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Alan and Barbara play a game in which they take turns filling entries of an initially empty $ 2008\times 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 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?

This question has been posted before.

I saw a solution recently this way:

- Show that $\sum_{i=0}^{n-1} 2^i = 2^n - 1$
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- Random $0-1$ matrices

When Alan places a number in any spot in the array, Barbara places the same number in the column, but one row up or down. That was she forces the rows to be linearly dependent, and the resulting determinate is zero.

This method works fine and I tried with arbitrary values in matrices of smaller rank.

But I need help in proving that this indeed forms linearly dependent rows which leads the determinant of the matrix to be zero?

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Group the rows into pairs as $\{1,2\},\ldots \{2007,2008\}$. In this way, if Alan places a number, Barbara will place the same number into the other paired row. This means that one row is identical to another row (in fact, by construction, every row is a duplicate of another row). Since you end up with two equal rows, the rows are linearly dependent (hence determinant zero).

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