JEE Main 2023 — Mathematics Algebra

Given that $A=[\begin{matrix}1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2\end{matrix}]$A=[1α1​231​312​]

$$\Rightarrow |A|=2$$⇒∣A∣=2

$$\Rightarrow 1(6-1)-\alpha -7=2$$⇒1(6−1)−α−7=2

$$\Rightarrow -\alpha -2=2$$⇒−α−2=2

$$\Rightarrow \alpha =-4$$⇒α=−4

$$\Rightarrow |2adj(2adj(2A))|={32}^{n}$$⇒∣2adj(2adj(2A))∣=32n

We know that $|kA|={k}^{n}|A|$∣kA∣=kn∣A∣, $adj(adj(A))={|A|}^{{(n-1)}^{2}}$adj(adj(A))=∣A∣(n−1)2 and $\mathrm{adj}(kA)={k}^{n-1}\mathrm{adj}(A)$adj(kA)=kn−1adj(A).

$$={2}^{3}|adj(2adj(2A))|$$=23∣adj(2adj(2A))∣

$$={2}^{3}|{2}^{3-1}adj(adj(2A))|$$=23∣23−1adj(adj(2A))∣

$$={2}^{3}\cdot (4{)}^{3}|adj(adj(2A))|$$=23⋅(4)3∣adj(adj(2A))∣

$$={2}^{9}|2A{|}^{(2{)}^{2}}$$=29∣2A∣(2)2

$$={2}^{9}\cdot |2A{|}^{4}$$=29⋅∣2A∣4

$$={2}^{9}\cdot {2}^{12}|A{|}^{4}$$=29⋅212∣A∣4

$$={2}^{21}\cdot {2}^{4}={2}^{25}=(32{)}^{n}=(2{)}^{5n}$$=221⋅24=225=(32)n=(2)5n

$$\therefore n=5$$∴n=5

$$\Rightarrow 3n+\alpha =15-4=11$$⇒3n+α=15−4=11

Hence this is the correct option.