Adjacency matrix of tournament I met a question in Bondy’s Graph theory (§1.5.13)

Let $\mathbf{A}$ be the adjacency matrix of a tournament on $n$ vertices. Show that $\operatorname{rank}\mathbf{A}=n-1$ if $n$ is odd and $\operatorname{rank}\mathbf{A}=n$ if $n$ is even.

which I think is completely wrong. However, are there some similar conclusions related to the rank of $\mathbf{A}$? For example, when is the rank of $\mathbf{A}$ equal to $n-1$?
 A: The following two papers give the lower bound $n-1$ on the rank of $n\times n$ tournament matrices over fields of characteristic zero. Here a tournament matrix $M$ is a $\{0,1\}$-matrix, with zeros on the diagonal, satisfying $M+M^T=J-I$.
D. de Caen, The Ranks of Tournament Matrices, The American Mathematical Monthly, Vol. 98, No. 9 (Nov., 1991), pp. 829-83. https://doi.org/10.2307/2324270
D. de Caen and D. G. Hoffman, Impossibility of decomposing the complete graph on n points into (n - 1) isomorphic complete bipartite subgraphs, SIAM J. on Discrete Math., 2 (1989) 48-50. https://epubs.siam.org/doi/10.1137/0402005
A: Here is a proof for a $(1,-1)$-version. In other words, an antisymmetric $n\times n$ matrix with not-diagonal entries $1$ and $-1$ has rank $n$ if $n$ is even and $n-1$ if $n$ is odd.
First of all, if $n$ is odd, the rank is at most $n-1$ since antisymmetric matrix of odd order is always singular, and its principal minor of order $n-1$ is an antisymmetric matrix of even order $n-1$. This reduces the odd case to the even case, so let further $n$ be even. I claim that the rank is full even modulo 2, in other words, the determinant is odd. Indeed, if you add all other columns to the $n$-th column, it becomes (modulo 2) the all-1's column. Now adding it to the first $n-1$ columns you get a unitriangular matrix.
