A Certain Semisimple $k$-algebra is Commutative

Suppose that $k$ is a field and that $R$ is a semisimple ring that is also a $k$-algebra with $\dim_k R$ at most $3$. Show that $R$ is commutative and that the semisimplicity is required.

The case when $R$ has dimension $1$ is simple as it is routine to verify that then $R \cong k$. As for the remaining two dimensions, I am unsure of how to continue. It is, however, clear to me that this must make use of the Artin-Wedderburn Theorem and the reason this fails for higher $n$ is that the larger the $n$, the 'more' non-commutative the matrix rings become. I assume the conditions on $R$ give just enough structure to force commutativity in the small matrix rings. How do I go about proving the result for $\dim_k R=2$ and $\dim_k R=3$?

You should be able to see that the smallest $k$ dimension you can get from $M_n(K)$ for a finite dimensional division ring extension $K$ of $k$ and $n>1$ is $4$, attained by $k=K$ and $n=2$.
Then what are the sizes of the matrix rings in the decomposition of $R$?
As for seeing that semisimplicity is necessary, think about the ring of $2\times 2$ upper triangular matrices over $k$.