Let A be a 3 times 3 real matrix. If det(2 op | vedclass

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(B) Given $A$ is a $3 	imes 3$ matrix,so $|A| = \Delta$. Using the property $\operatorname{adj}(kA) = k^{n-1} \operatorname{adj}(A)$,where $n=3$,we h

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$4$

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(B) Given $A$ is a $3 	imes 3$ matrix,so $|A| = \Delta$. Using the property $\operatorname{adj}(kA) = k^{n-1} \operatorname{adj}(A)$,where $n=3$,we have $\operatorname{adj}(2A) = 2^{2} \operatorname{adj}(A) = 4 \operatorname{adj}(A)$. Next,$\operatorname{adj}(\operatorname{adj}(2A)) = \operatorname{adj}(4 \operatorname{adj}(A)) = 4^{3-1} \operatorname{adj}(\operatorname{adj}(A)) = 16 \operatorname{adj}(\operatorname{adj}(A))$. Using $\operatorname{adj}(\operatorname{adj}(A)) = |A|^{n-2} A = |A| A$,we get $16 |A| A$. Now,the expression is $\det(2 \operatorname{adj}(2 \operatorname{adj}(\operatorname{adj}(2A)))) = \det(2 \operatorname{adj}(2(16 |A| A))) = \det(2 \operatorname{adj}(32 |A| A))$. Using $\operatorname{adj}(kA) = k^{n-1} \operatorname{adj}(A)$,we have $\operatorname{adj}(32 |A| A) = (32 |A|)^{2} \operatorname{adj}(A)$. So,$\det(2 \cdot (32 |A|)^{2} \operatorname{adj}(A)) = \det(2^{1} \cdot 2^{10} |A|^{2} \operatorname{adj}(A)) = \det(2^{11} |A|^{2} \operatorname{adj}(A))$. Since $\det(kM) = k^{n} \det(M)$,we have $(2^{11} |A|^{2})^{3} \det(\operatorname{adj}(A)) = 2^{33} |A|^{6} |A|^{2} = 2^{33} |A|^{8}$. Given $2^{33} |A|^{8} = 2^{41}$,so $|A|^{8} = 2^{8}$,which implies $|A| = \pm 2$. Thus,$\det(A^{2}) = |A|^{2} = (\pm 2)^{2} = 4$.

Let $A$ be a $3 \times 3$ real matrix. If $\det(2 \operatorname{Adj}(2 \operatorname{Adj}(\operatorname{Adj}(2 A))))=2^{41}$,then the value of $\det(A^{2})$ is equal to ..... .

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