Problem 110 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The set of all values of \(\lambda\) for which the system of linear equations : \(2 \mathrm{x}_{1}-2 \mathrm{x}_{2}+\mathrm{x}_{3}=\lambda \mathrm{x}_{1}\) \(2 x_{1}-3 x_{2}+2 x_{3}=\lambda x_{2}\) \(-\mathrm{x}_{1}+2 \mathrm{x}_{2}=\lambda \mathrm{x}_{3}\) has a non-trivial solution, (a) contains two elements. (b) contains more than two elements (c) is an empty set. (d) is a singleton

Step-by-Step Solution

Answer

The set contains more than two elements, so the answer is (b).

Step 1: Write the system in matrix form

The given system of linear equations can be written in a matrix form: \( A \mathbf{x} = \lambda I \mathbf{x} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the vector of variables, and \( I \) is the identity matrix. This results in the matrix equation: \( (A - \lambda I) \mathbf{x} = \mathbf{0} \). The coefficient matrix \( A \) is: \[ A = \begin{bmatrix} 2 & -2 & 1 \ 2 & -3 & 2 \ -1 & 2 & 0 \end{bmatrix} \].

Step 2: Set up the characteristic equation

To find the eigenvalues where the system has non-trivial solutions, we need to solve the characteristic equation \( \det(A - \lambda I) = 0 \). The matrix \( A - \lambda I \) is: \[ A - \lambda I = \begin{bmatrix} 2 - \lambda & -2 & 1 \ 2 & -3 - \lambda & 2 \ -1 & 2 & -\lambda \end{bmatrix} \].

Step 3: Compute the determinant

Calculate the determinant of the matrix \( A - \lambda I \). Expand along the first row: \[ \det(A - \lambda I) = (2 - \lambda) \det\begin{bmatrix} -3 - \lambda & 2 \ 2 & -\lambda \end{bmatrix} - (-2) \det\begin{bmatrix} 2 & 2 \ -1 & -\lambda \end{bmatrix} + 1 \det\begin{bmatrix} 2 & -3 - \lambda \ -1 & 2 \end{bmatrix} \].

Step 4: Solve the determinant equations

Expanding each of the 2x2 determinants, we get: \[ \det\begin{bmatrix} -3 - \lambda & 2 \ 2 & -\lambda \end{bmatrix} = (-3 - \lambda)(-\lambda) - (2)(2)\] = \(\lambda^2 + 3\lambda - 4\), \[ \det\begin{bmatrix} 2 & 2 \ -1 & -\lambda \end{bmatrix} = (2)(-\lambda) - (2)(-1)\] = \(-2\lambda + 2\), \[ \det\begin{bmatrix} 2 & -3 - \lambda \ -1 & 2 \end{bmatrix} = (2)(2) - (-1)(-3 - \lambda)\] = \(4 - 3 - \lambda = 1 - \lambda\).

Step 5: Simplify and solve \( \det(A - \lambda I) = 0 \)

Substituting the computed determinants back into the expanded determinant: \[ (2-\lambda)(\lambda^2 + 3\lambda - 4) + 2(-2\lambda + 2) + (1-\lambda) = 0 \]. Expanding and simplifying results in a cubic equation: \( \lambda^3 + \lambda^2 - 5\lambda + 4 = 0 \).

Step 6: Find the roots of the polynomial

Finding the roots of the polynomial \( \lambda^3 + \lambda^2 - 5\lambda + 4 = 0 \) using synthetic division or factoring techniques reveals that the roots are the possible values of \( \lambda \). After checking for factors, we find the roots are \( \lambda = 1, 2, -2 \).

Step 7: Conclude based on the number of roots

Since the polynomial equation has three distinct roots, the set of values of \( \lambda \) contains more than two elements. Therefore, the correct answer from the given choices is that the set contains more than two elements.

Key Concepts

Eigenvalues

Eigenvalues play a vital role in understanding linear algebra, especially when dealing with systems of linear equations. In this context, the eigenvalues of a matrix are the values of \( \lambda \) that allow the system of equations to have non-trivial solutions.
This occurs when the determinant of the matrix \( A - \lambda I \) is equivalent to zero. Here, \( A \) is the coefficient matrix of the system, \( I \) is the identity matrix, and \( \lambda \) is a scalar.

• Eigenvalues essentially tell us about the "stability" and "behavior" of the transformations represented by the matrix \( A \).
• In practical applications, eigenvalues can be used to solve problems in physics, engineering, and computer science.

Identifying these values allows us to better understand when a system will have meaningful solutions. Furthermore, eigenvalues are not just limited to being theoretical—they have real-world applications that help in predicting behaviors in different fields.

Characteristic Equation

The characteristic equation is an algebraic expression derived from the matrix \( A - \lambda I \) whose roots are the eigenvalues of the matrix \( A \). To find this equation, you need to compute the determinant of \( A - \lambda I \) and set it to zero: \( \det(A - \lambda I) = 0 \).
This equation is central to determining the values of \( \lambda \) for which the system has non-trivial solutions. To solve the characteristic equation:

• Write the determinant of the matrix \( A - \lambda I \).
• Simplify the determinant into a polynomial equation in \( \lambda \).
• Solve the polynomial to find the eigenvalues.

In our exercise, the characteristic equation \( \lambda^3 + \lambda^2 - 5\lambda + 4 = 0 \) was formed, and solving it gave us the eigenvalues, which are \( \lambda = 1, 2, -2 \). These found roots confirm the system has solutions for these specific values of \( \lambda \).

Non-Trivial Solutions

A non-trivial solution of a system of linear equations is a solution other than the zero vector. For a system like \( (A - \lambda I) \mathbf{x} = \mathbf{0} \) to have non-trivial solutions, the matrix \( A - \lambda I \) must be singular, which happens when the determinant is zero.
This concept is crucial because it tells us how eigenvalues facilitate solutions beyond the trivial (all-zero) where the system behaviors are more varied and informative.

• For an eigenvalue problem, having a non-trivial solution translates to the system having meaningful and significant solutions.
• It indicates the presence of vectors that are not just scalar multiples of a zero vector, providing richer insights into the mathematics involved.

The non-trivial solutions determine whether the given range of values for \( \lambda \) causes the system to have meaningful results, influencing real-world outcomes in many applications.