If $a d j B = A, | P | = | Q | = 1$ , then $a d j (Q^{- 1} B P^{- 1})$ equals

P Q

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Q A P

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P A Q

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P A^{- 1} Q

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Solution

The correct option is D $P A Q$
$a d j B = A | P | = | Q | = 1 a d j (Q^{- 1} {B P}^{- 1}) = a d j (P^{- 1}) \cdot a d j (B) \cdot a d j (Q^{- 1}) = a d j (\frac{a d j P}{| P |}) \cdot a d j B \cdot a d j (\frac{a d j Q}{| Q |}) = a d j (a d j P) \cdot A . a d j (a d j (Q)) = {| P |}^{n - 2} \cdot P \cdot A \cdot {| Q |}^{n - 2} \cdot Q [∵ a d j B = A, | A | = | B | = 1] = P \cdot A \cdot Q [a d j (a d j (X)) = {| X |}^{n - 2} \cdot X]$

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