### the first two Maxwell's equations in tensor form

dμFνλ+dνFλμ+dλFμν=0
given that Fμν=dμAνdνAμ=dAνdxμdAμdxν$F_{\mu\nu} = d_{\mu}A_{\nu}-d_{\nu}A_{\mu}=\frac{dA_{\nu}}{dx^{\mu}}-\frac{dA_{\mu}}{dx^{\nu}}$
assuming μ=0$\mu = 0$ and ν=j=(1,2,3)$\nu=j=(1,2,3)$ in the first equation, we get
d0Fjλ+djFλ0+dλF0j=0

now let’s write λ=(0,k)$\lambda =(0,k)$ where k = (1,2,3)
d0Fj0+d0Fjk+djF00+djFk0+d0F0j+dkF0j=0

d0(Ej)+d0(Bk)+dj(Ek)+d0(+Ej)+dk(+Ej)=0

d0(Bi)+dj(Ek)+dk(+Ej)=0

d0(Bi)+(×E)i=0

similarly when ν=0$\nu=0$ and λ=0$\lambda=0$, we get two more equations
d0(Bj)+(×E)j=0

d0(Bk)+(×E)k=0

in general we can write
d0B+×E=0

1cdBdt+×E=0

when d=(d0,di)=(1cddt,)$d = (d_0, d_i)=(\frac{1}{c}\frac{d}{dt},-\nabla)$
now, assuming μ=i$\mu=i$ and ν=j$\nu = j$
diFjλ+djFλi+dλFij=0

now, for λ=(0,k)$\lambda = (0,k)$ where k=(1,2,3)$k = (1,2,3)$ as before
diFj0+diFjk+djF0i+djFki+d0Fij+dkFij=0

diFj0+djF0i+d0Fij+diFjk+djFki+dkFij=0

di(Ej)+dj(+Ei)+d0(Bk)+di(Bi)+dj(Bj)+dk(Bk)=0

di(Bi)+dj(Bj)+dk(Bk)=0

B=0