Problem 122 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The system of equations $$ \begin{aligned} &\alpha x+y+z=\alpha-1 \\ &x+\alpha y+z=\alpha-1 \\ &x+y+\alpha z=\alpha-1 \\ &\text { has infinite solutions, if } \alpha \text { is } \end{aligned} $$ (a) \(-2\) (b) either \(-2\) or 1 (c) not \(-2\) (d) 1

Step-by-Step Solution

Answer

The system has infinite solutions if \( \alpha = 1 \), answer (d).

Step 1: Analyze the System of Equations

Examine the given system of equations: \[ \begin{aligned} \alpha x + y + z &= \alpha - 1 \ x + \alpha y + z &= \alpha - 1 \ x + y + \alpha z &= \alpha - 1 \end{aligned} \] We are looking for values of \( \alpha \) that lead to infinitely many solutions. For the system to have infinite solutions, the determinant of the coefficient matrix must be zero.

Step 2: Write the Coefficient Matrix

The coefficient matrix \( A \) for the given system is: \[ A = \begin{bmatrix} \alpha & 1 & 1 \ 1 & \alpha & 1 \ 1 & 1 & \alpha \end{bmatrix} \] The condition for infinitely many solutions is \( \det(A) = 0 \).

Step 3: Calculate the Determinant

Calculate the determinant of matrix \( A \): \[ \det(A) = \alpha(\alpha^2 - 1) + 1(\alpha - 1) + 1(1 - \alpha) \] This simplifies to \[ \det(A) = \alpha^3 - \alpha + \alpha - 1 + 1 - \alpha \] further simplifying gives \[ \det(A) = \alpha^3 - \alpha \] which simplifies to \[ \det(A) = \alpha(\alpha^2 - 1) \].

Step 4: Solve for Infinite Solutions Condition

Set the determinant equal to zero: \[ \alpha(\alpha^2 - 1) = 0 \] Solve \( \alpha = 0 \) or \( \alpha^2 - 1 = 0 \). For \( \alpha^2 - 1 = 0 \), we have \( \alpha = \pm 1 \). This means the solutions for \( \alpha \) are 0, 1, or -1.

Step 5: Evaluate the Given Options

Compare the solutions to the options provided. The system has infinite solutions for \( \alpha = 1 \), as 0 is not an option. Check other conditions: Option (a) is \(-2\), option (b) either \(-2\) or 1, option (c) not \(-2\), and option (d) 1. Since \(-2\) does not satisfy the determinant condition, only 1 works, matching option (d).

Key Concepts

System of Equations

A system of equations consists of multiple equations that we solve together because they share common variables. In this problem, we are dealing with three equations that are connected:

• \( \alpha x + y + z = \alpha - 1 \)
• \( x + \alpha y + z = \alpha - 1 \)
• \( x + y + \alpha z = \alpha - 1 \)

These equations have three variables: \(x\), \(y\), and \(z\). Each equation represents a plane in three-dimensional space. Solving the system means finding the values of \(x\), \(y\), and \(z\) that satisfy all equations simultaneously. When these planes intersect in a line or coincide entirely, the system has infinite solutions. This condition exists when the determinant of the coefficient matrix is zero.

Infinite Solutions

A system of equations has infinite solutions when there are more solutions than there are equations, or equivalently, when at least two of the equations describe the same plane or intersect along a line. It indicates that many combinations of the variables satisfy all equations in the system at once.
In practical terms, infinite solutions occur when the equations effectively describe the same geometric plane, pressing on the need to check the determinant of the coefficient matrix. If this determinant is zero, it implies dependency among the equations, signifying that they cannot define a single unique solution.

Coefficient Matrix

The coefficient matrix is a matrix that contains all the coefficients of the variables from the system of equations. For the system given in this exercise, the coefficient matrix \( A \) is:

• \( A = \begin{bmatrix} \alpha & 1 & 1 \ 1 & \alpha & 1 \ 1 & 1 & \alpha \end{bmatrix} \)

Each row in the matrix represents the coefficients of each respective equation in the system. Using the coefficient matrix, we can perform several operations, such as calculating the determinant, which helps us understand the system's solvability characteristics, such as determining if there are infinite solutions.

Determinant Calculation

Calculating the determinant of the coefficient matrix is crucial when assessing the solvability of a system of equations. In our given problem, we found:

• \( \det(A) = \alpha(\alpha^2 - 1) \)

Simplifying further, we get the expression \( \alpha^3 - \alpha \).

For the system to have infinite solutions, we set this determinant equal to zero. Solving \( \alpha(\alpha^2 - 1) = 0 \) gives us:

• Possible solutions where \( \alpha = 0 \)
• \( \alpha^2 - 1 = 0 \), leading to \( \alpha = \pm1 \)

These values tell us when the equations describe coincident planes, making infinite solutions possible. In other words, the determinant condition ensures these possibilities, deeply linked to matrix theory fundamentals in linear algebra.