= Solution
The <matrix trace> is
$$
\operatorname{Tr}(A)=\sum_{i=1}^nA_{ii}.
$$
Using <matrix> multiplication and interchanging dummy indices,
$$
\operatorname{Tr}(AB)
=\sum_{i,j}A_{ij}B_{ji}
=\sum_{j,i}B_{ji}A_{ij}
=\operatorname{Tr}(BA).
$$
The proposed noncyclic interchange is false. With <matrix> units,
$$
A=E_{11},\qquad B=E_{12},\qquad C=E_{21},
$$
one has
$$
\operatorname{Tr}(ABC)=1,
\qquad
\operatorname{Tr}(ACB)=0.
$$
Trace is invariant under cyclic permutations, but an arbitrary swap need not be cyclic.
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