QJEEM25Maa5d9: JEE Main 2025 Mathematics - Class 12 JEEM25Maa5d9 Solution: Let , where . Then is equal to __________.

$\begin{aligned}&\begin{aligned}& A=\left[\begin{array}{cc}2 & -1 \\1 & 0\end{array}\right] \\& A^2=\left[\begin{array}{ll}3 & -2 \\2 & -1\end{array}\right], A^3=\left[\begin{array}{ll}4 & -3 \\3 & -2\end{array}\right], A^4=\left[\begin{array}{ll}5 & -4 \\4 & -3\end{array}\right]\end{aligned}\\&\text { and so on }\\&\begin{aligned}& \mathrm{A}^6=\left[\begin{array}{ll}7 & -6 \\6 & -5\end{array}\right] \\& \mathrm{A}^{\mathrm{m}}=\left[\begin{array}{cc}\mathrm{m}+1 & -\mathrm{m} \\\mathrm{~m} & -\mathrm{m}+1\end{array}\right], \\& \mathrm{A}^{\mathrm{m}^2}=\left[\begin{array}{cc}\mathrm{m}^2+1 & -\mathrm{m}^2 \\\mathrm{~m}^2 & -\left(\mathrm{m}^2-1\right)\end{array}\right] \\& \mathrm{A}^{\mathrm{m}^2}+\mathrm{A}^{\mathrm{m}}=3 \mathrm{I}-\mathrm{A}^{-6} \\& {\left[\begin{array}{cc}\mathrm{m}^2+1 & -\mathrm{m}^2 \\\mathrm{~m}^2 & -\left(\mathrm{m}^2-1\right)\end{array}\right]+\left[\begin{array}{cc}\mathrm{m}+1 & -\mathrm{m} \\\mathrm{~m} & -\mathrm{m}+1\end{array}\right]}\end{aligned}\end{aligned}$

$\begin{aligned}& =3\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]-\left[\begin{array}{ll}-5 & 6 \\-6 & 7\end{array}\right] \\& =\left[\begin{array}{ll}8 & -6 \\6 & -4\end{array}\right] \\& =\mathrm{m}^2+1+\mathrm{m}+1=8 \\& =\mathrm{m}^2+\mathrm{m}-6=0 \Rightarrow \mathrm{~m}=-3,2 \\& \mathrm{n}(\mathrm{~s})=2\end{aligned}$