JEE MAIN - Mathematics (2025 - 29th January Morning Shift - No. 22)
Explanation
$$\begin{aligned} &\begin{aligned} & A=\left[\begin{array}{cc} 2 & -1 \\ 1 & 0 \end{array}\right] \\ & A^2=\left[\begin{array}{ll} 3 & -2 \\ 2 & -1 \end{array}\right], A^3=\left[\begin{array}{ll} 4 & -3 \\ 3 & -2 \end{array}\right], A^4=\left[\begin{array}{ll} 5 & -4 \\ 4 & -3 \end{array}\right] \end{aligned}\\ &\text { and so on }\\ &\begin{aligned} & \mathrm{A}^6=\left[\begin{array}{ll} 7 & -6 \\ 6 & -5 \end{array}\right] \\ & \mathrm{A}^{\mathrm{m}}=\left[\begin{array}{cc} \mathrm{m}+1 & -\mathrm{m} \\ \mathrm{~m} & -\mathrm{m}+1 \end{array}\right], \\ & \mathrm{A}^{\mathrm{m}^2}=\left[\begin{array}{cc} \mathrm{m}^2+1 & -\mathrm{m}^2 \\ \mathrm{~m}^2 & -\left(\mathrm{m}^2-1\right) \end{array}\right] \\ & \mathrm{A}^{\mathrm{m}^2}+\mathrm{A}^{\mathrm{m}}=3 \mathrm{I}-\mathrm{A}^{-6} \\ & {\left[\begin{array}{cc} \mathrm{m}^2+1 & -\mathrm{m}^2 \\ \mathrm{~m}^2 & -\left(\mathrm{m}^2-1\right) \end{array}\right]+\left[\begin{array}{cc} \mathrm{m}+1 & -\mathrm{m} \\ \mathrm{~m} & -\mathrm{m}+1 \end{array}\right]} \end{aligned} \end{aligned}$$
$$\begin{aligned} & =3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{ll} -5 & 6 \\ -6 & 7 \end{array}\right] \\ & =\left[\begin{array}{ll} 8 & -6 \\ 6 & -4 \end{array}\right] \\ & =\mathrm{m}^2+1+\mathrm{m}+1=8 \\ & =\mathrm{m}^2+\mathrm{m}-6=0 \Rightarrow \mathrm{~m}=-3,2 \\ & \mathrm{n}(\mathrm{~s})=2 \end{aligned}$$
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