Let α\alpha be a solution of x2+x+1=0x^2 + x + 1 = 0, and for some a and b in R\mathbb{R}, [4  a  b][11613−1−12−2−14−8]=[0  0  0][4 \; a \; b] \begin{bmatrix}1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8\end{bmatrix} = [0 \; 0 \; 0]. If 4α4+mα+nα2=3\frac{4}{\alpha^4} + \frac{m}{\alpha} + \frac{n}{\alpha^2} = 3, then m + n is equal to

Answer: -10

Step-by-step solution

Step 1: Analyze the quadratic equation

The given quadratic equation is x2+x+1=0x^2 + x + 1 = 0. The roots of this equation are the complex cube roots of unity, denoted by ω\omega and ω2\omega^2. Since α\alpha is a solution, we can say α=ω\alpha = \omega. We know that ω3=1\omega^3 = 1 and 1+ω+ω2=01 + \omega + \omega^2 = 0.

Step 2: Simplify the expression involving alpha

Substitute α=ω\alpha = \omega into the given expression. Since ω3=1\omega^3 = 1, we have ω4=ω3⋅ω=1⋅ω=ω\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega. Also, 1ω=ω2ω3=ω2\frac{1}{\omega} = \frac{\omega^2}{\omega^3} = \omega^2 and 1ω2=ωω3=ω\frac{1}{\omega^2} = \frac{\omega}{\omega^3} = \omega. Substitute these simplified terms back into the expression.

Step 3: Solve for m and n using properties of roots of unity

From 1+ω+ω2=01 + \omega + \omega^2 = 0, we know that ω2=−1−ω\omega^2 = -1 - \omega. Substitute this into the equation. Since 1,ω,ω21, \omega, \omega^2 are linearly independent over real numbers, for an equation of the form A+Bω=0A + B\omega = 0 where A and B are real, we must have A=0A=0 and B=0B=0. Here, we have (−4−m−3)+(n−4−m)ω=0(-4-m-3) + (n-4-m)\omega = 0.

Step 4: Formulate and solve system of equations

Equating the real and imaginary parts to zero, we get two linear equations. Solving the first equation for mm gives m=−7m = -7. Substituting this value into the second equation allows us to solve for nn, yielding n=−3n = -3.

Step 5: Calculate m + n

Finally, we need to find the sum m+nm+n. Substitute the values of mm and nn we found.

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