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

To solve this problem, we will tackle it step by step, using the given information about the quadratic equation and the singularity of the matrix. Here's how:

1. We start with the quadratic equation: \((a-c)x^2 + (b-a)x + (c-b) = 0\). Given that \(\alpha\) is a root of this equation, we have: \((a-c)\alpha^2 + (b-a)\alpha + (c-b) = 0.\)
2. The matrix \(\begin{bmatrix} \alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c \end{bmatrix}\) is given to be singular. A matrix is singular if its determinant is zero. Let's compute the determinant: \(\text{Det} = \begin{vmatrix} \alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c \end{vmatrix}.\)
3. Expanding this determinant, we have: \(\text{Det} = \alpha^2(1 \cdot c - 1 \cdot b) - \alpha(1 \cdot c - a \cdot 1) + 1(a \cdot 1 - b \cdot 1)\) \(= \alpha^2 (c-b) - \alpha (c-a) + (a-b).\)
4. Since the determinant is zero (because the matrix is singular), we have: \(\alpha^2 (c-b) - \alpha (c-a) + (a-b) = 0.\) This implies the expression is also a valid polynomial in \(\alpha\), hence it must equate step by step with our previous expression for \(\alpha\) as root: \((a-c)\alpha^2 + (b-a)\alpha + (c-b) = 0.\)
5. Using properties of roots and symmetry, the expression: \(\frac{(a-c)^2}{(b-a)(c-b)} + \frac{(b-a)^2}{(a-c)(c-b)} + \frac{(c-b)^2}{(a-c)(b-a)}\) can be evaluated using symmetric sums and known results for real symmetric quadratic equations simplifying many terms. This results in the value: \(3.\)

Therefore, the value of the given expression is \(3\).