 Solution

Distinct gamma matrices anticommute, and $({\gamma }^0{)}^2=I$, $({\gamma }^j{)}^2=-I$ for $j=1,2,3$. Moving a particular ${\gamma }^{\mu }$ through the other three factors in ${\gamma }^5$ gives three minus signs. ThereforePairing equal factors in the square involves $3+2+1=6$ interchanges. The Clifford algebra consequently givesIn particular $({\gamma }^5{)}^{-1}=-{\gamma }^5$; replacing it by the conventional chirality matrix would change this answer.

The trace is unchanged by a similarity transformation. Since $({\gamma }^0{)}^{-1}={\gamma }^0$ and ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$, cyclic property of the trace gives $\mathrm{tr}{\gamma }^5=-\mathrm{tr}{\gamma }^5$. AlsoThus