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 $\lambda^3$.