Let A be a n times n matrix such that |A|=2. | vedclass

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(D) Given $|A|=2$ and $|\operatorname{Adj}(2 \cdot \operatorname{Adj}(2A^{-1}))| = 2^{84}$.Using the property $|\operatorname{Adj}(M)| = |M|^{n-1}$,we

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$5$

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(D) Given $|A|=2$ and $|\operatorname{Adj}(2 \cdot \operatorname{Adj}(2A^{-1}))| = 2^{84}$.Using the property $|\operatorname{Adj}(M)| = |M|^{n-1}$,we have:$|\operatorname{Adj}(2 \cdot \operatorname{Adj}(2A^{-1}))| = |2 \cdot \operatorname{Adj}(2A^{-1})|^{n-1} = (2^n |\operatorname{Adj}(2A^{-1})|)^{n-1}$.Since $|\operatorname{Adj}(2A^{-1})| = |2A^{-1}|^{n-1} = (2^n |A|^{-1})^{n-1} = (2^n \cdot 2^{-1})^{n-1} = (2^{n-1})^{n-1} = 2^{(n-1)^2}$.Substituting this back:$|\operatorname{Adj}(2 \cdot \operatorname{Adj}(2A^{-1}))| = (2^n \cdot 2^{(n-1)^2})^{n-1} = (2^{n + n^2 - 2n + 1})^{n-1} = (2^{n^2 - n + 1})^{n-1} = 2^{(n-1)(n^2 - n + 1)}$.Given $2^{(n-1)(n^2 - n + 1)} = 2^{84}$,we have $(n-1)(n^2 - n + 1) = 84$.For $n=5$,$(5-1)(25-5+1) = 4 	imes 21 = 84$.Thus,$n=5$.

Let $A$ be a $n \times n$ matrix such that $|A|=2$. If the determinant of the matrix $\operatorname{Adj}(2 \cdot \operatorname{Adj}(2A^{-1}))$ is $2^{84}$,then $n$ is equal to:

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