JEE Main 2025 — Matrices Question with Solution

$x^2+x+1=0$
$\alpha$ is root
$\begin{aligned}
& \therefore \alpha^2+\alpha+1=0 \\ & \Rightarrow \alpha=\omega \text { as } \omega^2 \text { [cube root of unity] }
\end{aligned}$<br/>also<br/>$\left.\begin{array}{l}
{\left[\begin{array}{ll}
4-a-2 b & 64-a-14 b
\end{array} 52+2 a-8 b\right.}
\end{array}\right]$<br/>$\begin{aligned} & =\left[\begin{array}{ll}0 & 0&0\end{array}\right] \\ \therefore & a+2 b=4 \\ & a+14 b=64 \\ \Rightarrow & 12 b=60 \Rightarrow b=5 \\ \Rightarrow & a=-6 \\ \therefore & \frac{4}{\alpha^4}+\frac{m}{\alpha^{-6}}+\frac{n}{\alpha^5}=3 \\ \Rightarrow & \frac{4}{\omega}+\frac{m}{1}+\frac{n}{\omega^2}=3 \\ \Rightarrow & 4 \omega^2+m+n \omega=3\end{aligned}$<br/>$\begin{aligned}
& \Rightarrow 4\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right)+\mathrm{m}+\mathrm{n}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)=3 \\ & \therefore-2+\mathrm{m}-\frac{\mathrm{n}}{2}=3....(1) \\ & \& \frac{-4 \sqrt{3}}{2}+\frac{\mathrm{n} \sqrt{3}}{2}=0 \\ & \therefore \mathrm{n}=4 \\ & \mathrm{~m}=7 \\ & \therefore \mathrm{~m}+\mathrm{n}=11
\end{aligned}$