Can such a matrix exist? During my work i came up with the following problem:

I am trying to find an $n \times n$ $(0,1)$-matrix $M$, for any $n &gt; 3$, with the following properties:


The determinant of $M$ is even.
For any non-empty subsets $I,J\subseteq\{1,2,3\}$ with $|I| = |J|$, the submatrix $M^I_J$ has odd determinant if and only if $I=J$.  


Here $M^I_J$ denotes the submatrix of $M$ created by removing the rows with indices in $I$ and the columns with indices in $J$.

So far, I tried to find such a matrix via random sampling but I am only able to find a matrix that has all properties except the first one, i.e., the matrix always has an odd determinant. I tried various dimensions and different input/output sets without any success. So this makes me think: 

Is that there is a dependency among the requirements, which prevents them from being simultaneously true?

or

Is it possible that such a matrix exists and can someone give me an example?

Thanks,
Etsch
No such matrix exists.

The Desnanot-Jacobi identity says that for $i \neq j$, 
$$
\det M_{ij}^{ij} \det M = \det M_i^i \det M_j^j -\det M_i^j \det M_j^i
$$
so using this, we get
$$
\det M_{12}^{12} \det M = \det M_{1}^{1} \det M_{2}^{2} - \det M_{1}^{2} \det M_{2}^{1}
$$
But your requirements force the left-hand-side to be 0 (mod 2) and the right-hand-side to be 1 (mod 2), showing they are incompatible.