JEE Main 2022 — Mathematics Algebra

Given, $AB=I$AB=I and $|A|=\frac{1}{8}$∣A∣=81​

Now by using property $\mathrm{adj}(PQ)=(adjQ)(adjP)$adj(PQ)=(adjQ)(adjP) we get, $|adj(B\cdot adj2A)|=|adj(adj2A)||adjB|...(i)$∣adj(B⋅adj2A)∣=∣adj(adj2A)∣∣adjB∣...(i)

Now $|\mathrm{adj}(\mathrm{adj}2A)|={|\mathrm{adj}2A|}^{2}$∣adj(adj2A)∣=∣adj2A∣2 $={|2A|}^{4}$=∣2A∣4

$$={|{2}^{3}|}^{4}{|A|}^{4}={({2}^{3})}^{4}{|A|}^{4}={8}^{4}{|A|}^{4}...(\mathrm{ii})$$=∣23∣4∣A∣4=(23)4∣A∣4=84∣A∣4...(ii)

{Property used $|adjA|={|A|}^{n-1}$∣adjA∣=∣A∣n−1 and $|kA|={k}^{n}|A|$∣kA∣=kn∣A∣ where $n$n is order of matrix}

Now $|adjB|={|B|}^{2}.............(iii)$∣adjB∣=∣B∣2.............(iii)

Now putting the value of equation (\mathrm{ii})&(\mathrm{iii}) in eq $(i)$(i)

we get $|adj(B\cdot adj2A)|={8}^{4}{|A|}^{4}{|B|}^{2}$∣adj(B⋅adj2A)∣=84∣A∣4∣B∣2

Now using$|A|=\frac{1}{|B|}$∣A∣=∣B∣1​as $AB=I$AB=I, we get

$|adj(B\cdot adj2A)|={8}^{4}{|A|}^{4}\times \frac{1}{{|A|}^{2}}$∣adj(B⋅adj2A)∣=84∣A∣4×∣A∣21​ $={8}^{4}\times {|A|}^{2}={8}^{4}\times \frac{1}{{8}^{2}}=64$=84×∣A∣2=84×821​=64