Question

If \( \alpha \) and \( \beta \) are non-real numbers satisfying \( x^3 - 1 = 0 \), then the value of \[ \left| \begin{matrix} \lambda+1 & \alpha & \beta \\ \beta & \lambda + \beta & 1 \\ 1 & \lambda + \alpha & \lambda + \alpha \end{matrix} \right| \] is:

• A. 0
• B. \(\lambda^3\)
• C. \(\lambda^3 + 1\)
• D. \(\lambda^3 - 1\)

Hint:

For determinants involving cubic equations, simplify using the roots of the equation and apply the determinant rules accordingly.

Answer 1

The Correct Option is B

The determinant of the matrix is calculated, and we are given that \( \alpha \) and \( \beta \) satisfy the equation \( x^3 - 1 = 0 \). This equation implies that both \( \alpha \) and \( \beta \) are roots of the equation \( x^3 = 1 \), meaning that they are cube roots of unity. These roots can be expressed as \( \alpha = 1 \), \( \beta = \omega \), and \( \omega^2 \), where \( \omega \) is a primitive cube root of unity.

Given this, the determinant simplifies to \( \lambda^3 \), which indicates that the determinant is a function of the eigenvalues of the matrix, and it is raised to the third power. This suggests that the matrix has a characteristic structure tied to the properties of cube roots of unity.

Therefore, the simplified determinant of the matrix is \( \lambda^3 \).