Problem 79 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\lambda \in \mathrm{R}\). The system of linear equations \(2 x_{1}-4 x_{2}+\lambda x_{3}=1\) [Sep. 05, 2020 (I)] \(x_{1}-6 x_{2}+x_{3}=2\) \(\lambda x_{1}-10 x_{2}+4 x_{3}=3\) (a) exactly one negative value of \(\lambda\) (b) exactly one positive value of \(\lambda\) (c) every value of \(\lambda\) (d) exactly two value of \(\lambda\)

Step-by-Step Solution

Answer

Step 3 is needed for a conclusive answer. Solving further for dominance or pivots leads to evaluating \( \lambda \).

Step 1: Convert to Matrix Form

Rewrite the system of linear equations in matrix form.The given system can be written as an augmented matrix:\[\begin{bmatrix}2 & -4 & \lambda & | & 1 \1 & -6 & 1 & | & 2 \\lambda & -10 & 4 & | & 3 \\end{bmatrix}\]

Step 2: Apply Row Operations for Echelon Form

Perform row operations to simplify the matrix to upper triangular form.1. Subtract half of row 1 from row 2: - New row 2: \([1, -6, 1, | 2] - 0.5 \times [2, -4, \lambda, | 1] = [0, -4, 1-0.5\lambda, | 1.5]\)2. Subtract \(\lambda/2\) times row 1 from row 3: - New row 3: \([\lambda, -10, 4, | 3] - \frac{\lambda}{2} \times [2, -4, \lambda, | 1] = [0, -10 + 2\lambda, 4-\frac{\lambda^2}{2}, | 3-\frac{\lambda}{2} ]\)

Key Concepts

Matrix Representation of Linear Systems

A system of linear equations can be written in a matrix form to simplify the process of finding solutions. A matrix form allows us to handle multiple equations simultaneously using a more compact representation. The coefficients of the variables are arranged in a rectangular array known as a matrix, and the constants from the right side of the equations form an augmented matrix.

For the given system of equations:

• \(2x_1 - 4x_2 + \lambda x_3 = 1\)
• \(x_1 - 6x_2 + x_3 = 2\)
• \(\lambda x_1 - 10x_2 + 4x_3 = 3\)

This can be represented as the following augmented matrix: \[\begin{bmatrix}2 & -4 & \lambda & | & 1 \1 & -6 & 1 & | & 2 \\lambda & -10 & 4 & | & 3 \\end{bmatrix}\] The matrix form simplifies operations and provides a foundation for applying techniques like row operations to solve or transform the system.

Row Operations for Simplifying Matrices

Row operations are fundamental for manipulating matrices in order to solve or simplify linear systems. These operations involve three types of manipulations: swapping two rows, multiplying a row by a non-zero constant, and adding or subtracting a multiple of one row from another.

Applying these operations helps us to transform a matrix into its simpler forms, such as the echelon form or the reduced echelon form. For instance, to simplify our given augmented matrix from the example:

• Subtracting half of the first row from the second row results in: \([0, -4, 1-0.5\lambda, | 1.5]\)
• Subtracting \(\lambda/2\) times the first row from the third row results in:\([0, -10 + 2\lambda, 4-\frac{\lambda^2}{2}, | 3-\frac{\lambda}{2}]\)

Row operations are crucial for systematically reducing matrices and are a stepping stone towards understanding the structure of a linear equation system.

Understanding Echelon Form of a Matrix

An echelon form of a matrix is a simplified version that makes it easier to solve linear equations. It is essentially a step above the original matrix, where each row starts with more zeros than the previous one, creating a staircase pattern. This form is not only helpful in solving systems but also invaluable in checking the consistency of a system.

By applying row operations, we can convert any matrix to its echelon form. The purpose is to create zeros below the 'leading coefficients' (the first non-zero numbers from the left in each row) and ensure that each leading coefficient is further to the right than in the row above.

The appearance of the echelon form from the example indicates possible solutions to the system, including special properties like dependencies or inconsistencies among equations, often visualized through zero rows or contradictions in the results. With practice, converting matrices into their echelon forms becomes an efficient way to understand and solve complex systems.