Given a system of $n$ linear equations $$ \displaystyle \sum_{j=1}^n a_{ij}x_j = 0, \quad i=1,\dots, n $$ for some finite $n$ and unknowns $x_1,\dots,x_n$, the statement that the system has a non-zero solution implies that the determinant of the coefficient matrix is $0$, that is $$ \det(A)=0,\quad A=(a_{ij})_{i,j=1}^n \,. $$
Question: Is there any analogous statement for non-linear systems?
For example, consider the system of $n-1$ equations $$ \displaystyle \sum_{j=1}^{n} a_{ij}x_j = 0, \quad i=1,\dots, n-1 $$ but with the definition $x_{n} = f(x_1)$ where $f$ is some function.
If my system has a non-zero solution for $x_1,\dots,x_{n-1}$ is there an equivalent statement that the determinant of the coefficient matrix must vanish?
Based on some experiments, it seems to be that if I define $A_1 = (a_{ij})_{i,j=1}^{n-1}$ and $A_2 = (a_{ij})_{i,j=2}^{n}$ then the existence of a non-trivial solution requires both $A_1$ and $A_2$ to have a non-vanishing determinant. Of course, from this I can guess that if I have a system of $n-k$ equations on $n$ unknowns but with $x_{n+1-k}=f_k(x_k)$ then I would require that all minors of $A$ of size $n-k$ have non-vanishing determinant.
However, I am not interested in non-vanishing determinants. I would like to know is there anything which must vanish.