If there are three square matrix A, B, C of same order satisfying the equation `A^2=A^-1 and B=A^(2^n) and C=A^(2^((n-2))`, then prove that `det .(B-C) = 0, n in N`.

Explanation

`because B = ^(2^(n)) = A^(2.2^(n-1)) = (A^(2))^(2^(n-1)) = (A^(-1)) ^(2^(n-1) ) [ because A^(2) = A^(-1)]`
`= (A^(2^(n-1)))^(-1) = (A^(2.2^(n-2)))^(-1) = [(A^(2))^(2^(n-2))]^(-1)`
`= [(A^(-1))^(2^(n-2))]^(-1)= ((A^(-1))^(-1))^(2^(n-2)) = A^(2^(n-2))=C`
`rArr B-C=0 rArr det (B-C) = 0`