Where have I gone wrong in deriving Faraday's Law of Induction from its manifestly covariant form? EDIT:  ISSUE SOLVED
This is simply an error in expanding tensor components on my part I'm sure but I am struggling to discover the error - where is the minus sign??
In expressing the laws of classical electromagnetism (Maxwell's Equations) in manifestly covariant form - that is to say, in a form that is fully consistent with tensor transformations - two of Maxwell's equations (Gauss's Law for Magnetism and Faraday's Law of Induction) can be expressed by the following tensor equation:  $$ \partial_\mu \tilde{F}^{\mu\nu} = 0 $$
where $ \partial_\mu $ is the four-gradient $\left(\frac{\partial}{\partial(ct)},\vec{\nabla}\right)$ and $\tilde{F}^{\mu\nu}$ is the contravariant form of the dual field strength tensor (dual Electromagnetic tensor).  $\mu, \nu = 0, 1, 2, 3 $.  I have presented the full matrix at the end for completeness and throughout I am using a (+---) signature.
Gauss's Law for magnetism can readily be realised by evaluating $$ \partial_i \tilde{F}^{i0} = 0 $$
where $i = 1, 2, 3$.
Similarly, Faraday's Law of Induction can be realised by evaluating $$ \partial_\mu \tilde{F}^{\mu i} = 0 $$
where herein lies the problem.  Upon expanding this tensor equation into its summed components, I get the following (I have gone into detail to properly explain my line of reasoning (or what is, in fact, my incorrect line of reasoning)) where each of the four equations corresponding to each value of $i$ can be summed because they all equal zero and the last three bracketed terms correspond to the three Cartesian components of the curl:  $$ -\frac{1}{c}\left(\frac{\partial B_x}{\partial t} + \frac{\partial B_y}{\partial t} + \frac{\partial B_z}{\partial t} \right) + \frac{1}{c} \left(\frac{\partial E_y}{\partial z} - \frac{\partial E_z}{\partial y} \right) + \frac{1}{c} \left(\frac{\partial E_z}{\partial x} - \frac{\partial E_x}{\partial z} \right) + \frac{1}{c} \left(\frac{\partial E_x}{\partial y} - \frac{\partial E_y}{\partial x} \right) = 0 $$
where the final three bracketed terms are the $x$, $y$ and $z$ components of the curl of the $E$-field, $\vec{\nabla} \times \vec{E}$.
Now, we finally arrive at my problem.  Where have I gone wrong in this equation because the minus sign on the first time-derivative term means that simplifying this equation yields: $$ \vec{\nabla} \times \vec{E} = \frac{\partial \vec{B}}{\partial t} $$
which is Faraday's law of Induction without the all-important minus sign which encapsulates Lenz's Law.
I would appreciate a keen eye who can point out my error in calculation.
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As stated, below is the full matrix form of $\tilde{F}^{\mu\nu}$:  $$ \begin{pmatrix} 0 & -B_x & -B_y & -B_z \\ B_x & 0 & \frac{E_z}{c} & -\frac{E_y}{c}  \\ B_y & -\frac{E_z}{c} & 0 & \frac{E_x}{c} \\ B_z & \frac{E_y}{c} & -\frac{E_x}{c} & 0 \end{pmatrix} $$
 A: You didn't make an error in your calculation.
Well, okay, so you made the obvious error where you tried to flatten everything together. But if we remove that error then you derived (multiplying through by $c$) that
$$ 
-\frac{\partial B_x}{\partial t} + \frac{\partial E_y}{\partial z} - \frac{\partial E_z}{\partial y} = 0.$$
This is a totally correct expression.
Your only error is that you got sloppy in reasoning about minus signs. In fact $$\nabla\times\mathbf E = \nabla\times\begin{bmatrix}E_x\\E_y\\E_z\end{bmatrix} = \begin{bmatrix}
\partial_y E_z - \partial_z E_y\\
\partial_z E_x - \partial_x E_z\\
\partial_x E_y - \partial_y E_x\end{bmatrix}$$
so that when we see $\partial_z E_y - \partial_y E_z$ we must immediately see it as $-(\nabla \times \mathbf E)_x.$ Thus the proper promotion of the above equation is $$-\dot{\mathbf B} - \nabla \times \mathbf E = 0$$whereas you had a stray $+$ sign in the middle of those two terms incorrectly.
A: I use $G^{\mu\nu}$ for the dual tensor. Using the convention $(+,-,-,-)$
$$G^{\mu\nu}=\begin{pmatrix}
0 & -B_x & -B_y & -B_z \\
B_x & 0 & E_z/c & -E_y/c \\
B_y & -E_z/c & 0 & E_x/c \\
B_z & E_y/c & -E_x/c & 0\end{pmatrix}$$
You seem to have several errors in your derivation. First of all, in the equation $\partial_\mu G^{\mu\nu}=0$, we only sum over repeated indexes, one covarient and one contravarient. Thus, $\nu$ determines four equations, not four parts of one equation. Secondly, you should not have any unit vectors; this equation is entirely component based. Next, when you find the component $G^{\mu'\nu'}$, you first go to the $\mu'$th row (starting at zero), and then the $\nu'$th column. So, $G^{01}=B_x$, not $-B_x$. Finally, you have mixed up the curl by a negative sign: the $x$ component of the vector $\mathbf{\nabla} \times \mathbf{E}$ is $\partial_y E_z - \partial_z E_y$, not $\partial_z E_y - \partial_y E_z$, as you have. Correcting for these three things gives the correct answer. So, for example, the $\nu = 1$ equation would give $$\begin{align}
0 & = \frac{1}{c}\left( \frac{\partial B_x}{\partial t} +\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right) \\
& = \frac{\partial B_x}{\partial t} + (\nabla \times E)_x
\end{align}$$
$\nu = 2, 3$ follows in the exact same manner. We can combine these three equations into one vector equation:
