Does $\mathrm{Mat}_{m \times n}$ have boundary? To me, $\mathrm{Mat}_{m \times n}$ is isomorphic to $\mathbb{R}^{mn}$, hence is boundaryless. But this disqualified the use of Sard's theorem in this question: An exercise on Regular Value Theorem. But it seems using Sard's theorem is on the right track. Thank you~

$\mathrm{Mat}_{m \times n}$ is no boundary for exactly the reason you stated: it is diffeomorphic to $\Bbb R^{mn}$.

With regards to your other question, you can still apply the Transversality Theorem on page 68 of Guillemin and Pollack. This is because a manifold $M$ in the usual sense is also a manifold with boundary; it just has empty boundary: $\partial M = \varnothing$. Just look at the definition of manifold with boundary and you'll see that there is no requirement that the manifold actually has boundary points, but instead the definition merely says that every point is either an interior point or a boundary point.