1. Use the matrix equation to find $a$ and $b$

We are given

This gives three linear equations by equating each component:

• From column 1: $4(1)+a(-1)+b(-2)=0$ $4-a-2b=0 \quad \Rightarrow \quad a+2b=4 \qquad (1)$

• From column 2: $4(16)+a(-1)+b(-14)=0$ $64-a-14b=0 \quad \Rightarrow \quad a+14b=64 \qquad (2)$

• From column 3: $4(13)+a(2)+b(-8)=0$ $52+2a-8b=0 \quad \Rightarrow \quad a-4b=-26 \qquad (3)$

Now solve using (1) and (2):

Subtract (1) from (2): $12b=60 \Rightarrow b=5.$

Then from (1): $a+2(5)=4 \Rightarrow a=-6.$

Check in (3): $-6-4(5)=-6-20=-26,$ which is correct.

So, $a=-6, \qquad b=5.$

2. Use the property of $\alpha$

Given $\alpha$ is a root of $x^2+x+1=0.$

Since $x^3-1=(x-1)(x^2+x+1),$ any root of $x^2+x+1=0$ satisfies $\alpha^3=1, \qquad \alpha\ne 1.$

Hence powers of $\alpha$ repeat modulo $3$.

3. Simplify the given expression

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

Using $a=-6$ and $b=5$: $\frac{4}{\alpha^4}+\frac{m}{\alpha^{-6}}+\frac{n}{\alpha^5}=3.$

This becomes $\frac{4}{\alpha^4}+m\alpha^6+\frac{n}{\alpha^5}=3.$

Now reduce powers using $\alpha^3=1$:

• $\alpha^4=\alpha$, so $\frac{1}{\alpha^4}=\frac{1}{\alpha}=\alpha^2$ because $\alpha^3=1 \Rightarrow \alpha^{-1}=\alpha^2$.

Thus $\frac{4}{\alpha^4}=4\alpha^2.$

• $\alpha^6=(\alpha^3)^2=1$, so $m\alpha^6=m.$

• $\alpha^5=\alpha^2$, so $\frac{1}{\alpha^5}=\alpha$ (because $\alpha^{-2}=\alpha$ when $\alpha^3=1$).

Thus $\frac{n}{\alpha^5}=n\alpha.$

Therefore the equation becomes $4\alpha^2+m+n\alpha=3.$

4. Use $\alpha^2+\alpha+1=0$

From $\alpha^2+\alpha+1=0,$ we get $\alpha^2=-\alpha-1.$

Substitute into the expression: $4\alpha^2+m+n\alpha=3$ $4(-\alpha-1)+m+n\alpha=3$ $-4\alpha-4+m+n\alpha=3$ $(n-4)\alpha+(m-4)=3.$

So, $(n-4)\alpha+(m-7)=0.$

Since $\alpha$ is non-real and cannot satisfy a real linear equation unless both coefficients are zero, we must have $n-4=0, \qquad m-7=0.$

Hence $n=4, \qquad m=7.$

Therefore, $m+n=7+4=11.$

5. Compare with stored answer

Derived answer: $11$

Stored correct answer: A = $11$

They agree.