This might be an elementary question. For simplicity, let's assume $G=GL(n,F)$, where $F$ is a local field. Let $U$ be the subgroup of upper triangular unipotents, $A$ the subgroup of diagonal matrices. $\omega_n$ the longest Weyl element. Then $\Omega=U\omega_nAU$ is the open Bruhat cell.
Now I want to ask how to determine if a given matrix $g=(g_{ij})\in \Omega$?
For example, an easy calculation shows that if $g=(g_{ij})\in \Omega$, then $g_{n1}\ne 0$. But I'm not sure if this is also sufficient.
Of course, other characterizations are also welcome.
The elements of the open Bruhat cell are precisely the matrices where the determinant of the $k\times k$ minor of the bottom left hand corner of the matrix is non-zero for all $k\leq n$. You noted one of these conditions for $k=1$. This condition is true for $\omega_nA$, and won't be changed by multiplication by upper triangular matrices. You can check this is sufficient by solving for the upper triangular matrices.