Players A en B play a game. They take an empty n-by-n matrix (n > 0) 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. Is there, for a given n, 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 is odd and n > 3?

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?