How to show that a group $G$ is isomorphic to a subgroup of $GL_2(\mathbb{R})$

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Let $a\in\mathbb{R}^∗$ and $b\in \mathbb{R}$. Consider the function $f_{a,b} \in \operatorname{Fun}(\mathbb{R},\mathbb{R})$ given by $f_{a,b}(x)= ax + b$.

a) Show that $f_{a,b}$ is a bijection, and find its inverse function.

b) Let $G$ be the set of functions $\{f_{a,b}\mid a \in \mathbb{R}^∗ , b \in \mathbb{R}\}$. Show that $G$ is a group, where the group operation is composition of functions. (Thus $G$ is a subgroup of $\operatorname{Bij}(\mathbb{R}, \mathbb{R})$.)

c) Show that the group $G$ is isomorphic to a subgroup of $GL_2(\mathbb{R})$

I managed to solve parts a) and b) but part c) is a bonus question (which we didn't cover yet in the lecture) and I don't know how to solve it. Please help?

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Consider the following :

$$\phi:f_{a,b} \mapsto \left(\begin{matrix} a& b\\0& 1 \end{matrix}\right)$$

On one hand, we have $f_{a,b}\circ f_{c,d}=f_{ac, ad+b}$.

On the other hand, $$\left(\begin{matrix} a& b\\0& 1 \end{matrix}\right)\left(\begin{matrix} c& d\\0& 1 \end{matrix}\right)=\left(\begin{matrix} ac& ad+b\\0& 1 \end{matrix}\right)$$

Therefore, $\phi$ is an (obviously injective) group morphism. Therefore $G$ is isomorphic to $\phi(G)\subset GL_2(\mathbb{R})$.