Binary Chess Returns!

The main diagonal will have $8$ slots of which $7$ are filled. 2 remaining ones are to be filled in any of the other slots such that the determinant is not zero. This is only possible if one of these is in the same row as the missing one. There would exist only one possible place for the second one for each position of the first one such that the matrix is symmetric. There would be $7$ places in the row to place the extra one for each slot missing in the diagonal. $7\times8=56$

Up to exchanging variables (i.e. rows and columns simultaneously), the two possible ways to place the ones and zeroes are $\left(\begin{array}{cccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right)$ and $\left(\begin{array}{cccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right).$ In the first case the determinant is $-1$ (invertible), in the second case the determinant is 0 (not invertible).

Basic combinatorics shows that there are $7\times 8$ ways to permute the first matrix. The answer is therefore 56.