Problem 83 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the system of equations \(x+y+z=2\) \(2 x+4 y-z=6\) \(3 x+2 y+\lambda z=\mu\) has infinitely many solutions, then: [Sep. 04, 2020 (II)] (a) \(\lambda+2 \mu=14\) (b) \(2 \lambda-\mu=5\) (c) \(\lambda-2 \mu=-5\) (d) \(2 \lambda+\mu=14\)

Step-by-Step Solution

Answer

Option (d) is correct: \(2\lambda+\mu=14\).

Step 1: Understand the Condition for Infinite Solutions

The system of equations has infinitely many solutions if the equations are dependent, i.e., one can be expressed as a combination of the others, or if there are consistent relations that make a plane. We can investigate this by checking conditions for the coefficients that make the determinant zero in a typical linear system matrix, or when the 3rd plane equation is a linear combination of the first two.

Step 2: Write the System in Matrix Form

The augmented matrix for the system of equations is:\[\begin{bmatrix}1 & 1 & 1 & | & 2 \2 & 4 & -1 & | & 6 \3 & 2 & \lambda & | & \mu \\end{bmatrix}\]

Step 3: Apply Row Operations

Perform row operations to simplify. Subtract twice the first row from the second:\[\begin{bmatrix}1 & 1 & 1 & | & 2 \0 & 2 & -3 & | & 2 \3 & 2 & \lambda & | & \mu \\end{bmatrix}\] Now subtract three times the first row from the third:\[\begin{bmatrix}1 & 1 & 1 & | & 2 \0 & 2 & -3 & | & 2 \0 & -1 & \lambda - 3 & | & \mu - 6 \\end{bmatrix}\]

Step 4: Make the Third Equation a Linear Combination

For infinite solutions, the third row must be a linear combination of the other two. It means:\[-1(2) + (\lambda - 3) = 0\]and\[-1(2) = \mu - 6\].

Step 5: Solve the Equations

From the first equation, \[\lambda - 5 = 0 \Rightarrow \lambda = 5\].From the second equation, \[-2 = \mu - 6 \Rightarrow \mu = 4\].

Step 6: Verify the Options

Substitute \(\lambda = 5\) and \(\mu = 4\) into the given options:- (a) \(5 + 2 \times 4 eq 14\)- (b) \(2 \times 5 - 4 = 10 - 4 = 6eq 5\)- (c) \(5 - 2 \times 4 = 5 - 8 = -3eq -5\)- (d) \(2 \times 5 + 4 = 10 + 4 = 14\Rightarrow \text{True}\) Hence, option (d) is correct.

Key Concepts

System of Linear Equations

A system of linear equations is a collection of two or more linear equations that involve the same set of variables. In our example, the given system consists of three equations with variables \( x, y, \) and \( z \). The goal is often to find the values of these variables that satisfy all the equations simultaneously.
Identifying whether a system has a unique solution, no solution, or infinitely many solutions is crucial. In some systems, the equations can be dependent, meaning that one equation can be expressed as a linear combination of the others. This dependency is one of the conditions for a system to have infinitely many solutions.
In our example, for the system to have infinitely many solutions, there must be consistent relationships between the coefficients, allowing all the equations to lie on the same plane, leading them to have overlapping solutions.

Augmented Matrix

When working with systems of linear equations, the augmented matrix is a useful tool that combines the coefficients of the variables and the constants from the equations into a single matrix. This matrix representation simplifies the process of using row operations to solve or assess the system.
The given system can be represented in augmented matrix form as follows:

• \([1 \ 1 \ 1 \ | \ 2]\)
• \([2 \ 4 \ -1 \ | \ 6]\)
• \([3 \ 2 \ \lambda \ | \ \mu]\)

This matrix clearly shows the relationship between the coefficients and constants, providing a more organized way to apply operations needed to reach a solution or check the conditions for infinite solutions.

Linear Combination

The concept of linear combination is central to understanding systems with infinite solutions. A linear combination involves creating a new equation by adding together different multiples of the original equations.
For a system to have infinite solutions, it may be necessary for one equation to be a linear combination of the others. This means if you can multiply some equations by certain numbers and add them, you get another existing equation, indicating dependency.
In the exercise, the third row in the matrix (representing the third equation) must become a linear combination of the previous two. This is shown in the condition that involves modifying the coefficients to make the plane represented by this equation overlap with the others, leading to infinite solutions.

Row Operations

Row operations are techniques used to simplify a matrix in order to solve systems of linear equations efficiently. They include actions such as swapping rows, multiplying a row by a non-zero constant, and adding or subtracting rows.
In our matrix:

• Subtracting twice the first row from the second simplifies the equations, helping to isolate variables or conditions.
• Similarly, subtracting three times the first row from the third reduces the complexity of the matrix, making it easier to identify necessary conditions for infinite solutions.

These operations are fundamental in matrix methods like Gauss-Jordan elimination. In our task, row operations helped us transform the equations and check the conditions essential for the system to have infinite solutions, specifically turning dependent equations into obvious relationships.