I need to prove that $GL(n,\mathbb{R})$ is orientable.
Since our manifold is differentiable, we need to find an atlas $\mathcal{A}$ such that for each transition mapping between charts $f_b^{-1} \circ f_a$ we have that $\det(d(f_b^{-1} \circ f_a)) > 0$ where $d$ is the differential operator.
Let $\mathcal{U}$ be an open cover of $\mathbb{R}^{n\times n}$. Then our atlas is a one chart atlas with the identity mapping $$\mathcal{A} = \left\{(\left\{ U\in\mathcal{U} : \forall P\in U \;\;\det(P)\neq 0 \right\},id)\right\}$$
Then the transition mapping is also the identity and the differential is the identity matrix with determinant $1$.
Does this make sense?
EDIT: My atlas was not injective
$$\mathcal{O} = \left\{ U\in\mathcal{U} : \forall P\in U \;\;\det(P)\neq 0 \right\}$$
$$\mathcal{A} = \{(O,id):O\in\mathcal{O}\}$$
now it's not a one chart atlas anymore but the rest of the reasoning is the same