JEE Main 2025 Apr 8 Shift 2, Mathematics Q2: System Of Linear Equations

Since $\alpha$ satisfies $x^2+x+1=0$, it is a primitive cube root of unity, so $\alpha =\omega$ with ${\omega }^3=1$ and $1+\omega +{\omega }^2=0$.

Row-multiplying $[4ab]$ with each column of the matrix and equating to zero gives the system: $4-a-2b=0$, $64-a-14b=0$, $52+2a-8b=0$.

From the first: $a+2b=4$; from the second: $a+14b=64$. Subtracting gives $12b=60$, so $b=5$ and $a=-6$.

Substituting into the given equation: $\frac{4}{{\alpha }^4}+\frac{m}{{\alpha }^{-6}}+\frac{n}{{\alpha }^5}=3$, which simplifies to $\frac{4}{\omega }+m{\omega }^6+\frac{n}{{\omega }^2}=3$.

Using ${\omega }^3=1$: ${\omega }^6=1$, $\frac{1}{\omega }={\omega }^2$, $\frac{1}{{\omega }^2}=\omega$. So $4{\omega }^2+m+n\omega =3$.

With $\omega =-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and ${\omega }^2=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$, separating real and imaginary parts: real part gives $-2+m-\frac{n}{2}=3$; imaginary part gives $-2\sqrt{3}+\frac{n\sqrt{3}}{2}=0$, so $n=4$.

Substituting back yields $m=7$, so $m+n=11$, matching option (2).