Which reasoning is used here to prove the zero value for each of the polynomial functions?

Question

I am reading a research paper which has a square matrix $\mathbb{G}$ defined as $$\mathbb{G}=[\mathbb{v}~\mathbb{V}_2 ~\mathbb{Q}]$$ where $\mathbb{v}$ is a column matrix with $n$ elements and is equal to $[g_1v_1~g_2v_2~\cdots g_nv_n]^T$. In this case if $\mathbb{G}$ is rank deficient then $\det[\mathbb{G}]=0$ which means $$\det[\mathbb{G}]=\sum_{i=1}^n{g_i}f_i(\mathbb{z,V_2,Q})=0$$ where $\mathbb{z}=[v_1~v_2\cdots v_n]^T$ and $f_i(\mathbb{z,V,Q})$, $i=1\cdots n$, are the co-factors that only depend on $\mathbb{z,V_2,Q}$. Now in that paper it is mentioned that if $\det[\mathbb{G}]$ is $0$ then it is $0$ regardless of the values of $g_i$'s are therefore we can choose all $g_i$'s to be $1$. With this change we have $$\det[\mathbb{G}]=\sum_{i=1}^nf_i(\mathbb{z,V_2,Q})=0$$ After this they say that $\det[\mathbb{G}]=0$ means every $f_i(\mathbb{z,V_2,Q})$ must be zero. I do not understand why every $f_i(\mathbb{z,V_2,Q})$ must be zero (especially when they assume that all $g_i$'s are $1$). Any help in understanding this point will be much appreciated. Thanks in advance.