Players $A$ and $B$ take an empty $n \times n$ matrix and place, one by one, an element (say, a rational number) in an unoccupied place of this matrix. Player $A$ starts. The game ends if there is no move left. Player $A$ wins if the matrix is invertible; player $B$ wins if it is not.
For a given $n > 0$, is there a winning strategy for one of the two players?
It is not hard to show that for $n = 3$, player $A$ can win. Also if $n$ is even player $B$ has a winning strategy. But what if $n > 3$ is odd?
Best Answer
Player A wins the trivial n=1 case by playing any non-zero number in (1,1). For all even n, player B wins by using symmetry a la a horizontal mirror. As Ben points out in the comments, if n = 3, player B can force a win. I had a long demonstration written out, but I decided against it (if you want, I can put it in later). Anyway, as for the general case, after a little searching, I found a paper called "A determinantal version of the Frobenius - König Theorem" by D. J. Hartfiel and Raphael Loewy, which can be purchased here.
The abstract, at least, says that given an n by n matrix A of, say, rational numbers, if the determinant is zero, then A must contain an r by s submatrix B such that r + s = n + p, and rank(B) ≤ p - 1 (no more than p - 1 linearly independent rows), for some positive integer p. This means that if we have, say, a 5x5 matrix whose determinant is zero, then there exists a submatrix B in A such that B is:
While it doesn't say so explicitly, I think that it's a biconditional, so if player B manages to get one of these in the matrix, then she will win. However, even if it isn't biconditional, if player A can prevent any of those forming, he will win.
Of these two, I believe it would be easier for player A to prevent any of these forming than it would be for player B to force one of these, but I haven't given that in particular a great deal of thought. I hope this is helpful.