Question about rank of matrix $2\times n$

Question

I have matrix $A$ with nonnegative integers elements with $2$ rows and $n$ columns ($n\geq 2$). I have $n+1$ linear equations between minors of matrix $A$ of the form $$ \sum_{0\leq i,j\leq n} c_{ij}^{(k)}A_{ij} = 0,\ k=1,2,\ldots,n+1 $$ where $c_{ij}^{(k)}$ some constants (in my case $\pm 1$ or zero) in $k$-th equation, $A_{ij}$ - minor with $i$-th and $j$-th columns of $A$. These equations are lineary independent.

I have a questions: Is these equations suffice to check that rank of $A$ is equal to $1$? If yes: how to proove it?

Answer 1

No.

For a counterexample take $n=4$, take $$ A= \begin{pmatrix} 1 & 0 & 0 &0\\ 0 & 1 & 0 & 0\\ \end{pmatrix}. $$

There are six minors $M_k$ say, $M_1$ of value $1$ and the others of value $0$. There are $4+1$ linearly independent equations $$ M_2=M_3=M_4=M_5=M_6=0. $$

Answer 2

I am interested in answering the same question as before but with additional assumption that each column is different than zero vector, i.e. $$ [a_{1i},a_{2i}]^T\neq [0,0]^T,\ i=1,2,\ldots n $$

Is the answer in this case also "no"?