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NDA 2022 Maths · Mathematics · Determinants · hard

If A is a 3×3 matrix with |A|=5, then |adj(adjA)| equals

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If AA is a 3×33\times3 matrix with ∣A∣=5|A|=5, then ∣adj⁡(adj⁡A)∣|\operatorname{adj}(\operatorname{adj}A)| equals
  1. A. 125125
  2. B. 2525
  3. C. 625625
  4. D. 55
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Answer: (C) 625625

For order nn, ∣adj⁡A∣=∣A∣n−1=25|\operatorname{adj}A|=|A|^{n-1}=25. Applying it again, ∣adj⁡(adj⁡A)∣=252=625=∣A∣(n−1)2|\operatorname{adj}(\operatorname{adj}A)|=25^{2}=625=|A|^{(n-1)^2}. Stopping after one adjoint gives 2525.
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