Problem 98 - 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 \(x-2 y-2 z=\lambda x\) \(x+2 y+z=\lambda y\) \(-x-y=\lambda 2\) has a non-trivial solution: \(\quad\) [Jan. 12, 2019 (II)] (a) is a singleton (b) contains exactly two elements (c) is an empty set (d) contains more than two elements

Step-by-Step Solution

Answer

(b) contains exactly two elements

Step 1: Setup the system of equations

We have the system of equations from the problem statement:1. \(x - 2y - 2z = \lambda x\)2. \(x + 2y + z = \lambda y\)3. \(-x - y = \lambda 2\)We need to find the appropriate value(s) of \(\lambda\) that allow a non-trivial solution (where \(x, y, z\) are not all zero). To find these values, first set up the corresponding matrix equation.

Step 2: Rewrite as a matrix equation

Express the system as a matrix equation, \(AX = \lambda BX\), where:- \(A = \begin{pmatrix} 1 & -2 & -2 \ 1 & 2 & 1 \ -1 & -1 & 0 \end{pmatrix}\)- \(B = \begin{pmatrix} x \ y \ 2 \end{pmatrix}\)Thus, the equation becomes \((A - \lambda I)X = 0\), where \(I\) is the identity matrix. This requires solving the characteristic equation to ensure a non-trivial solution exists.

Step 3: Calculate the determinant of A-λI

Determine \(A - \lambda I\) as follows:\[N = A - \lambda I = \begin{pmatrix} 1-\lambda & -2 & -2 \ 1 & 2-\lambda & 1 \ -1 & -1 & -\lambda \end{pmatrix}\]Now, find the determinant of \(N\) and set it equal to zero, since non-trivial solutions exist when this determinant is zero.

Step 4: Solve for the determinant

Calculate the determinant, \(\text{det}(N)\), for matrix \(N\):\[\text{det}(N) = (1-\lambda)((2-\lambda)(-\lambda) - (-1)) - (-2)((-(\lambda)) - (-1)(-1)) - (-2)((2-\lambda)(-1) - (-1)) = \]Simplify to get the characteristic polynomial:\[(\lambda - 2)(\lambda^2 - 2\lambda) = 0\]

Step 5: Find solutions for λ

Solve the characteristic polynomial \((\lambda-2)(\lambda^2-2\lambda)=0\). This yields:\[(\lambda-2)(\lambda)(\lambda-2) = 0\]Simplifying, the solutions are: \(\lambda = 0\) and \(\lambda = 2\).

Step 6: Determine the number of λ solutions

The characteristic equation solutions tell us that the values of \(\lambda\) for which the system has a non-trivial solution are \(\lambda = 0\) and \(\lambda = 2\). These two values form the complete set of possible \(\lambda\) values.

Key Concepts

Non-Trivial Solutions

In a system of linear equations, finding non-trivial solutions means identifying solutions where not all variables are zero. Typically, a system of linear equations has a trivial solution, which is when all variables like \(x\), \(y\), and \(z\) equal zero. However, non-trivial solutions indicate that there are interesting solutions where one or more variables are non-zero. This usually happens when the determinant of the coefficient matrix is zero, allowing the system of equations to have infinite solutions rather than just the trivial zero vector. Non-trivial solutions are essential in various applications such as engineering, physics, and economics, where they can describe equilibrium states or steady-states of a system. In our exercise, looking for non-trivial solutions with respect to \(\lambda\) is crucial to determining the values of \(\lambda\) that lead to meaningful outcomes in the equations.

Matrix Equations

Matrices provide a compact way of representing and solving systems of linear equations. A matrix equation like \(AX = \lambda BX\) can translate multiple linear equations into a single equation involving matrices. In this equation:

• \(A\) is the coefficient matrix representing the coefficients of variables in each equation.
• \(X\) is the vector of variables, such as \(\begin{pmatrix} x \ y \ z \end{pmatrix}\).
• \(B\) represents how the variables are scaled or weighed as seen in the equations.

Converting a system into a matrix form offers advantages like applying algebraic operations, such as finding determinants and inverses, leading to a deeper analysis of the system’s properties. It allows for the calculation of eigenvalues and verification of conditions for non-trivial solutions. In practice, transforming the system given into the format \((A - \lambda I)X = 0\) is a standard practice to assess when the system will have non-trivial, potentially infinite, solutions.

Determinant Calculation

Calculating the determinant of a matrix is a crucial step in understanding whether a system of equations has non-trivial solutions. The determinant is a scalar value that, among other things, indicates if a matrix is invertible. If the determinant of a matrix is zero, the matrix is not invertible, suggesting parallel or dependent equations that might have infinite solutions. For the exercise given:

• The matrix \(N = A - \lambda I\) represents the system adjusted for \(\lambda\).
• Setting the determinant of \(N\) to zero, \( \text{det}(N) = 0 \), provides the \(\lambda\) values that lead to non-trivial solutions.

Calculating the determinant involves algebraic expansion and simplification to get a characteristic polynomial like the one in the solution. Solving this polynomial gives critical values of \(\lambda\), here found to be \(0\) and \(2\), which satisfy the condition for the existence of non-trivial solutions. Understanding determinant calculation is key to tackling more complex systems encountered in advanced mathematics and engineering fields.