= Solution
The <tensor contraction> of a symmetric tensor with an antisymmetric one vanishes. Rename the two summed indices and use their opposite exchange symmetries:
$$
R_{ij}A_{ij}=R_{ji}A_{ji}=R_{ij}(-A_{ij})=-R_{ij}A_{ij}.
$$
Hence $R_{ij}A_{ij}=0$. Using the decomposition from part (iii) gives
$$
\boxed{R_{ij}B_{ij}=R_{ij}(S_{ij}+A_{ij})=R_{ij}S_{ij}.}
$$
This is orthogonality of the real symmetric and antisymmetric matrix subspaces under the <Frobenius inner product>. It is a summed scalar identity, not an assertion about the unsummed products of individual entries.
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