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Let A and B are non singular square matrices of order $3 \times 3$ such that B is adjoint of A and $A^{- 1} A^{T} = | A | B^{- 1}$ . Match List-I with List-II and select the correct answer using the code given below the list .
LIST-I LIST-II
A) $| A |$ is equal to P) 1
B) $| B |$ is equal to Q) A= I
C) If A is symmetric matrix then R) A³ = I
D) If A is orthogonal matrix then S) B = I
T) B³ = I
U) $- 1$
Where $| X |$ denotes determinant of matrix X and I is identity matrix.

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$A - U, B - U, C - Q, R, D - S, T$

$A - P, B - U, C - Q, S, D - R, T$

$A - U, B - U, C - S, T, D - Q, R$

$A - P, B - P, C - Q, R, S, T, D - R, T$

answer is D.

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Detailed Solution

$A^{- 1} A^{T} = | A | B^{- 1}, B = a d j . A$
taking determinant,
$| A^{- 1} | . | A^{T} | = {| A |}^{3} . | B^{- 1} | | A^{3} | . \frac{1}{| B |} = 1 \frac{{| A |}^{3}}{{| A |}^{2}} = 1 \Rightarrow | A | = 1 \Rightarrow | B | = 1$
Now, $A^{- 1} A^{T} = B^{- 1}$
If A is symmetric
$\Rightarrow A^{- 1} A = B^{- 1} \Rightarrow B = I \Rightarrow a d j A = I \Rightarrow A = I$
If A is orthogonal then $A^{T} = A^{1}$
$A^{- 1} A^{- 1} = B^{- 1} \Rightarrow B = A^{2} \Rightarrow a d j A = A^{2} \Rightarrow A a d j A = A^{3} \Rightarrow A^{3} = | A | I \Rightarrow A^{3} = I \Rightarrow {(a d j A)}^{3} . A^{3} = B^{3} \Rightarrow B^{3} = I$

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