Putnam 2011, Problem B-4

In a tournament, 2011 players meet 2011 times to play a multiplayer game. Every game is played by all 2011 players together and ends with each of the players either winning or losing. The standings are kept in two $2011 \times 2011$ matrices, $T = (T_{hk})$ and $W = (W_{hk})$. Initially, $T = W = 0$. After every game, for every $(h,k)$ (including for $h = k$), if players $h$ and $k$ tied (that is, both won or both lost), the entry $T_{hk}$ is increased by 1, while if player $h$ won and player $k$ lost, the entry $W_{hk}$ is increased by 1 and $W_{kh}$ is decreased by 1.

Prove that at the end of the tournament, $\det(T + iW)$ is a non-negative integer divisible by $2^{2010}$.

Solution:
Number the games \(1, \dots, 2011\), and let \(A = (a_{jk})\) be the \(2011 \times 2011\) matrix whose \(jk\) entry is \(1\) if player \(k\) wins game \(j\) and \(i = \sqrt{-1}\) if player \(k\) loses game \(j\). Then \(\overline{a_{hj}} a_{jk} = 1\) if players \(h\) and \(k\) tie in game \(j\); \(i\) if player \(h\) wins and player \(k\) loses in game \(j\); and \(-i\) if \(h\) loses and \(k\) wins. It follows that \(T + iW = \overline{A}^T A\).  
Now the determinant of \(A\) is unchanged if we subtract the first row of \(A\) from each of the other rows, producing a matrix whose rows, besides the first one, are \((1 - i)\) times a row of integers. Thus we can write \(\det A = (1 - i)^{2010}(a + bi)\) for some integers \(a,b\). But then \(\det(T + iW) = \det(\overline{A}^T A) = 2^{2010}(a^2 + b^2)\) is a nonnegative integer multiple of \(2^{2010}\), as desired.