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(B) Given $x^2+x+1=0$. Since $\alpha$ is a root,$\alpha^2+\alpha+1=0$. Thus,$\alpha = \omega$ or $\omega^2$,where $\omega$ is the complex cube root of

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$11$

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(B) Given $x^2+x+1=0$. Since $\alpha$ is a root,$\alpha^2+\alpha+1=0$. Thus,$\alpha = \omega$ or $\omega^2$,where $\omega$ is the complex cube root of unity. We know $\omega^3=1$ and $1+\omega+\omega^2=0$.Performing matrix multiplication: $\begin{bmatrix} 4 & a & b \end{bmatrix} \begin{bmatrix} 1 & 16 & 13 \ -1 & -1 & 2 \ -2 & -14 & -8 \end{bmatrix} = \begin{bmatrix} 4-a-2b & 64-a-14b & 52+2a-8b \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$.This gives the system: $a+2b=4$ and $a+14b=64$.Subtracting the equations: $12b=60 \Rightarrow b=5$. Substituting $b=5$ into $a+2b=4$ gives $a+10=4 \Rightarrow a=-6$.Now,substitute $a=-6, b=5$ into $\frac{4}{\alpha^4} + \frac{m}{\alpha^a} + \frac{n}{\alpha^b} = 3$.Since $\alpha^3=1$,$\alpha^4=\alpha$,$\alpha^{-6}=(\alpha^3)^{-2}=1$,and $\alpha^5=\alpha^2$.So,$\frac{4}{\alpha} + m + \frac{n}{\alpha^2} = 3$.Using $\frac{1}{\alpha} = \alpha^2$ and $\frac{1}{\alpha^2} = \alpha$,we get $4\alpha^2 + m + n\alpha = 3$.Since $\alpha^2 = -1-\alpha$,we have $4(-1-\alpha) + m + n\alpha = 3 \Rightarrow -4 - 4\alpha + m + n\alpha = 3$.Equating real and imaginary parts (or coefficients of $\alpha$ and constants): $(m-4) + (n-4)\alpha = 3$.Since $3 = 3(1) = 3(-\alpha^2-\alpha) = 3(1+\alpha+\alpha^2-1-\alpha^2-\alpha) = 3(0 - \alpha^2 - \alpha) = -3\alpha^2 - 3\alpha$. This approach is complex,so use $\alpha = \omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$.$4(-\frac{1}{2} - i\frac{\sqrt{3}}{2})^2 + m + n(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 3 \Rightarrow 4(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) + m - \frac{n}{2} + i\frac{n\sqrt{3}}{2} = 3$.Real part: $-2 + m - \frac{n}{2} = 3 \Rightarrow m - \frac{n}{2} = 5$.Imaginary part: $2\sqrt{3} + \frac{n\sqrt{3}}{2} = 0 \Rightarrow n = -4$.Then $m - (-2) = 5 \Rightarrow m = 3$.Wait,checking the equation $4\alpha^2 + n\alpha + m = 3$. Since $\alpha^2+\alpha+1=0$,$\alpha^2 = -1-\alpha$.$4(-1-\alpha) + n\alpha + m = 3 \Rightarrow (n-4)\alpha + (m-4) = 3$.For this to hold for $\alpha$,$n-4=0 \Rightarrow n=4$ and $m-4=3 \Rightarrow m=7$.Thus,$m+n = 7+4 = 11$.

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