 Solution

The multiplication gives $v_1=1_A$ and $v_2^2+3v_2+2v_1=0$. Since $v_1,v_2$ form a basis,by the Chinese remainder theorem. Thus there are exactly two nonisomorphic simple modules, both one-dimensional. With basis vectors $s_{-1},s_{-2}$ their actions areThe two orthogonal primitive idempotents are $e_{-1}=2v_1+v_2$ and $e_{-2}=-v_1-v_2$. They sum to $v_1$ and span the two simple ideals. Consequently the regular module splits as $A=\mathbb{C}{e}_{-1}\oplus \mathbb{C}{e}_{-2}$, so $A$ is semisimple.

Its complete list of composition series isIndeed every proper nonzero submodule has dimension one and is an invariant line for multiplication by $v_2$. The matrix $\left(\begin{array}{cc}0& -2\\ 1& -3\end{array}\right)$ has the two distinct eigenvalues $-1,-2$, so these are exactly its two invariant lines.