Solution
= Solution
\b[False.] The <cyclic property of the trace> permits $ABC\mapsto BCA$, but not an arbitrary interchange. Even with $A$ and $B$ both <invertible matrices>, take
$$
A=\begin{pmatrix}1&0\\0&2\end{pmatrix},\quad B=\begin{pmatrix}1&1\\0&1\end{pmatrix},\quad C=\begin{pmatrix}0&0\\1&0\end{pmatrix}.
$$
Then $\operatorname{Tr}(ABC)=1$ and $\operatorname{Tr}(BAC)=2$, providing a counterexample.