Both implications here are true.
In fact, assume S : = Mn(R)is left primitive (for some n). For the idempotent e = E11(matrix unit),
eSe is also left primitive. Since eSe = RE11∼ R (as rings), we conclude that R is left primitive.Conversely, assume R is left primitive, and fix a faithful simple left R-moduleV . Let U be the group of column n-tuples (v1, .. . , vn)T , where vi ∈ V . We make U into a left S-moduleby using “matrix multiplication” as the action. We are done if we can show that
U is a faithful simple S-module. If (rij)U =0, wehave rijV = 0, so rij = 0, for all i, j. To show that Uis simple, it suffices to show that, given any u = (v1, .. . , vn)T <> 0 and u' = (v_1, . .. , v_n )T ,
we have u’ ∈ Su. Say vi <> 0. Then there exist r1,. . . , rn ∈ R such that v_j = rjvi forall j. For the matrix σ ∈ S with ithcolumn (r1, . . . , rn)T and other columnszero, we have clearly σu = u'.
Alternatively, we can use thecharacterization of left primitive rings :“R is left primitive iff thereexists a left ideal A <> R which is comaximal with anynonzero ideal B.” Fix such a left ideal A in R. Any nonzero ideal in Shas the form Mn(B) for some ideal B in R. Since A + B =R, the left ideal Mn(A) is clearly comaximal with Mn(B).
Therefore, S is leftprimitive.