Consider the congruence subgroup $$\Gamma_0(11)=\left\{\begin{pmatrix}a&b\\11c&d\end{pmatrix}\in M(2,\mathbb{Z}): ad-11bc=1\right\}$$ I want to prove that the family $$\Omega=\{-\text{Id}\}\cup\left\{\begin{pmatrix}a&b\\11c&d\end{pmatrix}\in \Gamma_0(11): c>0,\ gcd(c,6)=1\right\}$$ generates the whole group.
Here is my attempt: let's start by a generic $$\gamma=\begin{pmatrix}a&b\\11c&d\end{pmatrix}\in\Gamma_0(11)$$ My idea is to act with matrices in $\Omega$ until I find an element of this set. If $c\leq 0$ I can act with $-\text{Id}$ to get a matrix with $c\geq 0.$ If I have $c=0$ I also have $ad=1,$ hence $a=d=\pm1.$ Using $-\text{Id}$ I can suppose $a=d=1,$ and acting on the right, for instance, with the matrix $$\begin{pmatrix}2&-1\\11&-5\end{pmatrix}\in\Omega$$ I get $$\begin{pmatrix}1&b\\0&1\end{pmatrix}\begin{pmatrix}2&-1\\11&-5\end{pmatrix}=\begin{pmatrix}2+11b&-1-5b\\11&-5\end{pmatrix}$$ The issue is with the other condition. Suppose $gcd(c,6)\neq1:$ then $2$ or $3$ divide $c:$ how can I manage these cases?
EDIT Writing $c$ in the form $c=2^m\cdot 3^n\cdot C,$ if $m,n>0$ (i.e. $6\mid c$) I can write $\gamma$ using only elements of $\Omega$ in the following way: the determinant condition implies $$(11\cdot2^m\cdot3^n c,a)=1$$ and in particular $a$ is odd and is not a multiple of $3.$ The matrix $$\begin{pmatrix} d&2^m3^n b\\ 11c& a \end{pmatrix}\in\Omega,$$ and $$\begin{pmatrix} d&2^m3^n b\\ 11c& a \end{pmatrix}\begin{pmatrix} a&b\\ 11\cdot 2^m\cdot 3^n c& d \end{pmatrix}=\begin{pmatrix} ad+2^{m}3^n(11c+b)&bd(1+2^m3^n)\\ 11ac(2^m3^n+1)&11bc+ad \end{pmatrix}\in\Omega $$
Maybe I can generalize this approach even for the other cases
A calculation taking a tiny fraction of a second in Sage reveals that $$\{ -\mathrm{Id}\} \cup \left\{\left(\begin{array}{rr} 18 & -5 \\ 11 & -3 \end{array}\right), \left(\begin{array}{rr} 7 & -2 \\ 11 & -3 \end{array}\right), \left(\begin{array}{rr} 8 & -3 \\ 11 & -4 \end{array}\right)\right\}$$ is a generating set of $\Gamma_0(11)$. Since these all have $c = 1$ you are done.