I know that $$M_{m,n,r} = \{ A \in {\rm Mat}(m \times n,\Bbb R) \mid {\rm rank}(A)= r\}$$is a submanifold of $\Bbb R^{mn}$ of codimension $(m-r)(n-r)$. For example, we have that $M_{2, 3, 1}$ is non-orientable, while some others are, such as $M_{3,3,1}$ and $M_{3, 3,2}$.

Is there a way to decide in general if $M_{m,n,r}$ is orientable, in terms of $m,n$ and $r$?

For example, to see that $M_{2,3,1}$ is non-orientable, we parametrize it using two maps. Call $V_i$ the open set in $M_{2,3,1}$ on which the $i-$th row is a multiple of the other non-zero one. We have $M_{2,3,1} = V_0 \cup V_1$. Put $$\alpha_1:(\Bbb R^3 \setminus\{{\bf 0}\})\times \Bbb R\to V_1, \quad \alpha_1(v,t) = (tv,v),$$and similarly for $\alpha_2$, where we look at these pairs as rows of the matrix. These $\alpha_i$ are good parametrizations that cover $M_{2,3,1}$. One then checks that $\alpha_1^{-1}(V_1\cap V_2)$ has two connected components, namely, $(\Bbb R^3\setminus\{{\bf 0}\}) \times \Bbb R_{>0}$ and $(\Bbb R^3\setminus\{{\bf 0}\}) \times \Bbb R_{<0}$. And $\det D(\alpha_2^{-1}\circ \alpha_1)(v,t)$ changes sign there, so $M_{2,3,1}$ is non-orientable.

The strategy for proving that $M_{m,n,r}$ is a submanifold is different and writes it locally as an inverse image of regular value. This is an early exercise in Guillemin & Pollack's Differential Topology book.

It seems difficult to attack the general case using parametrizations this way (computing inverses is hard).