JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Complex Numbers

Since ${\alpha }^2+\alpha +1=0$, ${\alpha }^3=1$. The matrix equation gives $4-a-2b=0$, $64-a-14b=0$ and $52+2a-8b=0$; the first two yield $12b=60$, so $b=5$ and $a=-6$ (consistent with the third). Then $\frac{4}{{\alpha }^4}+\frac{m}{{\alpha }^{-6}}+\frac{n}{{\alpha }^5}$ reduces, using ${\alpha }^3=1$, to $\frac{4}{\alpha }+m+\frac{n}{{\alpha }^2}=4{\alpha }^2+m+n\alpha$. Equating this to $3$ and substituting ${\alpha }^2=-\alpha -1$ gives $(m-4)+(n-4)\alpha =3$; as $\alpha$ is non-real, $n-4=0$ and $m-4=3$, so $m=7,n=4$ and $m+n=11$.