1. Key Concepts and Formulas

• Properties of Cube Roots of Unity: If $\alpha$ is a non-real solution of $x^2+x+1=0$, then it is a complex cube root of unity. This implies two fundamental properties:
  • $\alpha^3 = 1$
  • $1+\alpha+\alpha^2=0$ (or equivalently, $\alpha^2 = -1-\alpha$) These properties are essential for simplifying expressions involving powers of $\alpha$.
• Matrix Multiplication and Equality: When a row vector is multiplied by a matrix, the result is another row vector. For two matrices to be equal, their corresponding elements must be identical. This principle allows us to convert a matrix equation into a system of linear equations.
• Equality of Complex Numbers: If $P + Q\alpha = R$, where $P, Q, R$ are real numbers and $\alpha$ is a non-real complex number (like $-\frac{1}{2} + i\frac{\sqrt{3}}{2}$), then the coefficient of $\alpha$ ($Q$) must be zero, and the constant term ($P$) must be equal to $R$. This is because $\alpha$ has a non-zero imaginary part.

2. Step-by-Step Solution

Step 1: Understanding the Properties of $\alpha$ We are given that $\alpha$ is a solution of the quadratic equation $x^2 + x + 1 = 0$. Why: Identifying the nature of $\alpha$ is the first critical step, as it provides powerful simplification tools for any expression involving $\alpha$. Multiplying the equation $x^2 + x + 1 = 0$ by $(x-1)$, we get: $(x-1)(x^2+x+1) = 0$ $x^3 - 1 = 0$ This means $\alpha^3 = 1$. Additionally, from the original equation, we have $\alpha^2 + \alpha + 1 = 0$. These two properties ($\alpha^3=1$ and $1+\alpha+\alpha^2=0$) are fundamental for simplifying powers of $\alpha$.

Step 2: Solving the Matrix Equation for $a$ and $b$ We are given the matrix equation: $\begin{bmatrix} 4 & a & b \end{bmatrix} \begin{bmatrix} 1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$ Why: This step determines the values of $a$ and $b$, which are unknown exponents in the complex expression we need to evaluate later. Performing the matrix multiplication, we equate the resulting elements to zero:

1. First element: $(4)(1) + (a)(-1) + (b)(-2) = 0$ $4 - a - 2b = 0 \implies a + 2b = 4 \quad \text{(Equation 1)}$
2. Second element: $(4)(16) + (a)(-1) + (b)(-14) = 0$ $64 - a - 14b = 0 \implies a + 14b = 64 \quad \text{(Equation 2)}$
3. Third element: $(4)(13) + (a)(2) + (b)(-8) = 0$ $52 + 2a - 8b = 0 \implies 2a - 8b = -52 \implies a - 4b = -26 \quad \text{(Equation 3)}$

Now, we solve this system of linear equations for $a$ and $b$. We can use Equations 1 and 2. Subtract Equation 1 from Equation 2: $(a + 14b) - (a + 2b) = 64 - 4$ $12b = 60$ $b = 5$ Substitute $b=5$ into Equation 1: $a + 2(5) = 4$ $a + 10 = 4$ $a = -6$ To ensure consistency, we verify these values with Equation 3: $a - 4b = (-6) - 4(5) = -6 - 20 = -26$. This matches Equation 3. So, we have $a = -6$ and $b = 5$.

Step 3: Simplifying the Complex Expression We are given the expression: $\frac{4}{\alpha^4} + \frac{m}{\alpha^a} + \frac{n}{\alpha^b} = 3$ Why: We substitute the values of $a$ and $b$ found in the previous step and then use the property $\alpha^3=1$ to simplify the powers of $\alpha$ into their lowest forms. Substitute $a=-6$ and $b=5$: $\frac{4}{\alpha^4} + \frac{m}{\alpha^{-6}} + \frac{n}{\alpha^5} = 3$ Now, simplify each term using $\alpha^3=1$:

• $\alpha^4 = \alpha^3 \cdot \alpha = 1 \cdot \alpha = \alpha$
• $\alpha^{-6} = (\alpha^3)^{-2} = (1)^{-2} = 1$
• $\alpha^5 = \alpha^3 \cdot \alpha^2 = 1 \cdot \alpha^2 = \alpha^2$

Substitute these simplified powers back into the expression: $\frac{4}{\alpha} + \frac{m}{1} + \frac{n}{\alpha^2} = 3$ To further simplify, we can eliminate the denominators using $\frac{1}{\alpha} = \frac{\alpha^3}{\alpha} = \alpha^2$ and $\frac{1}{\alpha^2} = \frac{\alpha^3}{\alpha^2} = \alpha$: $4\alpha^2 + m + n\alpha = 3$ Rearrange the terms: $m + n\alpha + 4\alpha^2 = 3$ Now, use the property $\alpha^2 = -1-\alpha$ (from $1+\alpha+\alpha^2=0$) to express the equation in a linear form of $\alpha$: $m + n\alpha + 4(-1-\alpha) = 3$ $m + n\alpha - 4 - 4\alpha = 3$ Group the constant terms and the terms containing $\alpha$: $(m-4) + (n-4)\alpha = 3$

Step 4: Solving for $m$ and $n$ using Complex Number Equality We have the equation $(m-4) + (n-4)\alpha = 3$. Why: Since $m$ and $n$ are real numbers and $\alpha$ is a non-real complex number, we can equate the "real" and "imaginary" parts (coefficients of 1 and $\alpha$) on both sides of the equation. Let's represent $\alpha$ in its standard complex form: $\alpha = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. Substitute this into the equation: $(m-4) + (n-4)\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 3 + 0i$ Expand and separate the real and imaginary parts: $\left( (m-4) - \frac{1}{2}(n-4) \right) + i\left( \frac{\sqrt{3}}{2}(n-4) \right) = 3 + 0i$ Equate the imaginary parts on both sides: $\frac{\sqrt{3}}{2}(n-4) = 0$ Since $\frac{\sqrt{3}}{2} \neq 0$, we must have $n-4 = 0$, which implies $n=4$.

Now, equate the real parts on both sides: $(m-4) - \frac{1}{2}(n-4) = 3$ Substitute the value $n=4$ into this equation: $(m-4) - \frac{1}{2}(4-4) = 3$ $(m-4) - 0 = 3$ $m-4 = 3$ $m = 7$ Thus, we have found $m=7$ and $n=4$.

Step 5: Calculate $m+n$ The problem asks for the value of $m+n$. $m+n = 7+4 = 11$

3. Common Mistakes & Tips

• Misinterpreting $\alpha$: Always remember that $\alpha$ is a non-real complex number. If it were real, then $x^2+x+1=0$ would have no real solutions, and the equality $(m-4) + (n-4)\alpha = 3$ would imply $n-4=0$ only if $\alpha \neq 0$. The non-real nature of $\alpha$ is crucial for equating coefficients.
• Algebraic Errors in System of Equations: A small mistake in solving the system of linear equations for $a$ and $b$ can lead to incorrect values and propagate errors throughout the rest of the problem. Always perform a quick check with the third equation if available.
• Careless Power Simplification: When simplifying powers of $\alpha$, ensure you correctly apply $\alpha^3=1$ and $\alpha^2=-1-\alpha$. For instance, $\frac{1}{\alpha}$ is often mistaken for $\alpha^{-1}$ and then simplified incorrectly if $\alpha^2$ is not used. Remember $\frac{1}{\alpha} = \alpha^2$ and $\frac{1}{\alpha^2} = \alpha$.

4. Summary This problem required a multi-faceted approach, integrating concepts from complex numbers, matrices, and systems of linear equations. We began by leveraging the fundamental properties of $\alpha$ as a non-real cube root of unity. Subsequently, we solved a matrix equation to precisely determine the values of $a$ and $b$. With these values, we simplified a complex expression involving $\alpha$ using its cyclic properties and the relation $1+\alpha+\alpha^2=0$. Finally, by equating the real and imaginary components of the resulting complex equation, we found the values of $m$ and $n$, allowing us to calculate their sum.

5. Final Answer The final answer is $\boxed{11}$, which corresponds to option (A).