Mathematics - Matrices Question with Solution | TestHub

Given,

$\left|A\right|=2$

Now simplifying,

$\left|\mathrm{Adj}\left(2.Adj\left(2 {\mathrm{A}}^{-1}\right)\right)\right|$

$={\left|2·Adj\left(2 {\mathrm{A}}^{-1}\right)\right|}^{n-1}$

$={\left(2^n\left|Adj\left(2 {\mathrm{A}}^{-1}\right)\right|\right)}^{n-1}$

$={\left(2^n{\left|2A^{-1}\right|}^{n-1}\right)}^{n-1}$

$=2^{n\left(n-1\right)}{\left({\left(2^n\left|A^{-1}\right|\right)}^{\left(n-1\right)}\right)}^{\left(n-1\right)}$

$\therefore \left|A^{-1}\right|=\frac{1}{\left|A\right|}=\frac{1}{2}$

$=2^{n\left(n-1\right)}{\left({\left(2^{\left(n-1\right)}\right)}^{\left(n-1\right)}\right)}^{\left(n-1\right)}$

$=2^{n\left(n-1\right)+{\left(n-1\right)}^3}=2^{84}$

Now comparing both side we get,

$n\left(n-1\right)+{\left(n-1\right)}^3=84$

$\Rightarrow \left(n-1\right)\left(n+n^2-2n+1\right)=84$

$\Rightarrow \left(n-1\right)\left(n^2-n+1\right)=4\times 21$

Now if $n-1=4\Rightarrow n=5$ now checking $n^2-3n+1=25-5+1=21$

Hence, $n=5$