Question
If $K$ is a field, fix $A = \left \{ \begin{pmatrix} a & 0\\ b & c \end{pmatrix} : a,b,c \in K \right \}$ and $e_{1,1} = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}$, $e_{2,2} = \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}$. Set $Ae_{1,1} = \left \{ \begin{pmatrix} a & 0\\ b & 0 \end{pmatrix} : a,b \in K \right \}$ and $Ae_{2,2} = \left \{ \begin{pmatrix} 0 & 0\\ 0 & c \end{pmatrix} : c \in K \right \}$.
I need help to show that $\operatorname{Hom}_A(Ae_{1,1}, Ae_{2,2}) = 0$.
I've tried to show that any morphism between $Ae_{1,1}$ and $Ae_{2,2}$ must be zero, using the $A$-linearity, but it did not work (at least, not for me).
Also, I've figured out that both $Ae_{1,1}$ and $Ae_{2,2}$ are projective $A$-modules, but I'm not sure if this helps or not.
Context
I am trying to show that the $A$-module $Ae_{2,2}$ is not injective. To do so, I've assumed that the following sequence is exact: $$0 \rightarrow Ae_{2,2} \xrightarrow{f} Ae_{1,1} \rightarrow X \rightarrow 0$$ in which $X$ an arbitrarily chosen $A$-module. My goal is to prove that such a sequence does not split. Particularly, this means that $f$ must not be a section (i.e., it must not be an invertible monomorphism), and my go-to strategy in those cases is to prove that there are no non-zero morphisms between $Ae_{1,1}$ and $Ae_{2,2}$. I am willing to change strategies if someone points out a better one.