Problem 92 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the system of linear equations \(x+y+z=5\) \(x+2 y+2 z=6\) \(\mathrm{x}+3 \mathrm{y}+\lambda \mathrm{z}=\mu,(\lambda, \mu \in \mathbf{R})\), has infinitely many solutions, then the value of \(\lambda+\mu\) is : [April 10, 2019 (I)] (a) 12 (b) 9 (c) 7 (d) 10

Step-by-Step Solution

Answer

\(\lambda + \mu = 10\).

Step 1: Understand the condition for infinite solutions

For the system to have infinitely many solutions, the three equations must be consistent and dependent, meaning one equation can be formed as a linear combination of the others.

Step 2: Prepare the matrix representation

Convert the equations into an augmented matrix form: \[\begin{bmatrix}1 & 1 & 1 & | & 5 \1 & 2 & 2 & | & 6 \1 & 3 & \lambda & | & \mu\end{bmatrix}\]Let's perform row operations to simplify.

Step 3: Perform row operations

Subtract the first row from the second row: \[\begin{bmatrix}0 & 1 & 1 & | & 1\end{bmatrix}\].Now subtract the first row from the third row: \[\begin{bmatrix}0 & 2 & \lambda-1 & | & \mu-5\end{bmatrix}\].

Step 4: Make rows dependent for infinite solutions

The second row \([0, 1, 1, |, 1]\) and the third row \([0, 2, \lambda-1, |, \mu-5]\) must be scalar multiples of each other for infinite solutions. Thus, \(2 = 2 \cdot 1\), \(\lambda - 1 = 2 \cdot 1\), and \(\mu - 5 = 2 \cdot 1\).

Step 5: Solve for \(\lambda\) and \(\mu\) using scalar multiples

From \(\lambda - 1 = 2\), solve for \(\lambda\): \[\lambda = 3\].From \(\mu - 5 = 2\), solve for \(\mu\): \[\mu = 7\].

Step 6: Calculate \(\lambda + \mu\)

Add \(\lambda\) and \(\mu\) to find their sum: \[\lambda + \mu = 3 + 7 = 10\].

Key Concepts

Augmented Matrix

An augmented matrix is a powerful tool for solving systems of linear equations. It combines both the coefficients of the variables and the constants from the equations into a single matrix. This format is very helpful in visualizing and simplifying the operations you need to perform.
For example, consider the system of equations:

• \(x + y + z = 5\)
• \(x + 2y + 2z = 6\)
• \(x + 3y + \lambda z = \mu\)

We can represent this system in an augmented matrix form as follows:\[\begin{bmatrix}1 & 1 & 1 & | & 5 \1 & 2 & 2 & | & 6 \1 & 3 & \lambda & | & \mu\end{bmatrix}\]This matrix combines the coefficients (1, 1, 1, etc.) on the left with the constants (5, 6, \(\mu\)) on the right after a separation bar. The bar indicates where the equations equal parts are placed, making it convenient for applying row operations.

Infinite Solutions

A system of equations has infinite solutions when the equations are consistent and dependent. This means one or more of the equations can be expressed as a linear combination of the others.
In practical terms:

• Consistency implies that there are solutions that satisfy all equations at the same time.
• Dependency indicates that the equations are not independent; instead, one can be derived or constructed from another.

In our problem, the goal is to make the second and third rows of the matrix scalar multiples of each other. When the system has infinite solutions, these rows align in such a way that they essentially "point" in the same direction in a geometric sense, allowing for multiple solutions to exist.

• Consistent solutions require setting the coefficients in such a relationship that all rows cannot be distinct equations.
• Each solution satisfies the conditions of being a linear combination of the others.

This approach helps ensure that the resulting matrix forms show the necessary constraints for an infinite number of solutions.

Row Operations

Row operations are vital in transforming an augmented matrix to determine the properties of the system of linear equations. These operations simplify the matrix to reveal critical relationships between the equations.
The three primary types of row operations are:

• Swapping two rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting a multiple of one row from another

In our example, the key step involved subtracting the first row from the second and third rows:

• Subtracting the first row from the second yields the new second row: \([0, 1, 1, |, 1]\).
• Subtracting the first row from the third gives \([0, 2, \lambda-1, |, \mu-5]\).

These operations help make the system easier to solve by simplifying the relationships between variables.

• They aim to bring the matrix to a form where more easily recognizable solutions, such as those indicating infinite solutions, can be identified.
• This form is often a row-echelon form, which aids in solving the system by back substitution or identifying dependency among the equations.

Through these transformations, it becomes feasible to identify and compute the necessary parameters for infinite solutions.