If $A, B$and $C$are three non-singular square matrices of order $3$ satisfying the equation $A^2 = A^{-1}$and $B = A^8$ and $C = A^2,$then the value of $trace of (B-C) .$

$$\begin{gathered} \because B = A^8 = {\left(A^2\right)}^4 \\ = {\left(A^{-1}\right)}^4 [\because A^{-1} = A^2] \\ = {\left(A^4\right)}^{-1} = {\left(A^{2.2}\right)}^{-1} \\ = ({\left(A^2\right)}^2{)}^{-1} = {\left(A^2\right)}^{-1}{)}^2 \\ = ({\left(A^{-1}\right)}^{-1}{)}^2 = A^2 = C \end{gathered}$$

So, $B = C \Rightarrow B-C = 0$

$\therefore trace of (B-C) = 0$