Problem 96 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \((b-c)^{2} \neq(a-b)(c-a)\), solve for \(x\left|\begin{array}{lll}a+x & b+x & c+x \\ b+x & c+x & a+x \\ c+x & a+x & b+x\end{array}\right|=0\)

Step-by-Step Solution

Answer

The roots of the equation are \(x = [-(a+b+c) \pm sqrt((a+b+c)^2-3.(ab+bc+ca))] / 3\).

Step 1: Expand the Determinant

First, expand the determinant using the formula for a 3x3 matrix. Doing so, we get:\[(a+x)[(b+x)(c+x)-(c+x)(b+x)] - (b+x)[(b+x)(c+x)-(a+x)(c+x)] + (c+x)[(b+x)(a+x)-(b+x)(c+x)]\]

Step 2: Simplify the Determinant

Simplifying the determinant, we find that all the terms get canceled out except ones involving \(x^2\), \(x\), and the constant term. This simplifies to the quadratic equation:\[3x^{2} + 2(a+b+c)x + (ab+bc+ca) = 0\]

Step 3: Solve the Quadratic Equation

The above quadratic equation can now be solved using the formula for the roots of a quadratic equation, given as:\[x = [-(b) ± sqrt(b^2-4ac)] / (2a)\]So, on substituting the values of a, b, and c, we get\[x = [-(2(a+b+c)) ± sqrt((2(a+b+c))^2-4.3.(ab+bc+ca))](2*3)^{-1}\]

Step 4: Simplify the Expression for Roots of x

On further simplification, we get\[x = [-(a+b+c) ± sqrt((a+b+c)^2-3.(ab+bc+ca))]/3\]So, the roots of the equation are \(x = [-(a+b+c) + sqrt((a+b+c)^2-3.(ab+bc+ca))] / 3 \) and \(x = [-(a+b+c) - sqrt((a+b+c)^2-3.(ab+bc+ca))] / 3\)

Key Concepts

Determinant of a Matrix

The determinant of a matrix plays a critical role in mathematics, specifically in linear algebra. It is a special number that is calculated from a square matrix. For a 3x3 matrix, the determinant provides valuable information about the matrix, such as whether the matrix is invertible and the volume scaling factor for transformations it represents.

To compute the determinant of a 3x3 matrix, you cross-multiply elements across the rows and columns following a specific pattern, then add and subtract these products. The process is known as expansion by minors and cofactors. In the given exercise, the determinant is expanded using this technique, which helps simplify the complex matrix into a more manageable quadratic equation.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, meaning the highest power of the variable, typically written as 'x', is 2. The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \) where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero. These equations are fundamental in algebra and appear frequently in various areas of science and engineering.

The quadratic formula, \( x = [-(b) \pm \sqrt{b^2-4ac}] / (2a) \), is a powerful tool derived from completing the square and allows us to solve any quadratic equation, even if the solutions are complex numbers. The formula provides the roots of the equation directly by substituting the coefficients a, b, and c.

Roots of a Quadratic Equation

The roots of a quadratic equation are the values of 'x' that satisfy the equation. They represent the points where the graph of the quadratic function crosses the x-axis. Depending on the value of the discriminant \( b^2 - 4ac \), a quadratic equation can have two distinct real roots, one real root (repeated), or two complex roots (if the discriminant is negative).

The discriminant also determines the nature of the roots. If it's positive, the roots are real and distinct; if it's zero, the roots are real and equal, known as a repeated or double root; and if it's negative, the roots are complex, which involve the imaginary unit 'i'. In solving the provided exercise, we see that the quadratic formula applied simplifies to express the roots in terms of known quantities 'a, b, c', giving a clearer insight into the algebraic structure of the solution.