Let A and B be two invertible matrices of order 3x3 such that $A B A = B A^{2} B, A^{3} = I a n d B^{2 n - 1} = I$ for some positive integer n, then

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A and B are commutative

$A B^{2} = B^{2} A$

B is idempotent but not involuntary

B is involutary but not idempotent

answer is A, B.

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Detailed Solution

$A B A = B A^{2} B = B A^{- 1} B . (∵ A^{3} = I)$
Now,
$A B^{2} = (A B A) A^{- 1} B = (B A^{- 1} B) A^{- 1} B = B A^{- 1} (B A^{- 1} B) = B A^{- 1} (A B A) = B^{2} A$
$A B^{4} = A B^{2} B^{2} = B^{2} A B^{2} = B^{2} (A B A) A^{- 1} B = B^{2} (B A^{- 1} B) A^{- 1} B = B^{2} B A^{- 1} (B A^{- 1} B) = B^{2} B A^{- 1} (A B A) = B^{4} A$
$A B^{2 r} = B^{2 r} A \Rightarrow A B = B A (∵ B^{2 r - 1} = I) A B A = B A^{2} B \Rightarrow A^{2} B = (A B) (A B) \Rightarrow A^{2} B^{2} \Rightarrow B = B^{2} \Rightarrow B = I$