Question

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

Answer 1

Answer: 5

$\left|\text{Adj}\left(2\text{Adj}\left(2A^{-1}\right)\right)\right|$
$={\left|2\text{Adj}\left(\text{Adj}\left(2A^{-1}\right)\right)\right|}^{n-1}$
$=2^{n(n-1)}{\left|\text{Adj}\left(2A^{-1}\right)\right|}^{n-1}$
$=2^{n(n-1)}{\left|\left(2A^{-1}\right)\right|}^{(n-1)(n-1)}$
$=2^{n(n-1)}{2}^{n(n-1)(n-1)}{\left|A^{-1}\right|}^{(n-1)(n-1)}$
$=2^{n(n-1)+n(n-1)(n-1)}\frac{1}{|A{|}^{(n-1{)}^2}}$
$=\frac{2^{n(n-1)+n(n-1)(n-1)}}{2^{(n-1{)}^2}}$
$=2^{n(n-1)+n(n+1{)}^2-(n-1{)}^2}$
$=2^{(n-1)\left(n^2-n+1\right)}$
$\text{ Now }{2}^{(n-1)\left(n^2-n+1\right)}$
$2^{(n-1)\left(n^2-n+1\right)}=2^{84}$
$\text{ So, }n=5$