If $A$ is an algebra, $M_{n}(A)$ denotes the algebra of all $n\times n$ matrices with entries in $A$. The operations are defined just as for scalar matrices. If $A$ is a *-algebra, so is $M_{n}(A)$, where the involution is given by $(a_{ij})^{\ast}=(a_{ji}^{\ast})$.
Thus $M_{n}(A)$ may be identified with (the algebraic tensor product) $A\otimes M_{n}(A)$.
My question is how to explain that $M_{n}(A)$ can be identified with (the algebraic tensor product) $A\otimes M_{n}(\mathbb{C})$?
Define a map $$ \varphi : A\times M_n(\mathbb{C}) \to M_n(A) $$ given by $$ \varphi(a,(\lambda_{i,j})) := (\lambda_{i,j}a) $$ This is bilinear, and so induces a linear map from the algebraic tensor product $$ \varphi : A\otimes M_n(\mathbb{C}) \to M_n(A) $$ It is easy to see that this map is bijective, and a *-homomorphism.