Let $A$ be a $n imes n$ matrix such that $|A|=2$. If the determinant of the matrix $\operatorname{

Question

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

Professor PC Thomas Classes · Answer

(5) : $\mid \operatorname{Adj}\left(2 \operatorname{Adj}\left(2 A^{-1}ight) \midight.$\ $=\left|2\left(\operatorname{Adj}\left(2 A^{-1}ight)ight)ight|^{n-1}=2^{n(n-1)} \quad\left|\operatorname{Adj}\left(2 A^{-1}ight)ight|^{n-1}$\ $=2^{n(n-1)}\left|\left(2 A^{-1}ight)ight|^{(n-1)(n-1)}$\ $=2^{n(n-1)+n(n-1)(n-1)} \frac{1}{|A|^{(n-1)^{2}}}$\ $=\frac{2^{n(n-1)+n(n-1)(n-1)}}{2^{(n-1)^{2}}} 2^{n(n-1)+n(n-1)^{2}-(n-1)^{2}}$\ $=2^{(n-1)\left(n^{2}-n+1ight)}=2^{84}$\ $\Rightarrow \quad(n-1)\left(n^{2}-n+1ight)=84$\ $\Rightarrow \quad(n-5)\left(n^{2}+3 n+17ight)=0 \Rightarrow n=5$\