JEE Main 2025 — 8 April, Evening Shift — Question 41

$x^{2}+x+1=0$

$$\begin{aligned} & {\left[\begin{array}{lll} 4 & a & b \end{array}\right]\left[\begin{array}{ccc} 1 & 16 & 13\\ -1 & -1 & 2\\ -2 & -14 & -8 \end{array}\right]_{3 \times 3}=\left[\begin{array}{lll} 0 \\ 0 \\ 0 \end{array}\right]} \\& \Rightarrow[4-a-2 b, 64-a-14 b, 52+2 a-8 b]=\left[\begin{array}{lll} 0 & 0 & 0 \end{array}\right] \\& a+2 b=4 …(i) \\& a+14 b=64 …(ii)\end{aligned}$$

Solving (i) and (ii)

We get $a=-6, b=5$

$$\begin{aligned} & \therefore \frac{4}{\alpha^{4}}+\frac{m}{\alpha^{a}}+\frac{n}{\alpha^{b}}=3 \\& \Rightarrow \frac{4}{w^{4}}+\frac{m}{w^{-6}}+\frac{n}{w^{5}}=3 \\& \Rightarrow 4 w^{2}+m+n w=3 …(i) \end{aligned}$$

For $\alpha=w^{2}$, $\begin{aligned} & \frac{4}{w^{8}}+\frac{m}{w^{-12}}+\frac{n}{w^{10}}=3 \\& \Rightarrow \frac{4}{w^{2}}+m+\frac{n}{w}=3 \\& \Rightarrow 4 w+m+n w^{2}=3 …(ii) \end{aligned}$

Adding (i) & (ii)

$\Rightarrow 4\left(w^{2}-w\right)+n\left(w-w^{2}\right)=0$

$\Rightarrow\left(w^{2}-w\right)(4-n)=0$

$\Rightarrow n=4$

$\therefore 4 w+m+4 w^{2}=3$

$\Rightarrow-4+m=3$

$\Rightarrow m=7$

$\therefore m+n=11$