Given that $\Gamma_0(6)$ is generated by matrices in the set
$$G = \left\{\left(\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right), \left(\begin{array}{cc} 7 & -3\\ 12 & -5 \end{array}\right),\left(\begin{array}{cc} 5 & -1\\ 6 & -1 \end{array}\right),\left(\begin{array}{cc} -1 & 0\\ 0 & -1 \end{array}\right) \right\}$$ Show that
$$ \eta(\tau) := e^{\frac{\pi iz}{12}}\prod_{n=1}^{\infty} (1-q^n) $$ where $q=e^{2\pi iz}$, is a modular form of weight $\frac{1}{2}$ for $\Gamma_0(6)$ through direct computation.
I have already proved the necessary analytic properties, my question is how to show the weight by computation. My approach has been the following: showing that $\eta(\tau)$ is weight $\frac{1}{2}$ is equivalent to showing that the following relation is true $$ \eta(\gamma \tau) = \epsilon(a,b,c,d)(-i(c\tau+d))^{\frac{1}{2}}\eta(\tau) $$ where $\gamma = \left(\begin{array}{cc} a & b\\ c & d \end{array}\right)\in \Gamma_0(6), \, \, \tau \in \mathbb{H}$ and $\epsilon(a,b,c,d)$ is an exponential that depends on $\gamma$. Since $\Gamma_0(6)$ is generated by $G$, we need only check the above relation for the elements of $G$. Is there a tractable way to do this through manipulation of the left hand side of the relation? I.e. manipulating $\eta\left(\frac{5\tau - 1}{6\tau - 1}\right)$ to look like $e^{\frac{\pi i}{3}}(-i(6\tau-1))^{\frac{1}{2}}\eta(\tau)$? I have done this for the first and last matrices in $G$ but can't seem to find a suitable method to do so for the second and third. Is there a relation with Euler's pentagonal number theorem that I can take advantage of? Please keep in mind I have only begun reading about modular forms and only have an undergraduate knowledge of analysis (I am currently a junior).