Problem 6 - 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. $$ \begin{array}{r} x_{1}+2 x_{2}-x_{3}=0 \\ 2 x_{1}+x_{2}+2 x_{3}=9 \\ x_{1}-x_{2}+x_{3}=3 \end{array} $$

Step-by-Step Solution

Answer

The solution is \(x_1 = 6\), \(x_2 = -3\), and \(x_3 = 0\).

Step 1: Write the Augmented Matrix

First, convert the system of equations into an augmented matrix. For the given system: \(\begin{bmatrix} 1 & 2 & -1 & | & 0 \ 2 & 1 & 2 & | & 9 \ 1 & -1 & 1 & | & 3 \end{bmatrix}\).

Step 2: Make the Leading Entry of First Row to 1

The leading entry of the first row is already 1. No changes are needed for Row 1.

Step 3: Eliminate First Column Entries Below the Leading 1

Use Row 1 to eliminate the below entries by performing Row 2 - 2*Row 1 and Row 3 - Row 1:- Row 2 becomes \([0, -3, 4, 9] \).- Row 3 becomes \([0, -3, 2, 3] \).

Step 4: Make the Leading Entry in the Second Row to 1

Currently, the leading entry in Row 2 is -3. Divide the entire row by -3 to turn the leading entry to 1: Row 2 becomes \([0, 1, -\frac{4}{3}, -3] \).

Step 5: Eliminate the Second Column Entries Above and Below the Leading Entry

Use Row 2 to eliminate the other entries in the second column:- Row 1 becomes [1, 0, \(\frac{1}{3}\), 6].- Row 3 becomes [0, 0, -\(\frac{2}{3}\), 0].

Step 6: Make the Leading Entry in the Third Row to 1

The leading entry in Row 3 is -\(\frac{2}{3}\). Divide the entire row by -\(\frac{2}{3}\) to get 1: Row 3 becomes \([0, 0, 1, 0] \).

Step 7: Eliminate the Third Column Entries Above the Leading Entry

Using Row 3, eliminate the third column entries above the leading 1:- Correct Row 1 to \([1, 0, 0, 6] \).- Row 2 remains \([0, 1, 0, -3] \).

Step 8: Write the Solution from the Row-Echelon Form

Read the solutions directly from the matrix:- \(x_1 = 6\), \(x_2 = -3\), \(x_3 = 0\).

Key Concepts

Gauss-Jordan elimination

Gauss-Jordan elimination is a method used in linear algebra to solve systems of linear equations. It is an extension of Gaussian elimination. This technique simplifies matrices to row-reduced echelon form. This specific form makes finding solutions straightforward, as each row corresponds to a simpler equation.

During Gauss-Jordan elimination, we perform row operations on the augmented matrix until we achieve row-echelon form and then continue to modify it to reduce further to just having zeros below and above the pivot positions. The key steps involve:

• Making leading entries (also known as pivot elements) of each row equal to 1.
• Eliminating all other entries in the pivot columns by adding or subtracting suitable multiples of the pivot row.

With its systematic approach, Gauss-Jordan elimination enables us to easily read off the solutions of the linear system from the final matrix.

augmented matrix

The concept of an augmented matrix offers a practical way to handle systems of linear equations. By creating an augmented matrix, we combine the coefficients of the variables and the constants of the equations into a single matrix.

For a system of equations:

• Each row in the augmented matrix corresponds to one equation from the system.
• Columns represent the coefficients of each variable and a separate column for the constants on the right-hand side.

It provides a compact form facilitating easier manipulation during elimination procedures, such as Gaussian or Gauss-Jordan elimination. Using an augmented matrix streamlines the transition from manipulating algebraic equations directly to performing matrix operations, which significantly speeds up the process of finding solutions.

row-echelon form

Row-echelon form is the result of applying Gaussian elimination to a matrix. In this form, the structure of the matrix will greatly simplify. Each leading entry of a row (the first non-zero number from the left, also called a pivot) must be to the right of the leading entry of any row above it. If a row is all zeros, it should be at the bottom of the matrix.

Characteristics of this form include:

• The first non-zero value in a row, called a pivot, is always 1, allowing easy eliminations in its column.
• All entries below each pivot are zeros.

This form is a stepping stone to further simplifications, particularly when moving towards reduced row-echelon form in Gauss-Jordan elimination. Achieving this form is crucial, as it simplifies back-substitution when solving for variables, leading ultimately to the direct extraction of solutions from the matrix.