= Solution
The <Jacobson radical> is $J=\mathbb Ce_{12}$. Indeed $J^2=0$, so $J$ is contained in the <Jacobson radical>, and $A/J\cong\mathbb C\times\mathbb C$ is a <semisimple algebra>, giving the reverse inclusion. The <radical of a module> of the <left regular module> is therefore
$$
\boxed{\operatorname{rad}({}_AA)=\mathbb Ce_{12}.}
$$
Its <socle> is the sum of its <simple modules>. The displayed left action shows that its $S_1$-<submodules> span $\mathbb Ce_{11}\oplus\mathbb Ce_{12}$. There is no $S_2$-<submodule>: if $av=tv$ for every $a$, applying $e_{11}$ forces the first row of $v$ to vanish, and applying $e_{12}$ then forces the remaining $e_{22}$ coefficient to vanish. Hence
$$
\boxed{\operatorname{soc}({}_AA)=\mathbb Ce_{11}\oplus\mathbb Ce_{12}.}
$$
A homomorphism from $S_1$ is determined by any image $\alpha e_{11}+\beta e_{12}$ of $s_1$. Thus $V_{S_1}$ is the two-dimensional row <right module> with action
$$
(\alpha,\beta)\begin{pmatrix}r&s\\0&t\end{pmatrix}
=(\alpha r,\alpha s+\beta t).
$$
Its line $\mathbb Ce_{12}$ is a nonzero proper <submodule>, so \b[$V_{S_1}$ is not simple]. Since $S_2$ has no copy in the <socle>, any homomorphism $S_2\to A$ is zero. Thus \b[$V_{S_2}=0$ is not simple either], since a <simple module> is required to be nonzero.
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