= Solution
{c}
Take the <tensor contraction> of the <Einstein field equations> with the <inverse metric> $g^{\mu\nu}$. In four-dimensional <spacetime>, $g^{\mu\nu}g_{\mu\nu}=4$, hence
$$
R-\frac12(4)R=CT,\qquad T=g^{\mu\nu}T_{\mu\nu},\qquad R=-CT.
$$
Here $T$ is the <tensor contraction> of the <stress-energy tensor>. Substituting this result back into the <Einstein field equations>, while keeping both free indices lowered, gives the <trace-reversed stress-energy tensor>:
$$
\boxed{R_{\mu\nu}=C\left(T_{\mu\nu}-\frac12g_{\mu\nu}T\right)
=CS_{\mu\nu}.}
$$
The factor $1/2$ uses the dimension four: in dimension $n>2$, the <Trace-reversed Einstein field equations in D dimensions> have $1/(n-2)$ in place of $1/2$.
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