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If A and B be two non-singular square matrices such that $B^{- 1} A B = A^{2}$ , find $B^{- 3} A B^{3}$

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$A^{2}$

$A^{4}$

$A^{8}$

answer is C.

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Detailed Solution

$Given B^{- 1} A B = A^{2} N o w B^{- 3} A B^{3} = B^{- 2} (B^{- 1} A B) B^{2}$

$B^{- 2} A^{2} B^{2} = B^{- 1} (B^{- 1} (A^{2}) B) B = B^{- 1} (A^{4}) B = {(A^{4})}^{2} = A^{8}$

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