JEE Main 2026 — 6 April, Morning Shift — Question 41

$|\mathrm{A}|=-1$ $\operatorname{adj}(\operatorname{adj}(\mathrm{A}))=|\mathrm{A}|^{\mathrm{n}-2} \mathrm{~A}=|\mathrm{A}|^{3-2} \cdot \mathrm{~A}=|\mathrm{A}| \cdot \mathrm{A}$ $\operatorname{adj}(\mathrm{A})(\operatorname{adj}(\operatorname{adj}(\mathrm{A})))=\operatorname{adj}(\mathrm{A})(|\mathrm{A}| . \mathrm{A})$ $=|\mathrm{A}|(\operatorname{adj}(\mathrm{A}) . \mathrm{A})$ since $\operatorname{adj}(\mathrm{A}) \mathrm{A}=|\mathrm{A}| \mathrm{I}$, then $\operatorname{adj}(\mathrm{A})(\operatorname{adj}(\operatorname{adj}(\mathrm{A})))=|\mathrm{A}|^{2} \mathrm{I}$ or $\operatorname{adj}(\mathrm{A})(\operatorname{adj}(\operatorname{adj}(\mathrm{A})))=\mathrm{I}$ solving the matrix equation $A^{2}+\alpha(-A)+\beta(I)=\left[\begin{array}{ccc}2 & -2 & 2\\ -2 & 0 & -1 \\ 0 & 0 & -1\end{array}\right]$ $A^{2}-\alpha A+\beta I=M$ $\left[\begin{array}{ccc}2 & -1 & 1 -1 & 1 & 0\\ 0 & 0 & 1\end{array}\right]-\alpha\left[\begin{array}{ccc}-1 & 1 & -1\\ 1 & 0 & 1\\ 0 & 0 & 1\end{array}\right]+\beta\left[\begin{array}{ccc}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}2 & -2 & 2\\ -2 & 0 & -1\\ 0 & 0 & 1\end{array}\right]$ $\alpha=1$ and $\beta=-1$ $\therefore(\alpha-\beta)^{2}=4$