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JEE Main 2022
Matrices & Determinants
Matrices and Determinants
Hard

Question

Let α\alpha be a root of the equation (a−c)x2+(b−a)x+(c−b)=0(a - c){x^2} + (b - a)x + (c - b) = 0 where a, b, c are distinct real numbers such that the matrix \left[ {\matrix{ {{\alpha ^2}} & \alpha & 1 \cr 1 & 1 & 1 \cr a & b & c \cr } } \right] is singular. Then, the value of (a−c)2(b−a)(c−b)+(b−a)2(a−c)(c−b)+(c−b)2(a−c)(b−a){{{{(a - c)}^2}} \over {(b - a)(c - b)}} + {{{{(b - a)}^2}} \over {(a - c)(c - b)}} + {{{{(c - b)}^2}} \over {(a - c)(b - a)}} is

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Solution

Key Concepts and Formulas Used:

  1. Singular Matrix: A square matrix is singular if and only if its determinant is zero.
  2. Roots of a Quadratic Equation: If x0x_0 is a root of the quadratic equation Ax2+Bx+C=0Ax^2 + Bx + C = 0, then Ax02+Bx0+C=0Ax_0^2 + Bx_0 + C = 0. A common observation is that if the sum of coefficients A+B+C=0A+B+C=0, then x=1x=1 is a root.
  3. Algebraic Identity: For any three numbers x,y,zx, y, z, if x+y+z=0x+y+z=0, then x3+y3+z3=3xyzx^3+y^3+z^3 = 3xyz.

Step-by-Step Elaborated Solution:

Step 1: Analyze the Given Quadratic Equation and Identify a Root

We are given the quadratic equation: (a−c)x2+(b−a)x+(c−b)=0(∗)(a - c){x^2} + (b - a)x + (c - b) = 0 \quad (*) Let's check if x=1x=1 is a root of this equation. We substitute x=1x=1 into the equation: (a−c)(1)2+(b−a)(1)+(c−b)(a - c)(1)^2 + (b - a)(1) + (c - b) =(a−c)+(b−a)+(c−b)= (a - c) + (b - a) + (c - b) =a−c+b−a+c−b= a - c + b - a + c - b =(a−a)+(b−b)+(c−c)=0= (a - a) + (b - b) + (c - c) = 0 Since substituting x=1x=1 makes the equation true (i.e., 0=00=0), x=1x=1 is a root

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