How to judge a determinant equals non-zero in cryptography?

Question

In the paper [1,p5], the authors construct a matrix to find out $n$ solutions of a equation set as follows:

where $c_i$ is generated by the adversary, $c$ is picked by the simulator. Note that $c^i$ means different forgeries instead of exponent.

My question is, they find out $n$ solutions iff the determinant equals non-zero, but they don't prove why the determinant equals non-zero. There is no need to prove it?

(I'm not true this is a crypto problem or not, it seems more like a math problem?)

[1] Liu, J. K., Au, M. H., Susilo, W., & Zhou, J. (2014). Linkable ring signature with unconditional anonymity. IEEE Transactions on Knowledge and Data Engineering, 26(1), 157–165. https://doi.org/10.1109/TKDE.2013.17

Answer 1

For linear system $Ax = \beta$ with a vector on $n$ unknowns $x$, $A$ is the Vandermonde matrix, produced with powers of $c_i$. For such a matrix, determinant is a product of $(c_i - c_j)$, and so is non-zero for different $c_j$ in the set of field elements.

Answer 2

Consider the matrix as follows

$$ \begin{pmatrix} 1 & c_1^{(1)} & c_2^{(1)} & ... & c_n^{(1)}\\ 1 & c_1^{(2)} & c_2^{(2)} & ... & c_n^{(2)}\\ ... & ... & ... & ... & ...\\ 1 & c_1^{(n+1)} & c_2^{(n+1)} & ... & c_n^{(n+1)} \end{pmatrix} $$

If all $c_i$ for $i\in[n]$ are determined after $H_Z$ query, we can say that they are generated from $c=H_Z$ which is randomly picked by $\mathcal{B}$. Then we know that the determinant of this matrix $det(M)=0$ with only negligible possibility.