Problem 120 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Consider the system of linear equations; \(x_{1}+2 x_{2}+x_{3}=3\) \(2 x_{1}+3 x_{2}+x_{3}=3\) \(3 x_{1}+5 x_{2}+2 x_{3}=1\) The system has (a) exactly 3 solutions (b) a unique solution (c) no solution (d) infinite number of solutions

Step-by-Step Solution

Answer

The system has no solution. (Answer: c)

Step 1/Identify the System

We are given a system of linear equations: \[\begin{align*}x_{1}+2x_{2}+x_{3}&=3\2x_{1}+3x_{2}+x_{3}&=3\3x_{1}+5x_{2}+2x_{3}&=1\end{align*}\]These equations will be represented in matrix form to simplify solving.

Step 2/Write the Augmented Matrix

Translate the system into an augmented matrix:\[\begin{bmatrix}1 & 2 & 1 & | & 3\2 & 3 & 1 & | & 3\3 & 5 & 2 & | & 1\end{bmatrix}\]This matrix represents the linear system, where each row corresponds to an equation.

Step 3/Perform Row Operations

Use row operations to simplify the matrix into row-echelon form. Begin by making the element below the first pivot (1) zero:Subtract 2 times the first row from the second row:\[\begin{bmatrix}1 & 2 & 1 & | & 3\0 & -1 & -1 & | & -3\3 & 5 & 2 & | & 1\end{bmatrix}\]Next, subtract 3 times the first row from the third row:\[\begin{bmatrix}1 & 2 & 1 & | & 3\0 & -1 & -1 & | & -3\0 & -1 & -1 & | & -8\end{bmatrix}\]

Step 4/Simplify to Row-Reachlon Form

Continue simplifying by making the pivot in the third column:Subtract the second row from the third row:\[\begin{bmatrix}1 & 2 & 1 & | & 3\0 & -1 & -1 & | & -3\0 & 0 & 0 & | & -5\end{bmatrix}\]

Step 5/Analyze the Matrix

Observe that the last row \[ \begin{bmatrix} 0 & 0 & 0 & | & -5 \end{bmatrix} \]corresponds to the equation\[0x_1 + 0x_2 + 0x_3 = -5\],indicating a contradiction as 0 cannot equal -5. This shows that the system is inconsistent.

Step 6/Determine the System's Solution Type

Since the system leads to a contradiction, it has no solutions. This implies that the correct answer to the exercise is that the system has no solution.

Key Concepts

Augmented Matrix

When dealing with a system of linear equations, transforming the equations into a matrix format can greatly simplify the solution process. This is where an augmented matrix comes into play. This type of matrix is constructed by writing down the coefficients of each variable from the equations and appending the constants from the right-hand side of the equations as an additional column.

For instance, consider our original system:

• Equation 1: \(x_1 + 2x_2 + x_3 = 3\)
• Equation 2: \(2x_1 + 3x_2 + x_3 = 3\)
• Equation 3: \(3x_1 + 5x_2 + 2x_3 = 1\)

This is represented as the augmented matrix:\[\begin{bmatrix}1 & 2 & 1 & | & 3\2 & 3 & 1 & | & 3\3 & 5 & 2 & | & 1\\end{bmatrix}\]The vertical bar divides the coefficient matrix and the constants, highlighting how each component of the equations translates directly into the matrix format. This notation is powerful as it sets the stage for using systematic row operations to solve for the variable values.

Row Operations

Row operations are techniques used to manipulate the rows of an augmented matrix to achieve a simpler form, ideally leading toward solutions of the system of equations. The primary types of row operations include:

• Swapping two rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting rows

These operations are performed to bring the augmented matrix into simpler forms like the row-echelon form or even the reduced row-echelon form. In the given solution:

• The goal is to use these operations to get leading ones (pivots) in the diagonal from the top-left to bottom-right, and zeros in all positions below these pivots.

Through operations in our example, the following transformations were made:

• Subtracting 2 times the first row from the second row.
• Subtracting 3 times the first row from the third row.
• Finally, subtracting the new second row from the third one.

These steps yield a row where all coefficients of variables are zero, which is crucial for determining whether the system is consistent or inconsistent.

Inconsistent System

A system of equations is classified as inconsistent if it has no solutions. In the context of matrices, this generally occurs when there's a row in the augmented form that leads to an impossible equation. For instance, a row that essentially states something like \(0x_1 + 0x_2 + 0x_3 = c\), where \(c\) is a non-zero constant, showcases a contradiction since the left side simplifies to zero, which cannot equal a non-zero constant.

In our solved system, the final row transformation resulted in:\[\begin{bmatrix}0 & 0 & 0 & | & -5\\end{bmatrix}\]This row corresponds to the equation \(0x_1 + 0x_2 + 0x_3 = -5\). Since it's impossible for zero to equal -5, this reveals the system's inconsistency. This contradiction confirms that there are no values for \(x_1, x_2,\) or \(x_3\) that can satisfy all the original system's equations simultaneously. Thus, the system has no solution.