Smooth maps on a manifold lie group $$
\operatorname{GL}_n(\mathbb R) = \{ A \in M_{n\times n} | \det A \ne 0 \} \\
\begin{align}
&n = 1, \operatorname{GL}_n(\mathbb R) = \mathbb R - \{0\} \\
&n = 2, \operatorname{GL}_n(\mathbb R) = \left\{\begin{bmatrix}a&b\\c&d\end{bmatrix}\Bigg| ad-bc \ne 0\right\}
\end{align}
$$
$(\operatorname{GL}_n(\mathbb R),\cdot)$ is a group.


*

*$AB$ is invertible if $A$ and $B$ are invertible.

*$A(BC)=(AB)C$

*$I=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

*$A^{-1}$ is invertible if $A$ is invertible.


$$(\operatorname{GL}_n(\mathbb R) := \det{}^{-1}(\{0\}))$$
$\det{}^{-1}(\{0\})$ is open in $M_{n\times n}(\mathbb R)$.
$\det : M_{n\times n}(\mathbb R) \to \mathbb R$ is continuous, why?
$\dim \operatorname{GL}_n(\mathbb R) = n^2 - 1$, why?
$(\operatorname{GL}_n(\mathbb R),\cdot)$ is a Lie group if:
$$
\mu : G\times G \to G \\
\mu(A,B) = A\cdot B \:\text{ is smooth} \\
I(A) = A^{-1} \:\text{ is smooth}
$$
How can I show this? 
I want to show that general special linear group is Lie group. I could not show the my last step. How can I show that $AB$ and $A^{-1}$ are smooth? please help me I want to learn this. Thanks
Here the my handwritten notes http://i.stack.imgur.com/tkoMy.jpg

Well first you have to decide what exactly the "topology" on matrices is. Suppose we considered matrices just as vectors in $\mathbb{R}^{n^2}$, with the usual metric topology. Matrix multiplication by say $A$ take a matrix $B$ to another $\mathbb{R}^{n^2}$ vector where the entries are polynomials in the components of $B$. 
The other bit is similar, if you know what crammer's rule is. 
