Let be a solution of x^2+x+1=0, and for some a and b in R,[gathered 4 a b gathered][[|gath…

Question

Let \alpha  be a solution of ${x}^{2}+x+1=0$, and for some a and b in
 $R,[\begin{gathered} 4 \quad  a \quad  b \end{gathered}][[|\begin{gathered} 1 \quad  16 \quad  13 \ -1 \quad  -1 \quad  2 \ -2 \quad  -14 \quad  -8 \end{gathered}]]|=[\begin{gathered} 0 \quad  0 \quad  0 \end{gathered}]$. If $\frac{4}{{\alpha }^{4}}+\frac{m}{{\alpha }^{a}}+\frac{n}{{\alpha }^{b}}=3$, 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:

$[\begin{gathered} 4 \quad  a \quad  b \end{gathered}][[|\begin{gathered} 1 \quad  16 \quad  13 \ -1 \quad  -1 \quad  2 \ -2 \quad  -14 \quad  -8 \end{gathered}]]|=[\begin{gathered} 0 \quad  0 \quad  0 \end{gathered}]$

Solving these equations:

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

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

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

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:

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

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

$\begin{gathered} 4(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)+m+n(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)=3 \ \Rightarrow -2+m-\frac{n}{2}=3 \ \frac{-4\sqrt{3}}{2}+\frac{n\sqrt{3}}{2}=0 \end{gathered}$

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

$\begin{gathered} n=4 \ m=7 \end{gathered}$

Thus, $m+n=11$.

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