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$If k \in R, then \det \{a d j (k I_{n})\} is equal to$

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kn−1

kn(n−1)

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Correct option is B

We have |adj⁡A|=|A|n−1⇒adj⁡KIn=KInn−1=KnInn−1 ∵|KA|=Kn|A|=Kn(n−1)(1) ∵In=1=Kn(n−1)

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If k∈R, then det⁡adjkIn is equal to

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$If k \in R, then \det \{a d j (k I_{n})\} is equal to$