If A and B are square matrices of the same order and A is non-singular, then for a positive integer n,(A−1BA)n is always equal to
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Solution
For, (A−1BA)2=(A−1BA)(A−1BA) =A−1B(AA−1)BA =A−1BIBA=A−1B2A
Similarly, (A−1BA)3=(A−1BA)2(A−1BA)=(A−1B2A)(A−1BA) =A−1B2(AA−1)BA =A−1B2IBA =A−1B3A and so on. ∴(A−1BA)n=A−1BnA