Blockquote
Let $k$ be a field. Let $I \subset k[x_1,...,x_n]$ be an ideal and let $f_1,...,f_s \in k[x_1,...,x_n]$.
Using the fact that the following are equivalent
i) $f_1,...,f_s \in I$
ii) $\langle f_1,...,f_s \rangle \subset I$
show that $\langle 2x^2+3y^2-11,x^2-y^2-3\rangle = \langle x^2-4,y^2-1\rangle$.
I have the following:
$2(x^2-4)+3(y^2-1)=2x^2+3y^2-11 \in \langle x^2-4,y^2-1\rangle$
and
$x^2-4-(y^2-1)=x^2-y^2-3 \in \langle x^2-4,y^2-1\rangle$
Thus by stated fact, I have the inclusion $\langle 2x^2+3y^2-11,x^2-y^2-3\rangle \subset \langle x^2-4,y^2-1\rangle$.
Hint $\,\ \left[\begin{array}{rr} \color{#0a0}2 & \color{#c00}{3} \\ 1 & -1\end{array}\right] \overbrace{\left[\begin{array}{r} x^2-4\\ y^2-1\end{array}\right]}^{\large f_{\large i}} = \overbrace{\left[\begin{array}{r} 2x^2-3y^2-5\\ x^2-y^2-3\end{array}\right]}^{\large g_{\large i}}\ $ and it is invertible, having determinant $\ne 0$.
The above shows $\,g_1 = \color{#0a0}2f_1+\color{#c00}{3}f_2 \in \langle f_1,f_2\rangle\,$ and $\,g_2 = f_1 - f_2 \in \langle f_1,f_2\rangle,\ $ so $\ \langle g_1,g_2\rangle\subseteq \langle f_1,f_2\rangle.\,$ Conversely, multiplying by the inverse of the matrix, shows that the $\,f_i$ are $k$-linear combinations of the $\,g_i,\,$ yielding the reverse containment $\,\langle f_1,f_2\rangle\subseteq \langle g_1,g_2\rangle.\,$ Therefore $\,\langle f_1,f_2\rangle = \langle g_1,g_2\rangle$
Remark $\ $ Generally, as above, ideals are preserved by an invertible linear basis transformation.