Variation on a matrix game 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.
 A: I also can't do the entire problem, but I can handle a more general case than the other answer: Barbara wins if Alan plays only 0s on a board of the form (4n)x(4n).  (This is more general because it considers all possible ways of having 0s force a determinant to be zero, not just rows and columns.  Admittedly, it's less general because it requires more of the size of the matrix.)
First, consider the 4x4 case.  Label the matrix as follows:
aacd
bbcd
efgg
efhh
This pairs up the entries of the matrix.  Then, Barbara's strategy is fairly easy: play in the matching pair of wherever Alan plays.  One can check that no matter how Alan plays, there will be four entries of Barbara's whose product contributes to the overall determinant.  If Barbara plays algebraically independent entries, then this implies the entire determinant is nonzero.
Extending this to (4n)x(4n) is fairly easy.  Make n 4x4 blocks along the diagonal of the big matrix.  If Alan plays in one of them, Barbara plays by the above strategy locally.  If he plays in an off-diagonal block, Barbara simply helps Alan by playing 0 in that block.  The end result will be a block diagonal matrix with nonzero determinant.
A: While I still can't answer in the general case, in the case where n > 2 and Alan moves only with 0 in attempt to fill a row or column with 0s, he cannot win. 
This proof is written semi-informally for ease of reading.

Call Alan's moves 0s and Barbara's moves Xs.
A "blocked" row or column is a row or column that Barbara has at least one X. A "unblocked" row or column is free of Xs.
For Alan to win on an nxn grid, after his move is complete there needs to be at least one row with n-1 unblocked 0s and at least one row with n-1 unblocked 0s.
Define a set R which contains, for each unblocked row, the number of 0s on that row.
Define a set C which contains, for each unblocked column, the number of 0s on that column.
For our the explanation that follows we will write set R followed by set C. For example, if R={2,1} and C={1,3} the sets will be written as {2,1} {1,3}.
The game begins with both R and C as the empty set.
    
 (source)


*

*(1st move) Alan moves and the sets become {1} {1}.

*(2nd move) Barbara moves to block a row and the sets become {} {1}.

*(3rd move) Alan has three choices, let us consider each:


*

*[Case 1]
Alan moves in the same column as the existing unblocked 0. The sets become {1} {2}
Barbara moves in the same column as the two unblocked 0s. The sets become {1} {}.

*[Case 2]
Alan moves in a square that is blocked in both row or column. Barbara blocks the unblocked column and the sets become {} {}.

*[Case 3]
Alan moves in a square that contains no 0s or Xs in both the row and column. The sets become {1} {1,1}.
    
 (source)
Now the situation is as in the diagram above. Suppose the 0s are in A and D, and B and D are empty. One of the rows must be blocked; suppose it is the same row as A. Then Barbara
moves at C and the sets become {} {1}. If the blocked row is the same row as D, Barbara moves at B and the sets become {} {1}.
The situation if the 0s are at B and C is symmetrical.
Now note that all the cases are either identical to an earlier position of the game or are symmetrical to an earlier position. Therefore Alan can never win the game.
A: 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.
A: Here's an almost-proof of parity which I believe makes the situation clearer.
Consider the last move. Call it x.
Starting with the equation where the determinant is set to zero,
$a_1 + ... + a_{(n-1)\Gamma(n)}+ xb_1 + ... + xb_{(n-1)!}= 0$
$a_1 + ... + a_{(n-1)\Gamma(n)}= - xb_1 - ... - xb_{(n-1)!}$
$a_1 + ... + a_{(n-1)\Gamma(n)}= -x (b_1 + ... + b_{(n-1)!})$
$-\frac{a_1 + ... + a_{(n-1)\Gamma(n)}}{b_1 + ... + b_{(n-1)!}}= x$
Therefore to win the person with the last move simply needs to play
$-\frac{a_1 + ... + a_{(n-1)\Gamma(n)}}{b_1 + ... + b_{(n-1)!}}$
If n is even, the last move is Barbara's so she wins, if n is odd, it is Alan's so he wins. 
The only thing foiling this strategy in general is that ${b_1 + ... + b_{(n-1)!}}$ might be zero.
