Suppose a $6 \times 4$ matrix satisfies the following
where $\alpha, \beta, \gamma, \theta, \sigma, \mu$ are non-zero. What are the conditions should be added so that any $4 \times 4$ submatrix has full rank?
I think the conditions are $\alpha_i \neq \beta_j \neq \gamma_k$ and $\theta_i \neq \sigma_j \neq \mu_k$ for $i,j,k=1,2,3$, in other words, all entries of both rank $3$ matrices are distinct. But I don't know whether the conditions are sufficient to conclude the statement. Can anyone help me?
EDIT: Okay, so the conditions stated above are not sufficient. Can I use an $6 \times 4$ Cauchy matrix instead? Because any square-submatrix of a cauchy matrix has full rank. But in this case, we have a few zeros in the matrix. So I don't know whether these zeros will affect the rank of submatrix or not. Also, what if I change to finite field?