Surprisingly, while their are similar but more advanced questions on this site, I don't see any answers to the basic version I am asking herein.
If I am taking the trace of a wedge product of matrices, can I still use the cyclic property? For instance, is it true that $$ \operatorname{Tr}(A B \wedge C) = \operatorname{Tr}(B \wedge C A) $$
Furthermore, is there a good way to think about the trace of a wedge product like there is for tensor product: $$ \operatorname{Tr}(A \otimes B) = \operatorname{Tr}(A) \operatorname{Tr}(B)? $$
It seems reasonable that the matrices $A, B, C$ are matrices with entries that are forms, and $A\wedge B$ is just matrix multiplication where we use $\wedge$ for the product. In this case "$AB\wedge C$" makes no sense unless $A$ is a matrix of 0-forms and we're omitting the wedge because it's the same as scalar multiplication in this case. I believe this is the case in the equations you linked to.
A simple computation in components will show $$ \mathrm{Tr}(B\wedge C) = B_{ij}\wedge C_{ji} = (-1)^{pq}C_{ji}\wedge B_{ij} = (-1)^{pq}\mathrm{Tr}(C\wedge B) $$ where I've assumed $B$ is made up of $p$-forms and $C$ of $q$-forms. Since we're assuming that $A$ is made up of 0-forms, the equation you want does indeed hold.