let’A’ be a 3×3 matrix such that |adj(adjA)|=124 then |A−1adjA| is equal to

|A|(n−1)3=124 and n=3|A|8=124⇒|A|2=12⇒|A|=23 |A−1.adjA|=|A−1||adjA|=1|A|.|A|2 |A|=23

let’A’ be a 3×3 matrix such that |adj(adjA)|=124 then |A−1adjA| is equal to

|A|(n−1)3=124 and n=3|A|8=124⇒|A|2=12⇒|A|=23 |A−1.adjA|=|A−1||adjA|=1|A|.|A|2 |A|=23

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let’A’ be a $3 \times 3$ matrix such that $| a d j (a d j A) | = 12^{4}$ then $| A^{- 1} a d j A |$ is equal to

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$2 \sqrt{3}$

$\sqrt{6}$

answer is C.

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${| A |}^{{(n - 1)}^{3}} = 12^{4}$ and n=3
${| A |}^{8} = 12^{4} \Rightarrow {| A |}^{2} = 12 \Rightarrow | A | = 2 \sqrt{3}$
$| A^{- 1} . a d j A | = | A^{- 1} | | a d j A | = \frac{1}{| A |} . {| A |}^{2} | A | = 2 \sqrt{3}$

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let’A’ be a 3×3 matrix such that |adj(adjA)|=124 then |A−1adjA| is equal to

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let’A’ be a $3 \times 3$ matrix such that $| a d j (a d j A) | = 12^{4}$ then $| A^{- 1} a d j A |$ is equal to