Question on cofactors If the row sums of a symmetric matrix of size 4 by 4 are all $0$, then why are all the cofactors of the matrix equal? 
Thanks in advance for any helpful answers.
 A: There is no reason to consider a $4\times 4$ matrix in particular. When $n=1$, there is only one cofactor so nothing to prove. The result is easy to figure out for $2\times 2$ matrices, and that's maybe a good starting point. 

Then one realizes that more generally, as observed by user1551, it suffices that $A$ be an $n\times n$ matrix over an arbitrary field such that the row sums and the column sums be null (the latter being equivalent to the former for symmetric matrices).

In short: we use the adjugate formula $A\cdot\mbox{adj}(A)=(\det A)I_n$ to prove that the rows and the columns of $C=\mbox{adj}(A)^T$, the matrix of cofactors, are all in the span of  $(1,\ldots,1)$, which is the nullspace of $A$ and $A^T$ in the only nontrivial case.
Proof: the row condition means that the sum of the columns of $A$ is null. Whence the columns the columns are linearly dependent and $\mbox{rank} A\leq n-1$. In particular, $\det A=0$. 
Let $C$ be the matrix of the cofactors of $A$. Then the transpose $C^T$ is the adjugate of $A$ and, as is well-know $AC^T=(\det A)I_n$. Therefore
$$
AC^T=0\quad\iff \quad \sum_{k=1}^na_{ik}c_{jk}=0\quad \forall 1\leq i,j\leq n.
$$
If $\mbox{rank} A\leq n-2$, then all the cofactors are zero and the result follows. So we assume that $\mbox{rank}\, A=n-1$. That is, the nullspace of $A$ has dimension $1$. But we already know that $v=(1,\ldots,1)$ is in $\ker A$ by assumption. So $\ker A$ is the one-dimensional span of $v$. Note that $\ker A^T$ also has dimension $1$ and is spanned by $v$ as-well by assumption.
Now looking at the above, we see that it means that every row $R_j$ of $C$ is in the nullspace of $A$, whence $R_j=r_jv=(r_j,\ldots,r_j)$. 
Since the adjugate matrix of $A^T$ is the transpose of the adjugate matrix of $A$, we get $A^TC=0$ (or simply by taking the tranpose of $C^TA=0$). From that, we deduce that the columns $C_j$ of $C$ are of the form $C_j=c_jv=(c_j,\ldots,c_j)$. 
Starting from $c=c_{1j}$ for every $j$ (first row), we see that $c_{ij}=c$ ($j$th column) for every $i$. That is, all the coefficients of the cofactor matrix are equal as desired. QED.
A: We can prove something more general:

If $A$ is an $n\times n$ matrix over any field such that all its row sums and column sums are zero, then its cofactors are identical to each other. That is, all entries of $\operatorname{adj}(A)$ are equal.

Denote by $M_{ij}$ the submatrix obtained by deleting the $i$-th row and $j$-th column of $A$ and let $m_{ij}=\det M_{ij}$. Now suppose $i<n$ and we want to calculate $m_{ik}$ for some index $k$. Add the first $n-2$ rows of $M_{ik}$ to the last row. Since all column sums of $A$ are equal to $0$, the last row becomes $(-a_{i1}, -a_{i2}, \ldots, -a_{i,j-1}, -a_{i,j+1}, \ldots,-a_{in})$. This resulting submatrix is almost identical to $M_{nj}$, except that the $i$-row of $M_{nj}$ is moved to the bottom and negated. It follows that
$$
m_{ij} = (-1)^{n-i}m_{nj}\tag{1}
$$
and this obviously also holds when $i=n$. Similarly, by using column operations, we have $m_{ij} = (-1)^{n-j}m_{in}$. In particular, when $i=n$ we get
$$m_{nj} = (-1)^{n-j}m_{nn}.\tag{2}$$
Combine $(1)$ and $(2)$, we have $(-1)^{i+j}m_{ij}=m_{nn}$. Hence the result.
A: Here is a simpler proof. In general, for any $n\times n$ singular matrix $A$, $\operatorname{adj}(A)$ is either zero (when the rank of $A$ is $<n-1$) or in the form of $uv^T$ (when the rank of $A$ is $n-1$), where $u,v$ are basis vectors in respectively the right and left null spaces of $A$. 
Now, in your case, $A$ is a singular matrix because its row sums and column sums are zero. If $\operatorname{rank}(A)<n-1$, its adjugate matrix is zero and we are done. If $\operatorname{rank}(A)=n-1$, its left or right null spaces are one-dimensional, and hence both null spaces are spanned by the all-one vector $e$. It follows that $\operatorname{adj}(A)$ is a multiple of $ee^T$ and all cofactors of $A$ are identical.
