I recently endeavored to re-derive the Einstein-Maxwell field equations, explicitly preserving signs.

First, the metric signature: let the Minkowski metric be $\eta_{\mu\nu}=\mathbb{S}_g\times {\rm diag}(1,-1,-1,-1)$.

Next, the Ricci-tensor: let $R_{\mu\nu}=\mathbb{S}_R\times \left(\partial_\alpha\Gamma^\alpha_{\mu\nu}-\partial_\nu\Gamma^\alpha_{\mu\alpha}+\Gamma^\alpha_{\mu\nu}\Gamma^\beta_{\alpha\beta}-\Gamma^\alpha_{\mu\beta}\Gamma^\beta_{\alpha\nu}\right)$.

Let the general relativistic action be

\begin{align}
S=\int d^4x\sqrt{-g}\left[\frac{1}{2\kappa}\left(\mathbb{S}_G\times R + \mathbb{S}_\Lambda\times 2\Lambda\right) +{\cal L}_M\right].
\end{align}

We assume that ${\cal L}_M$ depends on $g_{\mu\nu}$ but not its derivatives. The corresponding Euler-Lagrange equation reads

\begin{align}
\mathbb{S}_G\times\frac{1}{2\kappa}\sqrt{-g}\left[R^{\mu\nu}-\frac{1}{2}g^{\mu\nu}R\right]-\mathbb{S}_\Lambda\times\frac{1}{2\kappa}\sqrt{-g} g^{\mu\nu}\Lambda+\frac{\delta\sqrt{-g}{\cal L}_M}{\delta g_{\mu\nu}}=0.
\end{align}

After rearranging and substituting

\begin{align}
T^{\mu\nu}=\mathbb{S}_{T}\times\frac{-2}{\sqrt{-g}}\frac{\delta\sqrt{-g}{\cal L}_M}{\delta g_{\mu\nu}},
\end{align}

we get

\begin{align}
\mathbb{S}_G\times\left[R^{\mu\nu}-\frac{1}{2}g^{\mu\nu}R\right]-\mathbb{S}_\Lambda\times g^{\mu\nu}\Lambda=\mathbb{S}_{T}\times \kappa T^{\mu\nu}.
\end{align}

In the case of the free electromagnetic field, we have

\begin{align}
{\cal L}_M={\cal L}_{\rm EM}=-\mathbb{S}_{\rm EM}\times \frac{1}{4}F^{\alpha\beta}F_{\alpha\beta}.
\end{align}

To compute its variation with respect to $g_{\mu\nu}$, we note that ${\dfrac{\delta g^{\alpha\beta}}{\delta g_{\mu\nu}}=-g^{\alpha\mu}g^{\beta\nu}}$:

\begin{align}
\frac{\delta F^{\alpha\beta}F_{\alpha\beta}}{\delta g_{\mu\nu}}&=\frac{\delta}{\delta g_{\mu\nu}} g^{\alpha\kappa}g^{\beta\gamma}F_{\kappa\lambda}F_{\alpha\beta}=-g^{\alpha\mu}g^{\kappa\nu}g^{\beta\lambda}F_{\kappa\lambda}F_{\alpha\beta}-g^{\beta\mu}g^{\lambda\nu}g^{\alpha\kappa}F_{\kappa\lambda}F_{\alpha\beta}\\
&=-F^{\nu\beta}F^\mu{}_\beta-F^{\alpha\nu}F_\alpha{}^\mu=-2F^{\alpha\mu}F_\alpha{}^\nu.
\end{align}

And, of course, $\delta\sqrt{-g}/\delta g_{\mu\nu}=\frac{1}{2}g_{\mu\nu}\sqrt{-g}$. Therefore,

\begin{align}
T^{\mu\nu}_{\rm EM}=\frac{-2}{\sqrt{-g}}\frac{\delta\sqrt{-g}{\cal L}_{\rm EM}}{\delta g_{\mu\nu}}&=\mathbb{S}_T\times\mathbb{S}_{\rm EM}\times\frac{1}{2\sqrt{-g}}\left[\frac{1}{2}g_{\mu\nu}\sqrt{-g}F^{\alpha\beta}F_{\alpha\beta}-2\sqrt{-g}F^{\alpha\mu}F_\alpha{}^\nu\right]\\
&=\mathbb{S}_T\times\mathbb{S}_{\rm EM}\times\left[\frac{1}{4}g_{\mu\nu}F^{\alpha\beta}F_{\alpha\beta}-F^{\alpha\mu}F_\alpha{}^\nu\right],
\end{align}

and the Einstein-Maxwell field equations read

\begin{align}
\mathbb{S}_G\times\left[R^{\mu\nu}-\frac{1}{2}g^{\mu\nu}R\right]-\mathbb{S}_\Lambda\times g^{\mu\nu}\Lambda=\mathbb{S}_{\rm EM}\times \kappa \left[\frac{1}{4}g_{\mu\nu}F^{\alpha\beta}F_{\alpha\beta}-F^{\alpha\mu}F_\alpha{}^\nu\right].
\end{align}