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Suppose A=aij3×3, where aijR If det(adjA)=25 then |det (A)| equals:

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a
5
b
12.5
c
55
d
52/3

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detailed solution

Correct option is A

Using det⁡(adj⁡A)=(det⁡(A))2 we get(det⁡(A))2=25⇒|det⁡(A)|=5


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