If A is a square matrix of order 3 such that $| A | = 2$ then $∣ ∣ (a d j A^{- 1})^{- 1} ∣ ∣$ is

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Solution

The correct option is C 4
we know,
$a d j (A^{- 1}) = (a d j A)^{- 1}$
$∴ (a d j A^{- 1})^{- 1} = ((a d j A)^{- 1})^{- 1} = a d j A$
Also, $| a d j A | = (d e t (A))^{n - 1}$
Given, det(A) = 2, n = 3
$\Rightarrow | (a d j A^{- 1})^{- 1} | = (2)^{3 - 1}$
$= 2^{2} = 4$

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If A is a square matrix of order 3 such that $| A | = 2$ then the value of $| (a d j A^{- 1})^{- 1} |$ is ___ .

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