Let $A=\left[\begin{matrix}\alpha & 1 & 2 \ 2 & 3 & 0 \ 0 & 4 & 5\end{matrix}\right]$ and $B=\left[\begin{matrix}1 & 0 & 0 \ 0 & -5\alpha & 0 \ 0 & 4\alpha & -2\alpha \end{matrix}\right]+adj\left(A\right)$. If $\det (B)=66$, then $\det (adj(A))$ equals: [JEE Main 2026, 8 Apr (Shift 2)]

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YESPYQ · Accepted Answer

Correct answer: c) $441$. $A=\left[\begin{array}{lll}\alpha & 1 & 2 \ 2 & 3 & 0 \ 0 & 4 & 5\end{array}\right]$ $\Rightarrow|A|=15 \alpha-10+16=15 \alpha+6$ Now $\operatorname{adj}(\mathrm{A})=\left[\begin{array}{ccc}15 & 3 & -6 \ -10 & 5 \alpha & 4 \ 8 & -4 \alpha & 3 \alpha-2\end{array}\right]$ $\therefore \mathrm{B}=\left[\begin{array}{ccc}1 & 0 & 0 \ 0 & -5 \alpha & 0 \ 0 & 4 \alpha & -2 \alpha\end{array}\right]+\left[\begin{array}{ccc}15 & 3 & -6 \ -10 & 5 \alpha & 4 \ 8 & -4 \alpha & 3 \alpha-2\end{array}\right]$ $\Rightarrow B=\left[\begin{matrix}16 & 3 & -6 \ -10 & 0 & 4 \ 8 & 0 & \alpha -2\end{matrix}\right]$ $|\mathrm{B}|=30 \alpha+36=66$ $\Rightarrow \alpha=1$ $\Rightarrow|\mathrm{A}|=21$ $\Rightarrow|\operatorname{adj}(A)|=|A|^2=(21)^2=441$

Let \(A=\left[\begin{matrix}\alpha & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5\end{matrix}\right]\) and \(B=\left[\begin{matrix}1 …

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