Does $R$ is Dedekind-finite imply $\mathbb{M}_n(R)$ is Dedekind-finite Following Lam's notation, a ring (with identity) $R$ is called Dedekind-finite if $ab=1\iff ba=1$ in $R$.
There are a lot of result about left invertible implies right invertible. But the results all require some finiteness property on the ring or the matrix ring. I am asking a proof or a counterexample of that that $R$ is Dedekind-finite impies that the matrix ring $\mathbb{M}_n(R)$ is Dedekind-finite.
 A: The answer is no, even for $n=2$.
One thng to note is that over the decades, different groups of authors have used different terminology. The property you call Dedekind-finite has been called directly finite, or von Neumann finite. Then given a unital ring $R$, people have asked when $M_n(R)$ is DF for all $n\geq 1$, and this property has been called weakly finite (I believe this terminology was introduced by P. M. Cohn) or fully von Neumann finite, or (in the context of ${\rm C}^*$-algebras) stably finite. I mention all this in case it helps to look things up in the literature.
It turns out that one can find a unital ring $R$ which has no non-trivial zero-divisors (so in particular no non-trivial idempotents, which in turn implies $R$ is Dedekind finite) yet has the property that $M_2(R)$ is not Dedekind finite. See
J. C. Shepherdson, Inverses and zero divisors in matrix rings.
Proc. London Math. Soc. (3) 1, (1951). 71–85. MR0041831
which was pointed out to me by Mr A. Fisher.
I can't resist pointing out another example. Much later, it was shown that there is a unital, finite ${\rm C}^*$-algebra $A$ (which can be realised as an extension of $C({\mathbb T}^3)$ by the compacts) such that $M_k(A)$ is not DF for some $k\geq 2$; however, at a quick glance I don't see how to obtain an actual value of $k$ from the proof. The paper is:
N. P. Clarke, A finite but not stably finite ${\rm C}^*$-algebra.
Proc. Amer. Math. Soc. 96 (1986), 85–88 | MR0813816
A: The question is answered in the negative in Exercise 1.18 of Lam's Exercises in Modules and Rings (2007 ed.): Start with the free $K$-algebra $K \langle X \rangle$ over a field $K$ generated by the $8$-element set $X = \{s, t, u, v, w, x, y, z\}$ and let $R$ be the quotient of $K \langle X \rangle$ by the two-sided ideal generated by the relations dictated by the equation $AB = I$ in the matrix ring $\mathcal M_2(K \langle X \rangle)$, where $I$ is the identity of $\mathcal M_2(K \langle X \rangle)$ and we put $A := \left[\begin{array}{cc} s & u \\ t & v \end{array}\right]$ and $B := \left[\begin{array}{cc} x & y \\ z & w \end{array}\right]$. It turns out that $R$ is a (non-commutative) domain and hence Dedekind-finite; but the matrix ring $\mathcal M_2(R)$ is not Dedekind-finite (essentially by design).
Lam explains in the comment after his solution that the construction first appeared in [J.C. Shepherdson, Inverses and zero-divisors in matrix rings, Proc. London Math, Soc. 1 (1951), 71-85]: This is the same paper mentioned by Yemon Choi in their answer.
