I am working with the dihedral group $D_{8}$ of order $8$, given by the presentation $$D_{8} = \left\langle a,b: a^{4}=b^{2}=1,\: b^{-1} ab = a^{-1}\right\rangle.$$
I am trying to find if this representation is faithful or not: $\rho: D_{8} \to GL(2, \mathbb{C})$ satisfying $$\rho(a)=\left(\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right) \quad \text { and } \quad \rho(b)=\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right).$$
Is there a better way to do this rather than check if each individual element maps to the identity? With $8$ elements, it's not too hard, but this technique quickly becomes more tedious with larger groups.
As already stated in the comments, $\rho(a)$ has order $2$ while $a$ has order $4$. This means that $a^4 = 1$ and $\rho(a)^2 = 1$; because $\rho$ is a homomorphism, we get that $\rho(a^2) = 1$. Assume that $\rho$ is injective, then $a^4 = a^2 = 1$. This contradicts the minimality of $4$ as the order of $a$. Therefore, $\rho$ is not injective and thus not faithful.