Let α be a solution of $x^2 + x + 1 = 0$, and for some a and b in

$R, \begin{bmatrix} 4 & a & b \end{bmatrix} \begin{bmatrix} 1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$. If $\frac{4}{\alpha^4} + \frac{m}{\alpha^a} + \frac{n}{\alpha^b} = 3$, then m + n is equal to _______

<p>Let $ \alpha $ be a solution of the equation $ x^2 + x + 1 = 0 $. This implies that $ \alpha $ is a cube root of unity, denoted as $ \omega $, where $ \alpha^2 + \alpha + 1 = 0 $.</p> <p>We are given matrix equations with parameters $ a $ and $ b $ as follows:</p> <p>$ \begin{bmatrix} 4 & a & b \end{bmatrix} \begin{bmatrix} 1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix} $</p> <p>Solving these equations:</p> <p>$ \begin{aligned} & \left[4 - a - 2b, 64 - a - 14b, 52 + 2a - 8b\right] = \left[0, 0, 0\right] \\ & \Rightarrow a + 2b = 4 \\ & \Rightarrow a + 14b = 64 \end{aligned} $</p> <p>From these, solve for $ a $ and $ b $:</p> <p>$ \begin{aligned} & 12b = 60 \Rightarrow b = 5 \\ & a = -6 \end{aligned} $</p> <p>We now substitute $ a $ and $ b $ into the given function:</p> <p>$ \frac{4}{\alpha^4} + \frac{m}{\alpha^a} + \frac{n}{\alpha^b} = 3 $</p> <p>We substitute $ \alpha = \omega $ and simplify:</p> <p>$ \begin{aligned} & \frac{4}{\omega} + \frac{m}{1} + \frac{n}{\omega^2} = 3 \\ & \Rightarrow 4\omega^2 + m + n\omega = 3 \end{aligned} $</p> <p>Substitute $ \omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i $ and $ \omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i $:</p> <p>$ \begin{aligned} & 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 + m - \frac{n}{2} = 3 \\ & \text{and } \frac{-4\sqrt{3}}{2} + \frac{n\sqrt{3}}{2} = 0 \end{aligned} $</p> <p>From the above, solve for $ n $ and $ m $:</p> <p>$ \begin{aligned} & n = 4 \\ & m = 7 \end{aligned} $</p> <p>Thus, $ m + n = 11 $.</p>