Dimension of $GL(n, \mathbb{R})$
Solution 1:
It is an open submanifold of the set of $n$ by $n$ matrices, which is a vector space of dimension $n^2$.
Solution 2:
You should notice that the determinant is a continuous map $f:M(n,\mathbb{R})\equiv\mathbb{R}^{n^2}\to \mathbb{R}$, $f(X)=\det(X)$
Then note that
$$GL(n, \mathbb{R})=f^{-1}(\mathbb{R}\setminus\{0\})$$
How the pre-image by continuous map of open set is an open set then $GL(n, \mathbb{R})$ is an open set of $\mathbb{R}^{n^2}$ whose dimension is $n^2$.
(I am here using that an open set of $\mathbb{R}^{k}$ has $k$-dimension.)