Show that $SL(n, \mathbb{R})$ is a $(n^2 -1)$ smooth submanifold of $M(n,\mathbb{R})$

You want to consider the smooth map $\det\colon M_n(\mathbb{R})\to\mathbb{R}$. If you can show $1$ is a regular value of $\det$, then $SL(n,\mathbb{R})=\det^{-1}(1)$ is a smooth manifold of dimension $$\dim(M_n(\mathbb{R}))-\dim(\mathbb{R})=n^2-1$$ by the regular value theorem.

To show $1$ is a regular value of $\det$, first note that the domain and codomain are vector spaces, so they may be identified with their own tangent spaces.

If $A\in M_n(\mathbb{R})$, then $d(\det)_A(A)$ can be computed as

$$\lim_{t\to 0}\frac{\det(A+tA)-\det(A)}{t}=\lim_{t\to 0}\frac{(1+t)^n\det(A)-\det(A)}{t}=n\cdot\det(A) $$

This shows $d(\det)_A$ is nonzero linear functional, hence surjective, for invertible $A$. This implies in particular that $1$ is a regular value of $\det$.

Ben West's solution is extremely clean and simple, and it should be accepted as the answer.

This is the solution that I was thinking about.

$GL_n = \det^{-1}(\mathbb{R}\setminus\{0\})$

Since $\det$ is continuous, $GL_n$ is an open subset of $M_n$, meaning it has dimension $n^2$.

Restrict $\det$ to $GL_n$. Then, for any $A\in GL_n$ and any $B\in M_n \sim T_A GL_n$, we have:

\begin{align} d(\det)_A(B) &= \lim_{t\to0} \frac{\det(A+tB) - \det(A)}{t} \\ & =\det(A) \lim_{t\to0} \frac{\det(I+tA^{-1}B)-1}{t}\\ &= \det(A)\mathrm{tr}(A^{-1}B) \end{align}

Since $\det$ maps into a 1-dimensional space, we just need one $B$ that makes $\mathrm{tr}(A^{-1}B)≠0$. So, take $B=A$.

This shows that every non-zero number is a regular value of $\det$.