Problem 87 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The following system of linear equations \(7 x+6 y-2 z=0\) \(3 x+4 y+2 z=0\) \(x-2 y-6 z=0\), has \(\quad\) [Jan. 9,2020 (II)] (a) infinitely many solutions, \((x, y, z)\) satisfying \(y=2 z\). (b) no solution. (c) infinitely many solutions, \((x, y, z)\) satisfying \(x=2 z\). (d) only the trivial solution.

Step-by-Step Solution

Answer

(c) Infinitely many solutions, \(x = 2z\).

Step 1: Write the system of equations in matrix form

The given system of equations is: 1. \( 7x + 6y - 2z = 0 \) 2. \( 3x + 4y + 2z = 0 \) 3. \( x - 2y - 6z = 0 \)We can write this system in matrix form as \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the column vector of variables, and \( \mathbf{b} \) is the zero vector:\[A = \begin{bmatrix}7 & 6 & -2 \3 & 4 & 2 \1 & -2 & -6\end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix}x \y \z\end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix}0 \0 \0\end{bmatrix}\]

Step 2: Perform row operations to echelon form

Apply Gaussian elimination to the matrix \( A \). The operations aim to transform \( A \) to upper triangular form.First, make the first column's entries below the first row to be zero.- Replace the second row with \( (R_2 - \frac{3}{7} R_1) \).- Replace the third row with \( (R_3 - \frac{1}{7} R_1) \).The modified matrix becomes:\[\begin{bmatrix}7 & 6 & -2 \0 & \frac{10}{7} & \frac{20}{7} \0 & -\frac{20}{7} & -\frac{40}{7}\end{bmatrix}\]

Step 3: Further simplification

Further simplify the second column. Make the third row's first non-zero entry (which is currently in the second position) zero by:- Replace the third row with \( (R_3 + 2R_2) \).This action yields the following matrix:\[\begin{bmatrix}7 & 6 & -2 \0 & \frac{10}{7} & \frac{20}{7} \0 & 0 & 0\end{bmatrix}\]

Step 4: Interpret the reduced matrix

The system, now in echelon form, corresponds to two equations:1. \( 7x + 6y - 2z = 0 \)2. \( \frac{10}{7}y + \frac{20}{7}z = 0 \)The third row represents no new information (\(0 = 0\)) and can be disregarded.

Step 5: Solve the system for free variables

The second equation can be simplified to:\[ 10y + 20z = 0 \] \[ y + 2z = 0 \implies y = -2z \]Substitute \( y = -2z \) into the first equation:\[ 7x + 6(-2z) - 2z = 0 \]\[ 7x - 12z - 2z = 0 \]\[ 7x = 14z \]\[ x = 2z \]The solutions satisfy \( x = 2z \) and \( y = -2z \). Let \( z = t \), a parameter, giving the family of solutions \((x, y, z) = (2t, -2t, t)\).

Step 6: Conclusion

The reduced system has infinitely many solutions defined by \( (x, y, z) = (2z, -2z, z) \). This corresponds to option (c) in the problem statement where \( x = 2z \). The system does not have just the trivial solution, nor does it have no solution.

Key Concepts

Gaussian Elimination

At its core, Gaussian elimination is a method for solving systems of linear equations. It's a straightforward process that involves performing operations on the equations to systematically eliminate variables, eventually leading to a simpler form from which solutions can be easily identified.
The main steps include:

• Row swapping, to position equations in a beneficial sequence.
• Multiplication of a row by a non-zero scalar, allowing adjustments to simplify equations.
• Addition and subtraction of rows, to eliminate certain terms and uncover solutions.

Through these operations, the aim is to create zeros below the diagonal of an augmented matrix, transforming it into a triangular form. This simplifies the process of finding solutions and, when done correctly, can reveal a system's properties, such as if it has a single solution, no solutions, or infinity solutions.
In our example, Gaussian elimination was used to transform the system into a simpler form, making it clear that there are infinitely many solutions.

Matrix Form

The matrix form is an efficient tool in representing and solving systems of linear equations. In this method, the equations are expressed in a compact matrix notation which helps in applying systematic solution processes like Gaussian elimination.
The general idea is to arrange the coefficients of the variables from the system in a square matrix, which we call the coefficient matrix. Additionally, the variables themselves become a vector. This setup is represented as:

• The coefficient matrix \( A \): \\[A = \begin{bmatrix} 7 & 6 & -2 \ 3 & 4 & 2 \ 1 & -2 & -6 \end{bmatrix}.\]
• The vector of variables \( \mathbf{x} \):\[\mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix}.\]
• The zero vector \( \mathbf{b} \):\[\mathbf{b} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}.\]

This matrix equation \( A\mathbf{x} = \mathbf{b} \) is much simpler to manipulate than the original equations, allowing for efficient application of techniques such as row reduction, ultimately leading to a solved system.

Echelon Form

An echelon form of a matrix provides an easy-to-read way to interpret the solutions of the system of linear equations. When a matrix is transformed into echelon form, it reveals much about the nature of the system, like dependencies among equations.
The characteristics of a matrix in echelon form include:

• All non-zero rows are above rows of all zeros.
• The leading coefficient (the first non-zero number from the left, also called a pivot) in any non-zero row is to the right of the leading coefficient of the row above it.
• Leading entries in any column appear in descending order as you move down the matrix rows.

In our system, once converted to echelon form, it became evident that the last row did not add new information, emphasizing the system's dependency on two main equations. This form helped in recognizing that the system indeed has infinitely many solutions because the transformation results in a row of zeros, indicating free variables and leading to parametric solutions.

Infinitely Many Solutions

A system of linear equations is said to have infinitely many solutions when there is a whole family of solutions that fit the equations. This typically arises in cases where there are more variables than unique equations, or some equations are linear combinations of others.
In such cases, at least one variable is typically free, meaning it can take any value, and the other variables are expressed in terms of this free variable, creating a family of solutions.

• In our example, after simplifying, the system implied conditions like \( y = -2z \) and \( x = 2z \).
• Letting \( z = t \) turns these into \( x = 2t \) and \( y = -2t \), with \( t \) being any real number—a parameter expressing the solutions.

This method illustrates that indeed there are infinite solutions, echoing the nature of linear equations in a space where more freedom is given due to the lack of restrictive, independent conditions. This concept is foundational to understanding and applying higher-level mathematics to solve real-world problems in fields like engineering and economics.