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

Option: 1

84


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{ |A|=2 }
\mathrm{ \left|\operatorname{Adj}\left(2 \operatorname{Adj}\left(2 \mathrm{~A}^{-1}\right)\right)\right|=2^{84} }
\mathrm{ \left|2 \operatorname{Adj}\left(2 \mathrm{~A}^{-1}\right)\right|^{\mathrm{n}-1}=2^{84} }
\mathrm{ \left(2^{\mathrm{n}}\left|\operatorname{Adj}\left(2 \mathrm{~A}^{-1}\right)\right|\right)^{\mathrm{n}-1}=2^{84} }
\mathrm{ \left(2^{\mathrm{n}}\left|2^{\mathrm{n}-1} \operatorname{Adj}\left(\mathrm{A}^{-1}\right)\right|\right)^{\mathrm{n}-1}=2^{84} }
\mathrm{ \left(2^{\mathrm{n}} \times\left(2^{\mathrm{n}-1}\right)^{\mathrm{n}}\left|\operatorname{Adj}\left(\mathrm{A}^{-1}\right)\right|\right)^{\mathrm{n}-1}=2^{84} }
\mathrm{ \left(2^{\mathrm{n}} \times 2^{\mathrm{n}(\mathrm{n}-1)} \times \mid \mathrm{A}^{-1}\right)^{\mathrm{n}-1}=2^{84} }
\mathrm{ \left(2^{\mathrm{n}} \times 2^{\mathrm{n}(\mathrm{n}-1)} \times\left(\frac{1}{2}\right)^{\mathrm{n}-1}\right)^{\mathrm{n}-1}=2^{84} }
\mathrm{ \left(2^{\mathrm{n}+\mathrm{n}^2-\mathrm{n}-\mathrm{n}+1}\right)^{n-1}=2^{84} }
\mathrm{ \left(2^{n^2-\mathrm{n}+1}\right)^{n-1}=2^{84} }
\mathrm{ \left(n^2-n+1\right)(n+1)=84 }

 

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Rishi

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