JEE Main 2023 — Mathematics Algebra

Given that $A=[\begin{matrix}2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2\end{matrix}]$A=[210​12−1​0−12​] and $|adj(adj(adj2A))|={16}^{n}$∣adj(adj(adj2A))∣=16n.

We know that $|adjA|={|A|}^{n-1}$∣adjA∣=∣A∣n−1 and $|adj(adjA)|={|A|}^{{(n-1)}^{2}}$∣adj(adjA)∣=∣A∣(n−1)2

Similarly, $|adj(adj(adjA))|={|A|}^{{(n-1)}^{3}}$∣adj(adj(adjA))∣=∣A∣(n−1)3 where $n$n is the order of the square matrix.

Let us find $|A|$∣A∣.

$$|A|=2(2\times 2-(-1)(-1))-1(1\times 2-0(-1))+0(1\times (-1)-0\times 2)$$∣A∣=2(2×2−(−1)(−1))−1(1×2−0(−1))+0(1×(−1)−0×2)

$$=2(3)-(2)+0=4$$=2(3)−(2)+0=4

$$\Rightarrow {|2A|}^{{(n-1)}^{3}}={|2A|}^{{(3-1)}^{3}}$$⇒∣2A∣(n−1)3=∣2A∣(3−1)3

$$={|2A|}^{8}={({2}^{8})}^{3}{|A|}^{8}$$=∣2A∣8=(28)3∣A∣8

$$\Rightarrow {2}^{24}\times {4}^{8}={16}^{n}$$⇒224×48=16n

$$\Rightarrow {16}^{6}\times {16}^{4}={16}^{n}$$⇒166×164=16n

$$\Rightarrow {16}^{10}={16}^{n}$$⇒1610=16n

$$\Rightarrow n=10$$⇒n=10

Therefore, the value of $n$n is $10$10.