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

Since $$\alpha$$ satisfies $$x^{2}+x+1=0$$, we have
$$\alpha^{2}+\alpha+1=0$$ and hence $$\alpha^{3}=1,\;\alpha\neq 1.$$

The condition
$$[4\;a\;b]\, \begin{bmatrix} 1 & 16 & 13\\ -1& -1 & 2\\ -2&-14&-8 \end{bmatrix} =[0\;0\;0] $$ means the row-vector $$[4\;a\;b]$$ is in the left-null-space of the matrix. Multiplying out gives three linear equations:

$$4(1)+a(-1)+b(-2)=0\;\;\Longrightarrow\;\;4-a-2b=0\;\;\Longrightarrow\;\;a+2b=4\; -(1)$$
$$4(16)+a(-1)+b(-14)=0\;\;\Longrightarrow\;\;64-a-14b=0\;\;\Longrightarrow\;\;a+14b=64\; -(2)$$
$$4(13)+a(2)+b(-8)=0\;\;\Longrightarrow\;\;52+2a-8b=0\;\;\Longrightarrow\;\;a-4b=-26\; -(3)$$

Solving $$(1)$$ and $$(2)$$:
$$\bigl(a+14b\bigr)-\bigl(a+2b\bigr)=64-4\;\;\Longrightarrow\;\;12b=60\;\;\Longrightarrow\;\;b=5.$$

Substituting $$b=5$$ in $$(1)$$: $$a+2(5)=4\;\;\Longrightarrow\;\;a=-6.$$ (Equation $$(3)$$ is also satisfied, so the solution is consistent.)

Now evaluate each reciprocal power of $$\alpha$$ needed in the expression $$\frac{4}{\alpha^{4}}+\frac{m}{\alpha^{a}}+\frac{n}{\alpha^{5}}=3.$$ Because $$\alpha^{3}=1,$$ we can reduce every exponent modulo $$3$$:

$$\frac{1}{\alpha^{4}}=\frac{1}{\alpha^{3}\alpha}=\frac{1}{\alpha}= \alpha^{2},$$
$$\frac{1}{\alpha^{a}}=\frac{1}{\alpha^{-6}}=\alpha^{6}= (\alpha^{3})^{2}=1,$$
$$\frac{1}{\alpha^{5}}=\frac{1}{\alpha^{3}\alpha^{2}}=\frac{1}{\alpha^{2}}=\alpha.$$

Substituting these values converts the given relation to a polynomial in $$\alpha$$:

$$4\alpha^{2}+m\cdot 1+n\alpha = 3 \;\;\Longrightarrow\;\; 4\alpha^{2}+n\alpha+(m-3)=0.$$

The above equation must hold for both roots of $$x^{2}+x+1=0.$$ A polynomial of degree $$2$$ that vanishes at both roots of another irreducible quadratic is necessarily a scalar multiple of that quadratic. Hence

$$4\alpha^{2}+n\alpha+(m-3)=k\bigl(\alpha^{2}+\alpha+1\bigr).$$

Comparing coefficients gives the common scalar $$k=4$$ and

$$n = k = 4, \qquad m-3 = k = 4 \;\;\Longrightarrow\;\; m = 7.$$

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

Option B (11)