Question

If \( x \) is a complex root of the equation
\[ \begin{vmatrix} 1 & x & x \\ x & 1 & x \\ x & x & 1 \end{vmatrix} + \begin{vmatrix} 1 - x & 1 & 1 \\ 1 & 1 - x & 1 \\ 1 & 1 & 1 - x \end{vmatrix} = 0, \] then \( x^{2007} + x^{-2007} \) is:

• A. \( 1 \)
• B. \( -1 \)
• C. \( -2 \)
• D. \( 2 \)

Hint:

If \( x \) is a cube root of unity, then it satisfies \( x^3 = -1 \), which helps simplify large exponents.

Answer 1

The Correct Option is C

Step 1: {Expand both determinants}
\[ (1 - 3x^2 + 2x^3) + (3x^2 - x^3) = 0. \] Step 2: {Solve for \( x \)}
\[ x^3 + 1 = 0. \] \[ x^3 = -1. \] \[ x = -\omega, -\omega^2, -1. \] Step 3: {Compute \( x^{2007} + x^{-2007} \)}
Since \( x^3 = -1 \), \[ x^{2007} = (-1)^{669} = -1. \] \[ x^{-2007} = -1. \] Step 4: {Conclusion}
\[ x^{2007} + x^{-2007} = -1 - 1 = -2. \]

Answer 2

Step 1: Understanding the equation
We are given the following equation: \[ \begin{vmatrix} 1 & x & x \\ x & 1 & x \\ x & x & 1 \end{vmatrix} + \begin{vmatrix} 1 - x & 1 & 1 \\ 1 & 1 - x & 1 \\ 1 & 1 & 1 - x \end{vmatrix} = 0, \] where \( x \) is a complex root. We need to find \( x^{2007} + x^{-2007} \).

Step 2: Simplifying the first determinant
Consider the first determinant: \[ D_1 = \begin{vmatrix} 1 & x & x \\ x & 1 & x \\ x & x & 1 \end{vmatrix}. \] We can expand along the first row to simplify this determinant: \[ D_1 = 1 \cdot \begin{vmatrix} 1 & x \\ x & 1 \end{vmatrix} - x \cdot \begin{vmatrix} x & x \\ x & 1 \end{vmatrix} + x \cdot \begin{vmatrix} x & 1 \\ x & x \end{vmatrix}. \] Now compute the 2x2 determinants: \[ \begin{vmatrix} 1 & x \\ x & 1 \end{vmatrix} = 1 \cdot 1 - x \cdot x = 1 - x^2, \] \[ \begin{vmatrix} x & x \\ x & 1 \end{vmatrix} = x \cdot 1 - x \cdot x = x - x^2, \] \[ \begin{vmatrix} x & 1 \\ x & x \end{vmatrix} = x \cdot x - 1 \cdot x = x^2 - x. \] Substitute these back into the expansion: \[ D_1 = 1 \cdot (1 - x^2) - x \cdot (x - x^2) + x \cdot (x^2 - x) = 1 - x^2 - x(x - x^2) + x(x^2 - x). \] Simplify further: \[ D_1 = 1 - x^2 - x^2 + x^3 + x^3 - x^2 = 1 - 3x^2 + 2x^3. \]

Step 3: Simplifying the second determinant
Now consider the second determinant: \[ D_2 = \begin{vmatrix} 1 - x & 1 & 1 \\ 1 & 1 - x & 1 \\ 1 & 1 & 1 - x \end{vmatrix}. \] We can expand along the first row to simplify this determinant: \[ D_2 = (1 - x) \cdot \begin{vmatrix} 1 - x & 1 \\ 1 & 1 - x \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 1 \\ 1 & 1 - x \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1 - x \\ 1 & 1 \end{vmatrix}. \] Now compute the 2x2 determinants: \[ \begin{vmatrix} 1 - x & 1 \\ 1 & 1 - x \end{vmatrix} = (1 - x)(1 - x) - 1 \cdot 1 = (1 - x)^2 - 1 = x^2 - 2x, \] \[ \begin{vmatrix} 1 & 1 \\ 1 & 1 - x \end{vmatrix} = 1 \cdot (1 - x) - 1 \cdot 1 = 1 - x - 1 = -x, \] \[ \begin{vmatrix} 1 & 1 - x \\ 1 & 1 \end{vmatrix} = 1 \cdot 1 - (1 - x) \cdot 1 = 1 - (1 - x) = x. \] Substitute these back into the expansion: \[ D_2 = (1 - x) \cdot (x^2 - 2x) - 1 \cdot (-x) + 1 \cdot (x) = (1 - x)(x^2 - 2x) + x + x. \] Now expand the first term: \[ D_2 = (1 - x)(x^2 - 2x) = x^2 - 2x - x^3 + 2x^2 = -x^3 + 3x^2 - 2x. \] So: \[ D_2 = -x^3 + 3x^2 - 2x + 2x = -x^3 + 3x^2. \]

Step 4: Combining the results
Now substitute \( D_1 \) and \( D_2 \) into the original equation: \[ D_1 + D_2 = 0, \] \[ (1 - 3x^2 + 2x^3) + (-x^3 + 3x^2) = 0, \] \[ 1 - 3x^2 + 2x^3 - x^3 + 3x^2 = 0, \] \[ 1 + x^3 = 0, \] \[ x^3 = -1, \] \[ x = -1. \]

Step 5: Final answer
We are asked to find \( x^{2007} + x^{-2007} \). Since \( x = -1 \), we have: \[ x^{2007} = (-1)^{2007} = -1 \quad \text{and} \quad x^{-2007} = (-1)^{-2007} = -1. \] Thus: \[ x^{2007} + x^{-2007} = -1 + (-1) = -2. \]

The correct answer is: -2