Problem 74 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^{3}+p x+q=0\) then the value of the determinant \(\left|\begin{array}{ccc}1+\alpha & 1 & 1 \\ 1 & 1+\beta & 1 \\ 1 & 1 & 1+\gamma\end{array}\right|\) is (A) \(p^{2}-2 q\) (B) \(3 p q\) (C) \(p-q\) (D) None of these

Step-by-Step Solution

Answer

C) The value of the determinant is \( p-q \).

Step 1: Definition of a Determinant

The determinant of a 3x3 matrix \( \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \) is calculated as \( a(ei - fh) - b(di - fg) + c(dh - eg) \).

Step 2: Define the Matrix Elements

In the given matrix \( \begin{bmatrix} 1+\alpha & 1 & 1 \ 1 & 1+\beta & 1 \ 1 & 1 & 1+\gamma \end{bmatrix}\), the elements are as follows: \( a = 1+\alpha, b = 1, c = 1, d = 1, e = 1+\beta, f = 1, g = 1, h = 1, \text{ and } i = 1+\gamma \).

Step 3: Calculate the Determinant

Substitute the matrix elements into the determinant formula:\[\begin{align*}&= (1+\alpha)((1+\beta)(1+\gamma) - 1) - 1(1(1+\gamma) - 1) + 1(1(1) - 1(1+\beta)) \&= (1+\alpha)(1 + \beta + \gamma + \beta \gamma - 1) - (1+\gamma - 1) + (1 - 1 - \beta) \&= (1+\alpha)(\beta + \gamma + \beta \gamma) - \gamma - \beta \&= \alpha \beta + \alpha \gamma + \alpha \beta \gamma + \beta + \gamma + \beta \gamma - \gamma - \beta\end{align*}\]

Step 4: Applying Vieta's formulas

For the cubic equation \(x^3 + px + q = 0\), by Vieta's formulas: \(\alpha + \beta + \gamma = 0\), \(\alpha \beta + \beta \gamma + \gamma \alpha = p\), \(\alpha \beta \gamma = -q\). Substitute these into the determinant:\[\begin{align*}= \alpha \beta + \alpha \gamma + \beta \gamma + q = p - q.\end{align*}\]

Step 5: Evaluate the Expression

Since all complex components \((\gamma + \beta) - \gamma - \beta\) cancel out and simplify using Vieta's relations, the final expression results in \(p - q\).

Key Concepts

Vieta's Formulas

Understanding Vieta's Formulas can greatly help in solving polynomial equations and relating them to their roots. These formulas provide shortcuts that relate the coefficients of a polynomial to sums and products of its roots.

For a cubic polynomial equation, like the one we deal with in our exercise, we have:

• The sum of the roots \(\alpha + \beta + \gamma\) is equal to 0. This is derived from the fact that there is no \(x^2\) term in the polynomial \(x^3 + px + q = 0\).
• The sum of the products of its roots taken two at a time, \(\alpha \beta + \beta \gamma + \gamma \alpha\), equals \(p\), which is the coefficient of \(x\).
• The product of the roots \(\alpha \beta \gamma\) is equal to \(-q\), where \(-q\) is the constant term in the polynomial.

Vieta's Formulas provide these connections without directly solving the polynomial, serving as an invaluable tool to simplify complex calculations.

Cubic Equation Roots

Finding the roots of a cubic equation involves different techniques compared to quadratic equations. A cubic equation like \(x^3 + px + q = 0\) might not factor easily, and its roots can be real or complex.

When we talk about roots in the context of cubic equations:

• To solve a cubic equation analytically, mathematicians often use methods such as factoring by grouping, synthetic division, or applying special formulas designed for cubics.
• Roots can be found using the Rational Root Theorem or by utilizing numerical methods for approximations, though exact solutions sometimes involve demanding calculations.
• In the context of the exercise, \(\alpha\), \(\beta\), and \(\gamma\) are assumed as the roots, allowing us to neatly apply Vieta's relationships to solve problems involving their relationships or derived formulas.

Comprehending the nature of these roots is key since they provide a foundation for further applications such as system solvers and analyzing polynomial behavior graphically.

Matrix Algebra

Matrix Algebra allows us to solve systems of equations, find determinants, and more. Determinants, in particular, are pivotal in understanding linear transformations and solving matrices.

A determinant of a 3x3 matrix is calculated from its elements, requiring precise arithmetic:

• To compute a determinant, you use the formula with the structured arrangement of matrix elements to calculate specific weighted sums, like in our matrix, where the expansion \((1+\alpha)((1+\beta)(1+\gamma) - 1)\), results from this calculation method.
• Determinants are used to determine the singularity of a matrix, impacting its invertibility. If the determinant is 0, the matrix is singular (non-invertible).
• In the given problem, you can see how matrix configurations directly relate to solving for specific determinants, leading to problem simplifications as observed in the solution \(p - q\).

Understanding matrix algebra and determinants unlocks a greater level of proficiency, equipping you to tackle various mathematical tasks with confidence.