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Question

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 equal to

A
AnBnAn
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B
AnBnAn
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C
A1BnA
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D
n(A1BA)
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Solution

The correct option is C A1BnA
For n=1,(A1BA)1=A1BA
For n=2,(A1BA)2=(A1BA)(A1BA)
=A1B(AA1)BA
=A1B2A
For n=3,(A1BA)3=(A1BA)2(A1BA)
=(A1B2A)A1BA
=A1B2(AA1)BA
=A1B3A
(A1BA)n=A1BnA

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