Which of the following statements is correct | vedclass

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(C) Let $A$ and $B$ be square matrices of order $n$. We know that $	ext{adj}(kA) = k^{n-1} 	ext{adj}(A)$. For the expression $	ext{trace}(	ext{adj

Vedclass · Accepted Answer

$	ext{trace}(	ext{adj}(|A| |B| AB)) - 	ext{trace}(	ext{adj}(|AB| BA)) = 0$

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(C) Let $A$ and $B$ be square matrices of order $n$. We know that $	ext{adj}(kA) = k^{n-1} 	ext{adj}(A)$. For the expression $	ext{trace}(	ext{adj}(|A||B|AB))$,we have: $	ext{trace}((|A||B|)^{n-1} 	ext{adj}(AB)) = (|A||B|)^{n-1} 	ext{trace}(	ext{adj}(B) 	ext{adj}(A))$. Similarly,for $	ext{trace}(	ext{adj}(|AB|BA))$,we have: $	ext{trace}((|AB|)^{n-1} 	ext{adj}(BA)) = (|A||B|)^{n-1} 	ext{trace}(	ext{adj}(A) 	ext{adj}(B))$. Since $	ext{trace}(XY) = 	ext{trace}(YX)$ for any two matrices $X$ and $Y$,it follows that $	ext{trace}(	ext{adj}(B) 	ext{adj}(A)) = 	ext{trace}(	ext{adj}(A) 	ext{adj}(B))$. Therefore,the difference is $0$. Thus,option $C$ is correct.

Which of the following statements is correct about two square matrices $A$ and $B$ of the same order $n$?

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