Calculate the "inverse" of the Ricci tensor Let $R_{\mu\nu}$ be the coordinates of the Ricci tensor. Using that $R^{\mu\alpha}R_{\alpha\nu}=\delta_\nu^\mu$ and $R_{\mu\nu}=R_{\nu\mu}$ it is possible to get the expressión of $R^{\mu\nu}$? 
Many thanks!
 A: First a comment on notation. In GR (and sometimes differential geometry more generally) the notation of moving a tensor's indices up is used to denote contraction with the metric. So $R^{\mu\nu}$ would refer to $g^{\mu\alpha}g^{\nu\beta}R_{\alpha\beta}$ rather than the inverse of $R_{\mu\nu}$. (A special case is the metric itself, where $g^{\mu\nu}$ represents both the inverse of $g_{\mu\nu}$ and what you get when you raise the two indices of $g_{\mu\nu}$. This is because we have chosen to define $g^{\mu\nu}$ in this way, so that the operation of raising an index will be the inverse to the operation of lowering it.)
So let's instead denote the inverse of $R_{\mu\nu}$ as $S^{\mu\nu}$. We want
$$R_{\mu\alpha}S^{\alpha\nu}=\delta_{\mu}^{\;\nu}\tag{$*$}$$
and
$$S^{\nu\alpha}R_{\alpha\mu}=\delta_{\;\mu}^{\nu}.\tag{$**$}$$
In fact any inverse can be calculated using the Levi-Civita symbol $\varepsilon^{\alpha\beta\gamma\delta}$. See here for more details. In particular we have
$$S^{\mu\nu}=4\frac{R_{\kappa\pi}R_{\lambda\rho}R_{\xi\sigma}\varepsilon^{\mu\kappa\lambda\xi}\varepsilon^{\nu\pi\rho\sigma}}{R_{\alpha\zeta}R_{\beta\eta}R_{\gamma\theta}R_{\delta\iota}\varepsilon^{\alpha\beta\gamma\delta}\varepsilon^{\zeta\eta\theta\iota}}\tag{$\dagger$}$$
