The Ricci tensor is obtained by taking the trace of the curvature tensor: $R_{ab}=g^{cd}R_{acbd}$. For the coefficient to be non-zero, $a$ and $c$ must be of distinct type (in the decomposition $T^{1,0}M$, $T^{0,1}M$), and likewise $b$ and $d$, $c$ and $d$, so $a$ and $b$ must be of distinct type too. So due to the symmetry, the only coefficient to be computed is essentially

On the other hand,

from where

and the scalar curvature is