Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 2 / Section B / 7 / a / ii / Solution

False. The cyclic property of the trace permits $ABC\mapsto BCA$, but not an arbitrary interchange. Even with $A$ and $B$ both invertible matrices, takeThen $\mathrm{Tr}(ABC)=1$ and $\mathrm{Tr}(BAC)=2$, providing a counterexample.