If $A$ is skew-symmetric, its determinant remains unchanged when the same number is added to all its entries If we have a skew-symmetric matrix, $A^t = -A$, of size $2n\times 2n$ and we add the same number to every entry in the matrix and take the determinant, i'm told we get the same determinant as $\det A$. 
A couple of small examples have confirmed this to be true for me, but I can't fully see why. By making use of the multilinearity of the determinant function we get that the new determinat is equal to $\det A$ + the determinants of $2n$ matrices, each a permutation of the original matrix with one row swapped with a  row of the added constant. I can't see the symmetry as to why these all cancel to 0. 
 A: Let $E$ be the all-one matrix. We want to show that $\det(A+xE)=\det(A)$ for every scalar $x$. Let
$$
P=\pmatrix{1&-1&\cdots&-1\\ &1\\ &&\ddots\\ &&&1},
\quad P^TEP=E_{11}=\pmatrix{1&0&\cdots&0\\ 0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&0}
$$
and let $B=P^TAP=\pmatrix{0&v^T\\ -v&K}$, where $K$ is $(2n-1)\times(2n-1)$. Then it suffices to prove that
$$
\det(B+xE_{11})=\det(B).
$$
Since $K$ is a skew symmetric matrix of odd dimension, $\det(K)=0$. Therefore, by Laplace expansion along the first row of $B+xE_{11}$, we get $\det(B+xE_{11})=\det(B)+x\det(K)=\det(B)$. The proof is now complete.
A: That follows is valid over any field $K$ of characteristic not $2$ (as the user1551's proof).
Let $A=[a_{i,j}]$. The required formula is a formal polynomial equality in the parameters $(a_{i,j})_{i,j},x$ (that we may considered as transcendental over $K$). Since $n$ is even, $\det(A)$ is the square of a polynomial in the $(a_{i,j})$ that is not identically $0$ (as it is the case when $n$ is odd). Then it suffices to show the result when $A$ is invertible.
$\det(A+xE)=\det(A(I+xA^{-1}E))=\det(A)\det(I+xA^{-1}E)$, that is, $=\det(A)$ when $A^{-1}E$ is nilpotent. 
The previous condition is fulfilled since $rank(E)=1$, $E$ is symmetric and $tr(A^{-1}E)=0$.
