Let A be any 3 times 3 invertible matrix. The | vedclass

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(D) For a $3 	imes 3$ matrix $A$,we have the following properties:$1$. $adj(adj(A)) = |A|^{n-2} A$. Since $n=3$,$adj(adj(A)) = |A|^{3-2} A = |A| A$.

Vedclass · Accepted Answer

$adj(adj(A)) = |A| \cdot (adj(A))^{-1}$

Vedclass · Answer

(D) For a $3 	imes 3$ matrix $A$,we have the following properties:$1$. $adj(adj(A)) = |A|^{n-2} A$. Since $n=3$,$adj(adj(A)) = |A|^{3-2} A = |A| A$. Thus,option $B$ is true.$2$. We know that $adj(A) = |A| A^{-1}$. Replacing $A$ with $adj(A)$,we get $adj(adj(A)) = |adj(A)| (adj(A))^{-1}$.$3$. Since $|adj(A)| = |A|^{n-1} = |A|^{3-1} = |A|^2$,we have $adj(adj(A)) = |A|^2 (adj(A))^{-1}$. Thus,option $C$ is true.$4$. Comparing the results,$adj(adj(A)) = |A| A$ and $adj(adj(A)) = |A|^2 (adj(A))^{-1}$.$5$. Option $D$ states $adj(adj(A)) = |A| (adj(A))^{-1}$,which is generally false.$6$. Option $A$ states $adj(A) = |A| (adj(A))^{-1}$. Since $adj(A) = |A| A^{-1}$,this implies $A^{-1} = (adj(A))^{-1}$,which is true. Therefore,option $D$ is the one that is not always true.

Let $A$ be any $3 \times 3$ invertible matrix. Then which one of the following is not always true?

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