Let be a solution of x^2+x+1=0, and for some a,bR, pmatrix4&a&bpmatrixpmatrix1&16&13-1&-1&…

Question

Let $\alpha$ be a solution of $x^2+x+1=0$, and for some $a,b\in\mathbb{R}$, $\begin{pmatrix}4&a&b\end{pmatrix}\begin{pmatrix}1&16&13\-1&-1&2\-2&-14&-8\end{pmatrix}=\begin{pmatrix}0&0&0\end{pmatrix}$. If $4\alpha^4+m\alpha^a+n\alpha^b=3$, then $m+n$ is equal to:

Accepted Answer

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 $.

We are given matrix equations with parameters $ a $ and $ b $ as follows:

$ \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} $

Solving these equations:

$ \begin{aligned} & \left[4 - a - 2b, 64 - a - 14b, 52 + 2a - 8bight] = \left[0, 0, 0ight] \ & \Rightarrow a + 2b = 4 \ & \Rightarrow a + 14b = 64 \end{aligned} $

From these, solve for $ a $ and $ b $:

$ \begin{aligned} & 12b = 60 \Rightarrow b = 5 \ & a = -6 \end{aligned} $

We now substitute $ a $ and $ b $ into the given function:

$ \frac{4}{\alpha^4} + \frac{m}{\alpha^a} + \frac{n}{\alpha^b} = 3 $

We substitute $ \alpha = \omega $ and simplify:

$ \begin{aligned} & \frac{4}{\omega} + \frac{m}{1} + \frac{n}{\omega^2} = 3 \ & \Rightarrow 4\omega^2 + m + n\omega = 3 \end{aligned} $

Substitute $ \omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i $ and $ \omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i $:

$ \begin{aligned} & 4\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}iight) + m + n\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}iight) = 3 \ & \Rightarrow -2 + m - \frac{n}{2} = 3 \ & 	ext{and } \frac{-4\sqrt{3}}{2} + \frac{n\sqrt{3}}{2} = 0 \end{aligned} $

From the above, solve for $ n $ and $ m $:

$ \begin{aligned} & n = 4 \ & m = 7 \end{aligned} $

Thus, $ m + n = 11 $.

Solutions · 1