Let A and B are square matrices of same order which commute whith each other such that B is orthogonal and $A^{- 1} = A, B^{- 1} = B .$ Suppose $X = a d j {(A B^{T})}^{n}$ then which of the following is false.(Where det (AB)>0)

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$X = I, i f n i s e v e n$

$X = A B i f n i s e v e n$

X= BA if n is odd

$X = A^{2} B^{2}$ if n is even

answer is B.

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Detailed Solution

$B B^{T} = I, B^{- 1} = B \Rightarrow B^{T} = B^{- 1} = B$ and $B^{2} = I, A^{2} = I$
$X = a d j {(A B^{T})}^{n} = a d j {(A B)}^{n} = a d j (A^{n} B^{n}) (∵ A a n d B a r e c o m m i t e) = (a d j B^{n}) (a d j A^{n})$
If n is even $\Rightarrow A^{n} = B^{n} = I \Rightarrow X = I,$ If n is odd $\Rightarrow A^{n} = A, B^{n} = B$
$X = a d j B . a d j A = a d j (A B) = {(A B)}^{- 1} | A B | = B^{- 1} A^{- 1} = B A (| A B | = 1 > 0)$