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If AA is square matrix of order 3 and A⋅(Adj.(A))=10IA \cdot (Adj.(A)) = 10I, then the value of 125∣Adj.(A)∣\frac{1}{25}|Adj.(A)| is

Solution

✅ Correct Option: 4

We know A⋅adj(A)=∣A∣⋅IA \cdot adj(A) = |A| \cdot I. Comparing with A⋅adj(A)=10IA \cdot adj(A) = 10I gives ∣A∣=10|A| = 10. For a matrix of order nn, ∣adj(A)∣=∣A∣n−1|adj(A)| = |A|^{n-1}. So ∣adj(A)∣=103−1=100|adj(A)| = 10^{3-1} = 100. Therefore 125∣adj(A)∣=10025=4\frac{1}{25}|adj(A)| = \frac{100}{25} = 4.

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