones. In general, for $n\times n$ matrix over a general field, if $\operatorname{rank}(A)<n-1$, then $\operatorname{adj}(A)=0$. If $\operatorname{rank}(A)=n-1$, then $\operatorname{adj}(A)$ is a nonzero scalar multiple of $uv^T$, where $\ker(A)=\operatorname{span}\{u\}$ and $\ker(A^T)=\operatorname{span}\{v\}$. Therefore, in your case, $\operatorname{adj}(A)$ is always a (possibly zero) scalar multiple of $ee^T$ (and this scalar multiple must be nonnegative, as every $(n-1)$-rowed principal submatrix of $A$ is an M-matrix). Hence all elements of $\operatorname{adj}(A)$ are equal to each other.