Let A and B be any two n times n matrices suc | vedclass

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(B) Given that $A B = B A$,$A^k = I$,and $B^l = 0$ for some positive integers $k$ and $l$.Since $B^l = 0$,taking the determinant on both sides,we get

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$\operatorname{det}(A B)=0$

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(B) Given that $A B = B A$,$A^k = I$,and $B^l = 0$ for some positive integers $k$ and $l$.Since $B^l = 0$,taking the determinant on both sides,we get $\operatorname{det}(B^l) = \operatorname{det}(0) = 0$.Using the property $\operatorname{det}(B^l) = (\operatorname{det}(B))^l$,we have $(\operatorname{det}(B))^l = 0$,which implies $\operatorname{det}(B) = 0$.Now,consider the determinant of the product $A B$:$\operatorname{det}(A B) = \operatorname{det}(A) 	imes \operatorname{det}(B)$.Since $\operatorname{det}(B) = 0$,it follows that $\operatorname{det}(A B) = \operatorname{det}(A) 	imes 0 = 0$.Thus,the correct option is $(b)$.

Let $A$ and $B$ be any two $n \times n$ matrices such that the following conditions hold: $A B=B A$ and there exist positive integers $k$ and $l$ such that $A^k=I$ (the identity matrix) and $B^l=0$ (the zero matrix). Then,

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