= Solution
Write $A=D+L+R$, where $D$ is the <diagonal matrix> of diagonal entries of $A$, and $L,R$ are its strictly lower- and upper-triangular parts. Since $A$ is <symmetric>, $R=L^T$. Subtract the two factors in the <SOR> iteration:
$$
(D+\omega L)-[(1-\omega)D-\omega R]=\omega A.
$$
Therefore, for $\omega\ne0$,
$$
C(\omega)=I-\omega(D+\omega L)^{-1}A
=\boxed{I-B(\omega)^{-1}A,\qquad B(\omega)=\omega^{-1}D+L.}
$$
<Positive definiteness> makes each $D_{ii}>0$, so the <triangular matrix> $B(\omega)$ is an <invertible matrix> for every $\omega\ne0$. The original formula gives $C(0)=I$; the expression for $B(\omega)$ itself is undefined there. The affine term ensuring that the desired solution is a <fixed point> is $c=B(\omega)^{-1}b$.
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