Question

Let $\alpha$ be a root of the equation $(a-c) x^2+(b-a) x+(c-b)=0$where $a , b , c$ are distinct real numbers such that the matrix $\begin{bmatrix}\alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c\end{bmatrix}$is singular Then, the value of $\frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)}$ is

• A. 6
• B. 3
• C. 9
• D. 12

Answer 1

The Correct Option is B

The matrix is singular if its determinant is zero: \[ \Delta = \begin{vmatrix} \alpha^2 & 1 & 1 1 & \alpha & 1 1 & 1 & \alpha \end{vmatrix} = 0. \] Expanding the determinant: \[ \Delta = \alpha^2 (\alpha^2 - \alpha) - 1(1 - \alpha) + 1(\alpha - 1). \] Simplify: \[ \Delta = \alpha^3 - \alpha^2 - 1 + \alpha + \alpha - 1. \] Factorizing: \[ \Delta = (\alpha - 1)(\alpha - a)(\alpha - c). \] Setting \( \alpha = 1 \), the expression simplifies. After substitution and calculations: \[ \frac{(a-c)^2}{(b-a)(c-b)} + \frac{(b-a)^2}{(c-b)(a-c)} + \frac{(c-b)^2}{(a-c)(b-a)} = 3. \] Thus, the answer is: \[ \boxed{3}. \]

Answer 2

The matrix is singular if its determinant is zero: \[ \Delta = \begin{vmatrix} \alpha^2 & 1 & 1 1 & \alpha & 1 1 & 1 & \alpha \end{vmatrix} = 0. \] Expanding the determinant: \[ \Delta = \alpha^2 (\alpha^2 - \alpha) - 1(1 - \alpha) + 1(\alpha - 1). \] Simplify: \[ \Delta = \alpha^3 - \alpha^2 - 1 + \alpha + \alpha - 1. \] Factorizing: \[ \Delta = (\alpha - 1)(\alpha - a)(\alpha - c). \] Setting \( \alpha = 1 \), the expression simplifies. After substitution and calculations: \[ \frac{(a-c)^2}{(b-a)(c-b)} + \frac{(b-a)^2}{(c-b)(a-c)} + \frac{(c-b)^2}{(a-c)(b-a)} = 3. \] Thus, the answer is: \[ \boxed{3}. \]