the first two Maxwell's equations in tensor form

$$d_{\mu}F_{\nu\lambda}+d_{\nu}F_{\lambda\mu}+d_{\lambda}F_{\mu\nu} = 0$$ given that $F_{\mu\nu} = d_{\mu}A_{\nu}-d_{\nu}A_{\mu}=\frac{dA_{\nu}}{dx^{\mu}}-\frac{dA_{\mu}}{dx^{\nu}}$
assuming $\mu = 0$ and $\nu=j=(1,2,3)$ in the first equation, we get
$$d_{0}F_{j\lambda}+d_{j}F_{\lambda 0}+d_{\lambda}F_{0j} = 0$$
now let’s write $\lambda =(0,k)$ where k = (1,2,3)
$$d_{0}F_{j0}+d_{0}F_{jk}+d_{j}F_{0 0}+d_{j}F_{k 0}+d_{0}F_{0j}+d_{k}F_{0j} = 0$$
$$d_{0}(-E_j)+d_{0}(-B_k)+d_{j}(-E_k)+d_{0}(+E_j)+d_{k}(+E_j) = 0$$
$$d_{0}(-B_i)+d_{j}(-E_k)+d_{k}(+E_j) = 0$$
$$d_{0}(-B_i)+ (\nabla\times E)_i= 0$$
similarly when $\nu=0$ and $\lambda=0$, we get two more equations
$$d_{0}(-B_j)+ (\nabla\times E)_j= 0$$
$$d_{0}(-B_k)+ (\nabla\times E)_k= 0$$
in general we can write
$$-d_{0}B+ \nabla\times E= 0$$
$$-\frac{1}{c}\frac{dB}{dt}+ \nabla\times E= 0$$
when $d = (d_0, d_i)=(\frac{1}{c}\frac{d}{dt},-\nabla)$
now, assuming $\mu=i$ and $\nu = j$
$$d_{i}F_{j\lambda}+d_{j}F_{\lambda i}+d_{\lambda}F_{ij} = 0$$
now, for $\lambda = (0,k)$ where $k = (1,2,3)$ as before
$$d_{i}F_{j0}+d_{i}F_{jk}+d_{j}F_{0 i}+d_{j}F_{k i}+d_{0}F_{ij} + d_{k}F_{ij}= 0$$
$$d_{i}F_{j0}+d_{j}F_{0 i}+d_{0}F_{ij}+d_{i}F_{jk}+d_{j}F_{k i} + d_{k}F_{ij}= 0$$
$$d_{i}(-E_j)+d_{j}(+E_i)+d_{0}(-B_k)+d_{i}(-B_i)+d_{j}(-B_j) + d_{k}(-B_k)= 0$$
$$d_{i}(-B_i)+d_{j}(-B_j) + d_{k}(-B_k)= 0$$
$$\nabla\cdot B=0$$