Let $R$ be a commutative Noetherian ring. Let $f: R^{\oplus a}\to R^{\oplus b}$ be an $R$-linear map. Let $\{e_i\}_{i=1}^a$ and $\{e_i\}_{i=1}^b$ be the standard bases of $R^{\oplus a}$ and $R^{\oplus b}$ respectively and let $A$ be the matrix of $f$ with respect to this bases. Let $\{\beta_i\}_{i=1}^a$ and $\{\alpha_i\}_{i=1}^b$ be some bases of $R^{\oplus a}$ and $R^{\oplus b}$ respectively and let $B$ be the matrix of $f$ with respect to this bases. Let $I_A$ and $I_B$ be the ideals generated by the entries of $A$ and $B$ respectively.
Then, is it true that $I_A=I_B$?
Let's denote by $I_r(A)$ the ideal generated by all $r$-minors of $A$. Note that $I_r(MN)\subseteq I_r(M)I_r(N)$. (Use the multilinearity of the determinant or see here for a reference.)
In your case there are two invertible matrices $P,Q$ such that $B=PAQ$. Then $$I_r(B)=I_r(PAQ)\subseteq I_r(P)I_r(A)I_r(Q)\subseteq I_r(A).$$ (The converse should be obvious now.)