Find the dimension of $A= \left \{ \begin{bmatrix}a&b\\0&c\end{bmatrix} \in M(2,\mathbb{R) \mid a,c \neq 0} \right\}$
I've proved that $A$ is a differential manifold ($GL(2,\mathbb{R})$ is a differential manifold, and $A$ is an open subset).
I suppose that $A$ have dimension $3$, because i can try separate $A$ in $4$ subsets $[(a>0,c>0),(a>0,c<0),(a<0,c>0),(a<0,c<0)]$, every of that subsets is homomeorphic to an subset of $\mathbb{R}^3$
I'm right?, i appreciate your help.
Yes, your reasoning is correct. Another way to see this is by noticing that $a,c \neq 0$ is equivalent to requiring that $\det(B) \neq 0$ for a matrix $B$ of the form $\left(\begin{matrix} a & b \\ 0 & c \end{matrix}\right)$. Thus if $f:\mathbb{R}^3\to \mathbb{R}$ is the function $$f(a,b,c) = \left|\begin{matrix} a & b \\ 0 & c \end{matrix}\right| = ac,$$ which is continuous since the determinant function is continuous on $GL(2,\mathbb{R})$, then $A = f^{-1}(\mathbb{R}\setminus\{0\})$ must be open in $\mathbb{R}^3$ and so is a submanifold of dimension 3.