It's easier just to give a coordinate-free proof that $GL(V)$ is a variety. The point is that the multiplication map

$$\text{End}(V) \times \text{End}(V) \to \text{End}(V)$$

is an algebraic map (because it's bilinear), and $GL(V)$ is the closed subvariety of $\text{End}(V) \times \text{End}(V)$ given by the inverse image of $\text{id}_V \in \text{End}(V)$ under this map.