If A is a 3 times 3 matrix and |A|=2,then |op | vedclass

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(A) We know that for a square matrix $A$ of order $n$,the adjoint of the adjoint is given by $\operatorname{adj}(\operatorname{adj} A) = |A|^{n-2} A$.

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(A) We know that for a square matrix $A$ of order $n$,the adjoint of the adjoint is given by $\operatorname{adj}(\operatorname{adj} A) = |A|^{n-2} A$. Given $n=3$ and $|A|=2$,we have $\operatorname{adj}(\operatorname{adj} A) = |A|^{3-2} A = |A| A = 2A$. Next,we need to find the determinant of this matrix: $|\operatorname{adj}(\operatorname{adj} A)| = |2A| = 2^n |A| = 2^3 	imes 2 = 8 	imes 2 = 16$. Now,we calculate the product: $|\operatorname{adj}(\operatorname{adj} A)| \operatorname{adj}(\operatorname{adj} A) = 16 	imes (2A) = 32A$. Thus,the correct option is $(a)$.

If $A$ is a $3 \times 3$ matrix and $|A|=2$,then $|\operatorname{Adj}(\operatorname{Adj} A)| \operatorname{Adj}(\operatorname{Adj} A) = $ (in $A$)

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