TITLE: C$^*$-algebras isomorphic after tensoring with $M_n(\mathbb C)$
QUESTION [20 upvotes]: In 1977, Joan Plastiras gave a striking example of two non $*$-isomorphic C$^*$-algebras $\mathcal A$ and $\mathcal B$ such that $$\mathcal A \otimes M_2(\mathbb C) \simeq \mathcal B\otimes M_2(\mathbb C)$$ (http://www.ams.org/journals/proc/1977-066-02/S0002-9939-1977-0461158-9/S0002-9939-1977-0461158-9.pdf).
My question is this: can you find two non $*$-isomorphic C$^*$-algebras $\mathcal A,\mathcal B$ such that $$\mathcal A\otimes M_n(\mathbb C) \simeq \mathcal B\otimes M_n(\mathbb C), \ n=2,3?$$ 
Note that Plastiras' example does not satisfy this condition. Furthermore, can you find non $*$-isomorphic $\mathcal A, \mathcal B$ such that
$$\mathcal A\otimes M_n(\mathbb C) \simeq \mathcal B\otimes M_n(\mathbb C), \ \forall n\geq 2$$
or does this condition imply that $\mathcal A\simeq \mathcal B$?

REPLY [16 votes]: Such examples do exist. In the paper "Stability of C*-algebras is not a stable property,  Doc. Math. J. DMV , 2, (1997), 375-386.", Rordam gives examples of simple C*-algebras $A$ such that $M_2(A)$ is a stable C*-algebra but $A$ is not. Now take the pair $A$ and $B=M_2(A)$. These C*-algebras are not isomorphic, since $A$ is not stable, but $B$ is. On the other hand, $M_n(A)$ is isomorphic to $M_n(B)$ which is isomorphic to $A\otimes \mathcal K$ for all $n\geq 2$. (Once $M_n(A)$ is stable for some $n$, it remains stable for all larger $n$. This is proven in the same paper.)
Part of the hard work in Rordam's paper is to find an example as above with $A$ simple. At the heart of Rordam's construction is the following non-simple example:   There exists a countable generated Hilbert C*-module $H$ over $C(X)$, with $X=\prod_{i=1}^\infty S^2$ (an infinite product of 2-spheres) such that $H$ is not isomorphic to $\ell^2(C(X))$
but $H\oplus H$ is. Then one can choose $A=K(H)$ and $B=K(H\oplus H)$. $A$ is not stable since this is equivalent to $H\cong \ell^2(C(X))$ but $M_2(A)\cong B$ is stable.
This is only a comment:
It seems to me that it should be possible to find (hermitian) complex vector bundles $\eta_1$ and $\eta_2$ over some space $X$ such that $\eta_1^{n\oplus}\cong \eta_2^{n\oplus}$ for all $n\geq 2$ but $\eta_1\ncong \eta_2$. Then $A=\mathrm{End}(\eta_1)$ and $B=\mathrm{End}(\eta_2)$ are candidate examples. From $\eta_1^{n\oplus}\cong \eta_2^{n\oplus}$ for all $n\geq 2$ we can already deduce that $M_n(A)\cong M_n(B)$ for all $n\geq 2$. If we choose them with some care (they can't be line bundles) then we should also have that $A\ncong B$.