JEE MAIN - Mathematics (2023 - 8th April Morning Shift - No. 3)
Let $$A=\left[\begin{array}{ccc}2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2\end{array}\right]$$. If $$|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 A))|=(16)^{n}$$, then $$n$$ is equal to :
9
8
10
12
Explanation
We have,
$$ \begin{aligned} & |\mathrm{A}|=\left|\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{array}\right|=2(4-1)-1(2-0)+0 \\\\ & =6-2=4 \\\\ & \text { So, }|2 \mathrm{~A}|=2^3|\mathrm{~A}|=8 \times 4=32 \\\\ & \text { Now, }|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 \mathrm{~A}))|=|2 \mathrm{~A}|^{(n-1)^3} \\\\ & =(32)^{2^3}=32^8 \\\\ & \Rightarrow 16^n=(32)^8=2^8 \times 16^8 \\\\ & \Rightarrow 16^n=16^{2+8} \Rightarrow n=10 \end{aligned} $$
Concepts :
(a) $|k \mathrm{~A}|=k^n|\mathrm{~A}|$
(b) $|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{n-1}$
$$ \begin{aligned} & |\mathrm{A}|=\left|\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{array}\right|=2(4-1)-1(2-0)+0 \\\\ & =6-2=4 \\\\ & \text { So, }|2 \mathrm{~A}|=2^3|\mathrm{~A}|=8 \times 4=32 \\\\ & \text { Now, }|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 \mathrm{~A}))|=|2 \mathrm{~A}|^{(n-1)^3} \\\\ & =(32)^{2^3}=32^8 \\\\ & \Rightarrow 16^n=(32)^8=2^8 \times 16^8 \\\\ & \Rightarrow 16^n=16^{2+8} \Rightarrow n=10 \end{aligned} $$
Concepts :
(a) $|k \mathrm{~A}|=k^n|\mathrm{~A}|$
(b) $|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{n-1}$
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