JEE Main 2025 — 29 January, Morning Shift — Question 66

$A=\left[\begin{array}{cc}2 & -1 \\ 1 & 0\end{array}\right]$

$A^{2}=\left[\begin{array}{ll}3 & -2 \\ 2 & -1\end{array}\right], A^{3}=\left[\begin{array}{ll}4 & -3 \\ 3 & -2\end{array}\right], A^{4}=\left[\begin{array}{ll}5 & -4 \\ 4 & -3\end{array}\right]$

and so on $A^{6}=\left[\begin{array}{ll}7 & -6 \\ 6 & -5\end{array}\right]$

$A^{m}=\left[\begin{array}{cc}m+1 & -m \\ m & -m-1\end{array}\right]$,

$A^{m^{2}}=\left[\begin{array}{cc}m^{2}+1 & -m^{2} \\ m^{2} & -\left(m^{2}-1\right)\end{array}\right]$

$A^{m^{2}}+A^{m}=3 I-A^{-6}$

$\left[\begin{array}{cc} m^{2}+1 & -m^{2} \\ m^{2} & -\left(m^{2}-1\right)\end{array}\right]+\left[\begin{array}{cc}m+1 & -m \\ m & -(m-1)\end{array}\right]$ $=3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]-\left[\begin{array}{ll}-5 & 6 \\ -6 & 7\end{array}\right]$

$=\left[\begin{array}{ll}8 & -6 \\ 6 & -4\end{array}\right]$

$=\mathrm{m}^{2}+1+\mathrm{m}+1=8$

$=\mathrm{m}^{2}+\mathrm{m}-6=0 \Rightarrow \mathrm{~m}=-3,2$

$\mathrm{n}(\mathrm{s})=2$