Question:

Given $x^2 + x + 1 = 0$ where $\alpha$ is a root. $\therefore \alpha^2 + \alpha + 1 = 0$ $\Rightarrow \alpha = \omega$ where $\omega$ is a cube root of unity. From the matrix equation: $$\begin{bmatrix} 4 & a & b \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 2 & -2 \\ -14 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 0 \end{bmatrix}$$ This gives us: $$\begin{bmatrix} -4 + 2a - 14b & -4 - 2a - 8b \end{bmatrix} = \begin{bmatrix} 0 & 0 \end{bmatrix}$$ $\therefore a + 2b = 4$ ... (1) $a + 14b = 64$ ... (2) Subtracting (1) from (2): $12b = 60 \Rightarrow b = 5$ From (1): $a = 4 - 2(5) = -6$ Now, $\frac{4}{\alpha^2} + \frac{m}{\alpha} + n = 3$ $\Rightarrow \frac{4}{\omega^2} + \frac{m}{\omega} + n = 3$ Since $\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$ and $\omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$: $\Rightarrow 4\omega + m + n\omega^2 = 3$ $\Rightarrow 4\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) + m + n\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = 3$ $\Rightarrow -2 + 2\sqrt{3}i + m - \frac{n}{2} - \frac{n\sqrt{3}}{2}i = 3$ Equating real parts: $-2 + m - \frac{n}{2} = 3 \Rightarrow m - \frac{n}{2} = 5$ ... (i) Equating imaginary parts: $2\sqrt{3} - \frac{n\sqrt{3}}{2} = 0 \Rightarrow n = 4$ From (i): $m = 5 + \frac{4}{2} = 7$ $\therefore m + n = 7 + 4 = 11$