My thought process went like this:
I relied on the permutation expansion of the determinant
In det A, we get two groups of factors: ones that include a_11 and ones that don't
We may ignore all the terms in which one of the numbers is zero
But since, apart from a_11, if a_ij is 1 then i+j is odd, this means that we only count the permutations in which sigma(i) has parity different than i
(except possibly for i=1)
And now I look at odd an even n separately
Oh, before that
This implies that sigma decomposes into a product of even length cycles
And no fixed points except possibly 1
For even n, all terms containing a_11 must vanish, since they would contain another fixed point or odd-length cycle
For odd n, all terms not containing a_11 similarly vanish
So for even n, we may assume a_11=0, since no terms containing it contribute anything
For odd n, we may pass to the submatrix we get by removing first row and column
SO
We may assume the matrix has even dimensions and a_11=0
And the way I did it was
If we look at (odd,even) entries and (even,odd) entries, we get two identical matrices, two which we can apply row and column operations independently
So we can transform them into a diagonal form
And then we get a matrix which has only 2x2 blocks on the diagonal
(others are zero)
And each block has the form [[0,a],[a,0]], which has determinant -a^2
And now we just multiply