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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) $A^{3} = I$
D) If A is orthogonal matrix then S) $B = I$
T) $B^{3} = I$

Where $|X|$ denotes determinant of matrix X and $I$ is identity matrix.

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A–P, B–U, C–Q,S, D–R,T

A–U, B–U, C–S,T, D–Q,R

A–U, B–U, C–Q,R, D–S,T

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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