TITLE: Variation on a matrix game
QUESTION [19 upvotes]: The original problem appeared on last year's Putnam exam:
"Alan and Barbara play a game in which they take turns filling entries of an initially empty 2008×2008 array. Alan plays ﬁrst. At each turn, a player chooses a real number and places it in a vacant entry. The game ends when all the entries are filled. Alan wins if the determinant of the resulting matrix is nonzero; Barbara wins if it is zero. Which player has a winning strategy?"
It's not hard to see that Barbara can win this game by reflecting Alan's moves over a vertical line. (In fact, you might say she "wins with multiplicity 1004".) My question is, what if the goals were reversed? That is, suppose Alan (the first player) wants the determinant to be zero, and Barbara wants it to be nonzero. Now who has the winning strategy?
If you expect the result to rely solely on parity, then you should note that Alan wins in the 2×2 case, because he can force a row or column to have only zeroes. Unfortunately, it's not at all clear (to me, anyway) that he can do anything similar to a 4×4 matrix, let alone a 2008×2008 one.

REPLY [3 votes]: Let me expand upon my comment. Let us say we are dealing with a 4x4 matrix Alan is restricted to integers, Barbara can use any rational numbers. Alan moves first and wants to force the matrix to have an integral determinant. Alan moves first Barbara moves second and her entry is 1/2 in the same row as Alan the look at the matrix with the row Alan moved first in and the column Barbara moved first in deleted. Suppose Alan moves next in that matrix then Barbara moves in the same row and then the row Alan moved in and the column Barbara moved in is deleted or Alan doesn't move in the matrix still Barbara moves in the derived matrix and again her entry is 1/2 in the same row as Alan. In either case the row Alan moved in and the column Barbara moved in are deleted and we get a new derived matrix in which the process is repeated until there are 4 elements in the matrix with entry 1/2 whose product is in the determinant adding or subtracting 1/16. Barbara makes sure that the rest of her entries are integral as a result the 1/16 cannot be canceled out since the all other contributions to the determinant can be expressed as fractions with denominator 8 or less and the determinant is not integral. This idea can be extended in various ways depending on the restiction on Alan's moves.