Gauss-Jordan Elimination Solving Linear Systems

In this article, we will delve into the Gauss-Jordan elimination method, a powerful technique for solving systems of linear equations. We will illustrate this method by solving the following system:

$$\begin{array}{l} x - 2z = 9 \\ 6x - 2y - 5z = 29 \\ -5x + 5y + 3z = -14 \end{array}$$

We'll explore each step of the process in detail, ensuring you understand how to apply this method effectively. The solutions provided are:

A. $(5, 3, -5)$ B. $(3, 6, -4)$ C. $(-5, -3, 0)$ D. $(3, 2, -3)$

Understanding Gauss-Jordan Elimination

Before diving into the solution, let's briefly discuss the Gauss-Jordan elimination method. This method is an extension of Gaussian elimination, which transforms a system of linear equations into an equivalent system in reduced row-echelon form. This form makes it straightforward to read off the solutions for the variables.

The key idea is to use elementary row operations to manipulate the augmented matrix of the system. These operations include:

1. Swapping two rows
2. Multiplying a row by a non-zero constant
3. Adding a multiple of one row to another row

The goal is to transform the augmented matrix into reduced row-echelon form, where:

• The leading entry (the leftmost non-zero entry) in each row is 1.
• The leading entry in each row is to the right of the leading entry in the row above it.
• Rows with all zero entries are at the bottom of the matrix.
• Each column containing a leading entry has all other entries equal to zero.

Once the matrix is in this form, the solutions to the system can be easily determined.

Step-by-Step Solution

Let's now apply the Gauss-Jordan elimination method to solve the given system of equations.

1. Write the Augmented Matrix

The first step is to write the augmented matrix for the system. This matrix consists of the coefficients of the variables and the constants on the right-hand side of the equations.

For our system, the augmented matrix is:

$$\left[ \begin{array}{ccc|c} 1 & 0 & -2 & 9 \\ 6 & -2 & -5 & 29 \\ -5 & 5 & 3 & -14 \end{array} \right]$$

This matrix represents the system of equations, with each row corresponding to an equation and each column (except the last) corresponding to a variable.

2. Perform Row Operations to Obtain Row-Echelon Form

The next step is to perform elementary row operations to transform the matrix into row-echelon form. We'll start by eliminating the 6 in the second row and the -5 in the third row of the first column.

• Operation 1: Replace Row 2 with Row 2 - 6 * Row 1 ($R_2 \rightarrow R_2 - 6R_1$)

  This operation will eliminate the 6 in the first column of the second row:

  $$\left[ \begin{array}{ccc|c} 1 & 0 & -2 & 9 \\ 0 & -2 & 7 & -25 \\ -5 & 5 & 3 & -14 \end{array} \right]$$

• Operation 2: Replace Row 3 with Row 3 + 5 * Row 1 ($R_3 \rightarrow R_3 + 5R_1$)

  This operation will eliminate the -5 in the first column of the third row:

  $$\left[ \begin{array}{ccc|c} 1 & 0 & -2 & 9 \\ 0 & -2 & 7 & -25 \\ 0 & 5 & -7 & 31 \end{array} \right]$$

Now, we'll focus on the second column. We want a leading 1 in the second row, so we'll divide the second row by -2.

• Operation 3: Replace Row 2 with Row 2 / -2 ($R_2 \rightarrow R_2 / -2$)

  This gives us:

  $$\left[ \begin{array}{ccc|c} 1 & 0 & -2 & 9 \\ 0 & 1 & -7/2 & 25/2 \\ 0 & 5 & -7 & 31 \end{array} \right]$$

Next, we'll eliminate the 5 in the third row of the second column.

• Operation 4: Replace Row 3 with Row 3 - 5 * Row 2 ($R_3 \rightarrow R_3 - 5R_2$)

  This operation results in:

  $$\left[ \begin{array}{ccc|c} 1 & 0 & -2 & 9 \\ 0 & 1 & -7/2 & 25/2 \\ 0 & 0 & 21/2 & -63/2 \end{array} \right]$$

Finally, we want a leading 1 in the third row, so we'll multiply the third row by 2/21.

• Operation 5: Replace Row 3 with Row 3 * (2/21) ($R_3 \rightarrow R_3 * (2/21)$)

  This yields:

  $$\left[ \begin{array}{ccc|c} 1 & 0 & -2 & 9 \\ 0 & 1 & -7/2 & 25/2 \\ 0 & 0 & 1 & -3 \end{array} \right]$$

3. Perform Row Operations to Obtain Reduced Row-Echelon Form

Now, we'll continue performing row operations to transform the matrix into reduced row-echelon form. This means we need to eliminate the -2 in the first row and the -7/2 in the second row of the third column.

• Operation 6: Replace Row 1 with Row 1 + 2 * Row 3 ($R_1 \rightarrow R_1 + 2R_3$)

  This eliminates the -2 in the first row:

  $$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 3 \\ 0 & 1 & -7/2 & 25/2 \\ 0 & 0 & 1 & -3 \end{array} \right]$$

• Operation 7: Replace Row 2 with Row 2 + (7/2) * Row 3 ($R_2 \rightarrow R_2 + (7/2)R_3$)

  This eliminates the -7/2 in the second row:

  $$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -3 \end{array} \right]$$

4. Read the Solution

The matrix is now in reduced row-echelon form. We can directly read off the solutions for the variables:

• $x = 3$
• $y = 2$
• $z = -3$

Therefore, the solution to the system of equations is $(3, 2, -3)$.

Final Answer

The correct answer is D. (3, 2, -3).

Conclusion

The Gauss-Jordan elimination method is a powerful tool for solving systems of linear equations. By systematically applying elementary row operations, we can transform the augmented matrix into reduced row-echelon form, making it easy to determine the solutions. This method is widely used in various fields, including mathematics, engineering, and computer science. Understanding the steps involved in Gauss-Jordan elimination is crucial for anyone working with linear systems.

This comprehensive walkthrough demonstrates how to apply the Gauss-Jordan elimination technique. By following each step carefully, you can solve a wide range of linear systems effectively. Remember, practice is key to mastering this method. Work through various examples, and you'll become proficient in using Gauss-Jordan elimination to solve linear equations. This method not only provides the solution but also offers a structured approach, making it easier to understand the underlying mathematical principles.