If K in R_0,then det(adj(KI_n)) is equal to: | vedclass

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(B) We know that for any square matrix $A$ of order $n$,$A(adj(A)) = |A|I_n$.Let $A = KI_n$,where $I_n$ is the identity matrix of order $n$.The determ

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$K^{n(n - 1)}$

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(B) We know that for any square matrix $A$ of order $n$,$A(adj(A)) = |A|I_n$.Let $A = KI_n$,where $I_n$ is the identity matrix of order $n$.The determinant $|A| = |KI_n| = K^n |I_n| = K^n(1) = K^n$.Using the property $adj(kA) = k^{n-1} adj(A)$,we have:$adj(KI_n) = K^{n-1} adj(I_n)$.Since $adj(I_n) = I_n$,it follows that $adj(KI_n) = K^{n-1} I_n$.Now,we calculate the determinant:$|adj(KI_n)| = |K^{n-1} I_n| = (K^{n-1})^n |I_n| = K^{n(n-1)} 	imes 1 = K^{n(n-1)}$.

If $K \in R_0$,then $\det(adj(KI_n))$ is equal to:

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