Problem 13 - Advanced Engineering Mathematics

Question

Use either Gaussian elimination or Gauss-Jordan elimination to solve the given system or show that no solution exists. \(x_{1}+2 x_{2}+2 x_{3}=2\) \(x_{1}+x_{2}+x_{3}=0\) \(x_{1}-3 x_{2}-x_{3}=0\)

Step-by-Step Solution

Answer

The solution is \(x_1 = 6\), \(x_2 = 4\), \(x_3 = -6\).

Step 1: Write the System as an Augmented Matrix

Convert the given system of equations into an augmented matrix. The system is: \[\begin{align*}x_1 + 2x_2 + 2x_3 &= 2, \x_1 + x_2 + x_3 &= 0, \x_1 - 3x_2 - x_3 &= 0.\end{align*}\]The augmented matrix is: \[\begin{bmatrix} 1 & 2 & 2 & | & 2 \1 & 1 & 1 & | & 0 \1 & -3 & -1 & | & 0\end{bmatrix}.\]

Step 2: Perform Row Operations

First, subtract the second row from the first row to eliminate the first element of the second row.\[\begin{bmatrix} 1 & 2 & 2 & | & 2 \0 & 1 & 1 & | & -2 \1 & -3 & -1 & | & 0\end{bmatrix}.\]Next, subtract the third row from the first row to eliminate the first element of the third row.\[\begin{bmatrix} 1 & 2 & 2 & | & 2 \0 & 1 & 1 & | & -2 \0 & -5 & -3 & | & -2\end{bmatrix}.\]

Step 3: Continue Row Operations for Row Echelon Form

Change the third row by adding 5 times the second row to make the second coefficient zero.\[\begin{bmatrix} 1 & 2 & 2 & | & 2 \0 & 1 & 1 & | & -2 \0 & 0 & 2 & | & -12\end{bmatrix}.\]Now, we have a row-echelon form.

Step 4: Back Substitution

Solve for the variables using back substitution starting from the last row.From the third row: \(2x_3 = -12\) gives \(x_3 = -6\).From the second row: \(x_2 + x_3 = -2\) gives \(x_2 + (-6) = -2\) or \(x_2 = 4\).From the first row: \(x_1 + 2x_2 + 2x_3 = 2\) gives \(x_1 + 2(4) + 2(-6) = 2\), so \(x_1 = 6.\)

Step 5: Solution Verification

Verify the solution \(x_1 = 6\), \(x_2 = 4\), \(x_3 = -6\) by substituting it back into the original equations. Verify:1. \(6 + 2(4) + 2(-6) = 2\). True.2. \(6 + 4 - 6 = 0\). True.3. \(6 - 3(4) - (-6) = 0\). True.All equations are satisfied.

Key Concepts

Gauss-Jordan Elimination

Gauss-Jordan elimination is a method to solve systems of linear equations. This technique is a detailed extension of the Gaussian elimination process. The goal is to transform the augmented matrix of a system into a specific form, called the reduced row-echelon form. This form makes it easy to identify and extract solutions for the variables in the system.

The process involves a series of row operations. These operations help make the leading coefficient of each row equal to one and ensure that all elements above and below the leading coefficients are zeros. Once you achieve this form, reading off the solutions becomes straightforward.

Gauss-Jordan is highly efficient because it gives the exact solution or indicates if no solution exists. This method is especially useful when dealing with large systems or looking for a systematic way to find solutions.

System of Equations

A system of equations consists of multiple equations with multiple variables. In this exercise, the system is composed of three equations involving three variables: \(x_1\), \(x_2\), and \(x_3\). Solving a system means finding a set of values for the variables that make all the equations true simultaneously.

When dealing with a system of equations, you might encounter different scenarios:

• One unique solution: The lines or planes represented by the equations intersect at a single point.
• No solution: The lines or planes are parallel and never intersect.
• Infinite solutions: The lines or planes are the same, resulting in an overlap of solutions.

Understanding the system's nature is crucial for selecting the appropriate method of solving it, such as Gaussian or Gauss-Jordan elimination.

Augmented Matrix

An augmented matrix is a convenient way to represent a system of equations. It combines the coefficients and constants into a single matrix, separating coefficients from constants with a vertical line. This matrix form is the foundation for performing elimination methods like Gaussian and Gauss-Jordan.

For example, an augmented matrix for the system:
\[\begin{align*}x_1 + 2x_2 + 2x_3 &= 2, \ x_1 + x_2 + x_3 &= 0, \ x_1 - 3x_2 - x_3 &= 0.\end{align*}\]
is represented as:
\[\begin{bmatrix} 1 & 2 & 2 & | & 2 \ 1 & 1 & 1 & | & 0 \ 1 & -3 & -1 & | & 0 \end{bmatrix}.\]

This format allows for easy application of row operations to transform the matrix into a simpler form, ultimately leading to the solution of the system.

Row-Echelon Form

Row-echelon form is a stage in matrix manipulation where the goal of Gaussian elimination is almost reached. In this form, all non-zero rows are above any rows of all zeros, and each leading coefficient of a non-zero row is strictly to the right of the leading coefficient of the row above.

Transforming a matrix into row-echelon form involves using row operations like row swapping, scaling rows, and adding/subtracting rows. This step is crucial as it simplifies the solving process and sets a clear path for back substitution.

For the matrix from our example, row-echelon form appears as:
\[\begin{bmatrix} 1 & 2 & 2 & | & 2 \ 0 & 1 & 1 & | & -2 \ 0 & 0 & 2 & | & -12 \end{bmatrix}.\]

From here, it is easier to find solutions through back substitution, systematically solving each variable starting from the bottom row upwards.