ધારોકે $\alpha \in(0, \infty)$ અને $\mathrm{A}=$ $=\left[\begin{array}{lll}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]$જો $\operatorname{det}\left(\operatorname{adj}\left(2 A-A^{\mathrm{T}}\right) \cdot \operatorname{adj}\left(A-2 A^{\mathrm{T}}\right)\right)=2^8$ હોય, તો $(\operatorname{det}(A))^2$....................
JEE MAIN 2024, Difficult
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$ \left|\operatorname{adj}\left(\mathrm{A}-2 \mathrm{~A}^{\mathrm{T}}\right)\left(2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}\right)\right|=28 $
$ \left|\left(\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right)\left(2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}\right)\right|=24 $
$ \left|\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right|\left|2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}\right|= \pm 16 $
$ \left(\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right)^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}}-2 \mathrm{~A} $
$ \left|\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right|=\left|\mathrm{A}^{\mathrm{T}}-2 \mathrm{~A}\right| $
$ \Rightarrow\left|\mathrm{A}-2 \mathrm{~A}^{\mathrm{T}}\right|^2=16 $
$ \left|\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right|= \pm 4$
$\begin{aligned} & {\left[\begin{array}{ccc}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]-\left[\begin{array}{ccc}2 & 2 & 0 \\ 4 & 0 & 2 \\ 2 \alpha & 2 & 4\end{array}\right]} \\ & \left|\begin{array}{ccc}-1 & 0 & \alpha \\ -3 & 0 & -1 \\ -2 \alpha & -1 & -2\end{array}\right|\end{aligned}$
$ 1+3 \alpha=4 $
$ 3 \alpha=3 $
$ \alpha=1$
$|A|=\left|\begin{array}{lll}1 & 2 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right|=-1-3=-4$
$|A|^2=16$
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