A transversal in an $n$ by $n$ Latin square is a set of $n$ element, each from a different row and a different column, in which every symbol (value) appears exactly once.
Prove that for even $n$, the Latin square below does not have any transversal: \begin{bmatrix}1&2&\dots&n-1&n\\2&3&\dots&n&1\\\vdots&\vdots&\ddots&\vdots&\vdots\\n&1&\dots&n-2&n-1\end{bmatrix}