Problem 26 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\alpha\) and \(\beta\) be the roots of the equation \(x^{2}+x+1=0\). Then for \(y^{\prime \prime \prime} 0\) in \(R,\left|\begin{array}{ccc}y+1 & \alpha & \beta \\ \alpha & y+\beta & 1 \\ \beta & 1 & y+\alpha\end{array}\right|\) is equal to: [April09, 2019 (I)] (a) \(y\left(y^{2}-1\right)\) (b) \(y\left(y^{2}-3\right)\) (c) \(\mathrm{y}^{3}\) (d) \(\mathrm{y}^{3}-1\)

Step-by-Step Solution

Answer

The answer is (b) \( y(y^2 - 3) \).

Step 1: Recognize the Problem

The given problem involves evaluating a determinant with roots of a quadratic equation. Recognize that \( \alpha \) and \( \beta \) are roots of \( x^2 + x + 1 = 0 \).

Step 2: Identify Key Properties

From the equation \( x^2 + x + 1 = 0 \), using Vieta's formulas, we have \( \alpha + \beta = -1 \) and \( \alpha \beta = 1 \).

Step 3: Calculate the Determinant

Set up the determinant to be calculated as \( \begin{vmatrix} y+1 & \alpha & \beta \ \alpha & y+\beta & 1 \ \beta & 1 & y+\alpha \end{vmatrix} \).

Step 4: Expand the Determinant

Expand along the first row to get:\[ (y+1) \begin{vmatrix} y+\beta & 1 \ 1 & y+\alpha \end{vmatrix} - \alpha \begin{vmatrix} \alpha & 1 \ \beta & y + \alpha \end{vmatrix} + \beta \begin{vmatrix} \alpha & y + \beta \ \beta & 1 \end{vmatrix} \].

Step 5: Solve Each Minor Determinant

1. The first minor: \( \begin{vmatrix} y+\beta & 1 \ 1 & y+\alpha \end{vmatrix} = (y+\beta)(y+\alpha) - 1 \).2. The second minor: \( \begin{vmatrix} \alpha & 1 \ \beta & y+\alpha \end{vmatrix} = \alpha(y+\alpha) - \beta \).3. The third minor: \( \begin{vmatrix} \alpha & y+\beta \ \beta & 1 \end{vmatrix} = \alpha - \beta(y+\beta) \).

Step 6: Substitute and Simplify

Substitute the minors into the determinant expression from Step 4, and simplify using \( \alpha + \beta = -1 \) and \( \alpha \beta = 1 \).

Step 7: Simplify Further Using Key Properties

Replace and simplify all terms, ensuring to use that for roots: \( \beta = -1 - \alpha \).This results in: \( (y+1)((y-1)^2 + y) = y(y^2 - 3) \) after simplification.

Step 8: Match to Given Options

After simplification, the expression matches option (b) \( y(y^2 - 3) \).

Key Concepts

Quadratic Roots

Understanding quadratic roots is crucial for solving quadratic equations. Quadratic equations are in the form of \( ax^2 + bx + c = 0 \). The solutions, known as roots, are the values of \( x \) that satisfy the equation. For the equation \( x^2 + x + 1 = 0 \), the roots are complex numbers, \( \alpha \) and \( \beta \).These roots are not real numbers because the discriminant, \( b^2 - 4ac \) (computed as \( 1^2 - 4 \cdot 1 \cdot 1 = -3 \)), is negative. Complex roots appear as conjugate pairs. For this equation, the roots can be expressed as:

• \( \alpha = \frac{-1 + i\sqrt{3}}{2} \)
• \( \beta = \frac{-1 - i\sqrt{3}}{2} \)

These roots are used to evaluate the determinant in the exercise problem.

Vieta's Formulas

Vieta's formulas establish a relationship between the coefficients of a polynomial and sums and products of its roots. For the quadratic equation \( ax^2 + bx + c = 0 \), the formulas are:

• Sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
• Product of the roots \( \alpha \beta = \frac{c}{a} \)

In the given quadratic equation \( x^2 + x + 1 = 0 \), the sum of the roots is \( \alpha + \beta = -1 \) and the product is \( \alpha \beta = 1 \).
These properties simplify the evaluation of determinants and reduce computation errors, making them a powerful tool in algebra.

Matrix Expansion

Matrix expansion allows us to calculate the determinant of a matrix by expanding along a chosen row or column. In this exercise, the determinant of a \(3 \times 3\) matrix is calculated by expanding along the first row. The formula involves creating smaller \(2 \times 2\) determinants, known as minors.For a matrix \( A \) with elements \( a_{ij} \), the determinant can be expanded as:

• \( \det(A) = a_{11}C_{11} - a_{12}C_{12} + a_{13}C_{13} \)

Here, \( C_{ij} \) represents the cofactor of the element \( a_{ij} \), which is obtained by multiplying the minor of \( a_{ij} \) by \( (-1)^{i+j} \).In our problem, expansion and substitution aided in simplifying the determinant into polynomial form. Understanding matrix expansion is crucial for evaluating determinants effectively.

Polynomial Simplification

Polynomial simplification involves reducing an expression to its simplest form by combining like terms and using algebraic properties. It's an essential step in solving equations and simplifying expressions for clearer insights.In the determinant exercise, once the minors were substituted and expanded, polynomial simplification was necessary. The expression:

• \((y+1)((y-1)^2 + y)\)

was simplified by expanding and combining like terms, thus reducing it to \( y(y^2 - 3) \). This process used the key properties of the roots obtained via Vieta's formulas to help simplify the terms.Practical algebra skills greatly enhance the simplification process, making complex expressions more manageable and leading to solutions that match given options.