I have been trying to write a bound quiver representation of each of the following algebras: $$A=\begin{bmatrix} K&0&0&0&0\\ K&K&0&0&0\\ K&0&K&0&0\\ K&0&K&K&0\\ K&K&K&K&K \end{bmatrix}, B=\begin{bmatrix} K&0&0&0&0\\ K&K&0&0&0\\ K&0&K&0&0\\ K&0&0&K&0\\ K&K&K&K&K \end{bmatrix}, C=\begin{bmatrix} K&0&0&0&0\\ 0&K&0&0&0\\ K&K&K&0&0\\ K&0&0&K&0\\ K&K&K&K&K \end{bmatrix}.$$
I would like to know whether my solution is correct or not.
I just consider the easiest quiver I can think of associated to each of the algebras:
There is a $K$-algebra homomorphism $\phi_*:KQ_*\to *$, $*=A,B,C$, for the three algebras each, defined in the natural way that maps $e_i$ to $E_{ii}$ and any arrow $i\to j$ to $E_{ij}$. Clearly $\phi_*$ is surjective since any matrix in the algebra has a preimage given by some path in the respective quiver. Now we decide the kernel of the homomorphisms.
For $A$, there are not paths $4\to 2$ or $3\to 2$, so the only relation generating the kernel is the commutativity: $\xi\alpha=\omega\tau\beta$. Clearly $\mathcal I_1:=\langle \xi\alpha-\omega\tau\beta\rangle\subseteq R_Q^2$ is admissible, since $Q_A$ is acyclic. Hence $A\cong KQ_A/\mathcal I_1$.
For $B$, similarly we only get the commutativity ones: $\eta\alpha=\xi\beta=\omega\gamma$. $\mathcal I_2:=\langle \eta\alpha-\xi\beta,\eta\alpha-\omega\gamma\rangle$ is also admissible. Hence $B\cong KQ_B/\mathcal I_2$.
For $C$, we also get the generating relation $\delta\alpha=\eta\gamma$. For the same reason $\mathcal I_3:=\langle \delta\alpha-\eta\gamma\rangle$ is admissible. Hence $C\cong KQ_C/\mathcal I_3$.
The relations seem to be just the commutativity, so I am wondering whether there is something wrong in the demonstration.
Thank you very much for your patience!