JEE Main 2023 — Mathematics Algebra

Given,

$$|A|=2$$∣A∣=2

Now simplifying,

$$|\mathrm{Adj}(2.Adj(2{A}^{-1}))|$$∣Adj(2.Adj(2A−1))∣

$$={|2\cdot Adj(2{A}^{-1})|}^{n-1}$$=∣2⋅Adj(2A−1)∣n−1

$$={({2}^{n}|Adj(2{A}^{-1})|)}^{n-1}$$=(2n∣Adj(2A−1)∣)n−1

$$={({2}^{n}{|2{A}^{-1}|}^{n-1})}^{n-1}$$=(2n∣2A−1∣n−1)n−1

$$={2}^{n(n-1)}{({({2}^{n}|{A}^{-1}|)}^{(n-1)})}^{(n-1)}$$=2n(n−1)((2n∣A−1∣)(n−1))(n−1)

$$\therefore |{A}^{-1}|=\frac{1}{|A|}=\frac{1}{2}$$∴∣A−1∣=∣A∣1​=21​

$$={2}^{n(n-1)}{({({2}^{(n-1)})}^{(n-1)})}^{(n-1)}$$=2n(n−1)((2(n−1))(n−1))(n−1)

$$={2}^{n(n-1)+{(n-1)}^{3}}={2}^{84}$$=2n(n−1)+(n−1)3=284

Now comparing both side we get,

$$n(n-1)+{(n-1)}^{3}=84$$n(n−1)+(n−1)3=84

$$\Rightarrow (n-1)(n+{n}^{2}-2n+1)=84$$⇒(n−1)(n+n2−2n+1)=84

$$\Rightarrow (n-1)({n}^{2}-n+1)=4\times 21$$⇒(n−1)(n2−n+1)=4×21

Now if $n-1=4\Rightarrow n=5$n−1=4⇒n=5 now checking ${n}^{2}-3n+1=25-5+1=21$n2−3n+1=25−5+1=21

Hence, $n=5$n=5