If A and B are square matrices of order ‘3’ such that A adj (2B) = $16 I_{3}$ and det (B)=2 then which of the following is/are TRUE? (Where adj (X) denotes adjoint of matrix (X), $X^{- 1}$ denotes inverse of matrix (X) and det (X) denotes determinant value of matrix (X) and $I_{n}$ denotes identity matrix of order ‘n’)

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${(A^{- 1} (a d j B))}^{- 1} = 4 A^{2}$

${(A^{- 1} (a d j B))}^{- 1} = B^{2}$

$B (a d j A) = 8 I_{3}$

$B (a d j A) = 4 I_{3}$

answer is B, C.

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Detailed Solution

A. adj (2B) = 16I

$A (4 a d j B) = 16 I \Rightarrow A . a d j B = 4 I \Rightarrow A | B | B^{- 1} = 4 I A B^{- 1} = 2 I \Rightarrow A - 2 B B . a d j A = B . a d j (2 B) = 4 B . a d j B = 4 | B | I_{3} = 8 I_{3} A . a d j B = 4 I \Rightarrow A^{- 1} A . a d j B \Rightarrow a d j B = 4 I \Rightarrow A^{- 1} A . a d j B = 4 A^{- 1} \Rightarrow a d j B = 4 A^{- 1} A^{- 1} . a d j B = A^{- 1} . (4 A^{- 1}) = 4 {(A^{- 1})}^{2} {(A^{- 1} (a d j B))}^{- 1} = {(4 A^{- 1} A^{- 1})}^{- 1} = \frac{1}{4} A^{2} = \frac{1}{4} {(2 B)}^{2} = B^{2}$