A <symmetric matrix> that is also an <orthogonal matrix> satisfies $C=C^{-1}$. Multiplying by the commuting invertible factors gives $(I-B)^2=(I+B)^2$, hence $4B=0$. Thus \b[there is no nonzero choice of $B$]. The only symmetric case is $B=0$, giving $C=I$.

Solution

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 1 / b / i / Solution

 Solution (source code)