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If AA is a square matrix of order n×nn \times n, then which of the following are correct?

(A) ∣kA∣=kn∣A∣|kA| = k^n|A|, kk is a constant

(B) ∣k adjA∣=kn∣adjA∣n−1|k\ adjA| = k^n|adjA|^{n-1}, kk is a constant

(C) ∣adj adjA∣=∣A∣(n−1)2|adj\ adjA| = |A|^{(n-1)^2}

(D) A(adjA)=∣A∣IA(adjA) = |A|I, II is an identity matrix of order nn

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

(A) is the standard scaling result. (C) ∣adj(adj A)∣=∣A∣(n−1)2|adj(adj\,A)| = |A|^{(n-1)^2} and (D) A(adj A)=∣A∣IA(adj\,A)=|A|I are standard identities.

(B) is incorrect: ∣k adj A∣=kn∣adj A∣=kn∣A∣n−1|k\,adj\,A| = k^n|adj\,A| = k^n|A|^{n-1}, not kn∣adj A∣n−1k^n|adj\,A|^{n-1}.

So (A), (C) and (D) are correct.

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