JEE MAIN - Mathematics (2024 - 4th April Morning Shift - No. 3)
Let $$\alpha \in(0, \infty)$$ and $$A=\left[\begin{array}{lll}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]$$. If $$\operatorname{det}\left(\operatorname{adj}\left(2 A-A^T\right) \cdot \operatorname{adj}\left(A-2 A^T\right)\right)=2^8$$, then $$(\operatorname{det}(A))^2$$ is equal to:
16
36
49
1
Explanation
$$\begin{aligned}
& \left|\operatorname{adj}\left(A-2 A^T\right) \cdot \operatorname{adj}\left(2 A-A^T\right)\right|=2^8 \\
& P=A-2 A^{\top} \\
& Q=2 A^T-A \Rightarrow Q^T=2 A^T-A=-P \\
& |\operatorname{adj}(P) \operatorname{adj}(Q)| \Rightarrow|P Q|=-2^4 \\
& \Rightarrow|P|(-|P|)=-2^4 \Rightarrow|P|=4 \text { and }|Q|=-4 \\
& \left|A-2 A^T\right|=4 \\
& A-2 A^T=\left[\begin{array}{lll}
1 & 2 & \alpha \\
1 & 0 & 1 \\
0 & 1 & 2
\end{array}\right]-2\left[\begin{array}{lll}
1 & 1 & 0 \\
2 & 0 & 1 \\
\alpha & 1 & 2
\end{array}\right]=\left[\begin{array}{ccc}
-1 & 0 & \alpha \\
-3 & 0 & -1 \\
-2 \alpha & -1 & -2
\end{array}\right] \\
& \Rightarrow\left|A-2 A^T\right|=1+3 \alpha=4 \Rightarrow \alpha=1 \Rightarrow|A|=-4 \\
& \Rightarrow|A|^2=16
\end{aligned}$$
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