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Let $n \in \Bbb N, n \gt 1$ and $X \in M_n(\Bbb R)$ a matrix whose elements are chosen by two players ($A$ and $B$) one after another, starting with $A$. The game is won by $A$ if $\det X \ne 0$ and by $B$ in the other case. 1. Show that if $n$ is even, $B$ has a winning strategy.
2. Study whether $A$ or $B$ has a winning strategy in the case that $n$ is odd.

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    How exactly the players choose the matrix element positions? Do they pick them freely or fill the matrix row by row and column by column starting from the top left corner? In the former case B obviously wins for any $n>1$. – 2017-02-04

1 Answers 1

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I'll do problem $1$ (the easier of the two).

Suppose $n$ is even.

Then $B$ can force rows $1$ and $2$ to be equal, using the following strategy . . .

Whenever $A$ places a value in row $1$, column $j$, $B$ responds by placing the same value in row $2$, column $j$.

Similarly, whenever $A$ places a value in row $2$, column $j$, $B$ responds by placing the same value in row $1$, column $j$.

If $A$ places a value somewhere other than rows $1$ or $2$, $B$ can also avoid rows $1$ and $2$. In this case, $B$ can use any value and place it in any free cell, as long as it's not in row $1$ or row $2$.

Since $n$ is even, the number of cells not in rows $1$ or $2$ is even, so $B$'s strategy is never blocked.

Thus at the end, the first two rows will be equal, so the determinant will be $0$, hence $B$ wins.

The case where $n$ is odd seems harder, and at this point, I don't have an answer, although my gut feeling is that $A$ has the advantage.