Problem 7 - Advanced Engineering Mathematics

Question

In Problems 1-20, use either Gaussian elimination or Gauss-Jordan elimination to solve the given system or show that no solution exists. $$ \begin{aligned} &x_{1}+x_{2}+x_{3}=0 \\ &x_{1}+x_{2}+3 x_{3}=0 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \((t, -t, 0)\), for any real number \(t\).

Step 1: Write the System as an Augmented Matrix

Express the given system of equations as an augmented matrix. This forms the starting point for elimination methods. The system \(\begin{aligned} x_1 + x_2 + x_3 &= 0 \ x_1 + x_2 + 3x_3 &= 0 \end{aligned}\) translates to the matrix: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 1 & 1 & 3 & | & 0 \end{bmatrix} \]

Step 2: Perform Gaussian Elimination

Subtract the first row from the second row to eliminate the \(x_1\) and \(x_2\) terms in the second equation: - Row 2 becomes (Row 2 - Row 1), resulting in: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 0 & 0 & 2 & | & 0 \end{bmatrix} \]

Step 3: Solve the Simplified System

From the matrix: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 0 & 0 & 2 & | & 0 \end{bmatrix} \]We derive the equations:1. \(x_1 + x_2 + x_3 = 0\)2. \(2x_3 = 0\) implying \(x_3 = 0\). Substitute \(x_3 = 0\) into the first equation resulting in:\(x_1 + x_2 = 0\).

Step 4: Express the Solution

With \(x_3 = 0\), the first equation \(x_1 + x_2 = 0\) can be solved in terms of a free variable. Let \(x_1 = t\), then \(x_2 = -t\). The solution can be expressed as a parameterized vector: \[ (x_1, x_2, x_3) = (t, -t, 0) \] where \( t \) is any real number.

Key Concepts

Gauss-Jordan Elimination

Gauss-Jordan Elimination is a method used to solve systems of linear equations. It is a systematic process for converting an augmented matrix to row-echelon form, then further reducing it to reduced row-echelon form. This means you'll end up with a matrix where each leading entry in a row is 1, and all elements above and below these leading 1s are zeros. The primary goal of this process is to simplify the system to a form where the solutions can be easily identified.

Here’s a quick overview of the procedure:

• Start by writing the system of equations as an augmented matrix.
• Use row operations to create zeros in columns below the leading 1.
• Adjust the matrix further by creating zeros above leading 1s until the identity matrix is formed on the left side of the augmented matrix.
• The matrix is now in reduced row-echelon form, and the solution to the system can be easily read from the matrix.

While Gauss-Jordan elimination can look intense, especially for larger systems, it is a powerful tool for finding solutions or determining that no solutions exist.

Augmented Matrix

An augmented matrix is an essential concept when dealing with systems of linear equations. It's a matrix that includes the coefficients of the variables on the left side and the constants from the equations on the right, separated by a vertical line. This way, it visually organizes the system, making it easier to apply elimination methods.

For example, if you have two equations:

• The first equation: \(x_1 + x_2 + x_3 = 0\)
• The second equation: \(x_1 + x_2 + 3x_3 = 0\)

You express this in augmented matrix form as:\[\begin{bmatrix} 1 & 1 & 1 & | & 0 \ 1 & 1 & 3 & | & 0 \end{bmatrix}\]

The matrix clearly organizes the information for each variable and the result of the equation, simplifying further manipulations like row reduction.

Parameterized Solution

In solving systems of equations, particularly when there are infinitely many solutions, a parameterized solution can succinctly express the solution set. This involves using a free variable to describe all possible solutions for the system.

For instance, in our current problem, after using Gaussian elimination, we end up with:

• \(x_3 = 0\)
• \(x_1 + x_2 = 0\)

This indicates there is a dependency between \(x_1\) and \(x_2\). By introducing a parameter, say \(t\), you can express these variables in terms of \(t\):

• Set \(x_1 = t\)
• Then \(x_2 = -t\)

This results in a parameterized vector solution:\[(x_1, x_2, x_3) = (t, -t, 0)\]Here \(t\) is any real number, providing a compact form to describe all possibilities in the solution space of the original system.