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If det (A_3x3) = 6, then det (Adj 2A) =

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3²x2⁸

3³x2⁴

2²x3⁸

144

answer is B.

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Detailed Solution

$| A |_{3 \times 3} = 6, | Adj | = ?$
Adj(KA) = K^n-1AdjA
Adj(2A) = 2^3-1Adj= 4AdjA
$| Adj (2 A) | = | 4 Adj A | = 4^{3} | Adj A | = 4^{3} | 6 |^{2}$
= 2⁶ x 2² x 3² = 2⁸ x 3²

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