Gauss-Jordan: Solving Systems Of Equations

Nov 18, 2025 by Alex Johnson 43 views

Let's dive into the fascinating world of linear algebra and tackle the challenge of solving a system of equations using the Gauss-Jordan elimination method. This method is a powerful and systematic approach to finding solutions for systems of linear equations, whether they have a unique solution, infinitely many solutions, or no solution at all. In this article, we'll walk through the process step-by-step, and if the system has infinitely many solutions, we'll express the solution in terms of an arbitrary variable, typically z. Let's get started!

Understanding the Gauss-Jordan Method

The Gauss-Jordan method is an extension of Gaussian elimination. Both methods use elementary row operations to transform a system of equations into a simpler, equivalent system that is easier to solve. The key difference is that Gauss-Jordan elimination continues the process until the matrix is in reduced row-echelon form. This means that the leading coefficient (also called a pivot) in each row is 1, and all other entries in the column containing a leading 1 are 0. This makes the solution immediately apparent.

The elementary row operations we'll use are:

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

These operations do not change the solution set of the system of equations.

Setting up the Augmented Matrix

First, we need to represent the given system of equations as an augmented matrix. The augmented matrix consists of the coefficients of the variables and the constants on the right-hand side of the equations.

Given the system:

x + y - 5z = -14
3x - 3y + 2z = 5
x + 3y - 5z = -24

The augmented matrix is:

[ 1  1 -5 | -14 ]
[ 3 -3  2 |   5 ]
[ 1  3 -5 | -24 ]

Now, we'll perform row operations to transform this matrix into reduced row-echelon form.

Step-by-Step Gauss-Jordan Elimination

Eliminate x from the second and third rows:
- Subtract 3 times the first row from the second row (R2 = R2 - 3R1):
```
[ 1  1 -5 | -14 ]
[ 0 -6 17 |  47 ]
[ 1  3 -5 | -24 ]
```
- Subtract the first row from the third row (R3 = R3 - R1):
```
[ 1  1 -5 | -14 ]
[ 0 -6 17 |  47 ]
[ 0  2  0 | -10 ]
```
Make the leading coefficient in the second row 1:
- Divide the second row by -6 (R2 = R2 / -6):
```
[ 1  1 -5 | -14 ]
[ 0  1 -17/6 | -47/6 ]
[ 0  2  0 | -10 ]
```
Eliminate y from the first and third rows:
- Subtract the second row from the first row (R1 = R1 - R2):
```
[ 1  0 -13/6 | -37/6 ]
[ 0  1 -17/6 | -47/6 ]
[ 0  2  0 | -10 ]
```
- Subtract 2 times the second row from the third row (R3 = R3 - 2R2):
```
[ 1  0 -13/6 | -37/6 ]
[ 0  1 -17/6 | -47/6 ]
[ 0  0  17/3 |  44/3 ]
```
Make the leading coefficient in the third row 1:
- Multiply the third row by 3/17 (R3 = R3 * 3/17):
```
[ 1  0 -13/6 | -37/6 ]
[ 0  1 -17/6 | -47/6 ]
[ 0  0  1 |  44/17 ]
```

Eliminate z from the first and second rows:

Add 13/6 times the third row to the first row (R1 = R1 + (13/6)R3):

[ 1  0  0 | -37/6 + (13/6)*(44/17) ]
[ 0  1 -17/6 | -47/6 ]
[ 0  0  1 |  44/17 ]

Simplifying the first row:

[ 1  0  0 | (-629 + 572)/102 ]
[ 0  1 -17/6 | -47/6 ]
[ 0  0  1 |  44/17 ]

[ 1  0  0 | -57/102 ]
[ 0  1 -17/6 | -47/6 ]
[ 0  0  1 |  44/17 ]

[ 1  0  0 | -19/34 ]
[ 0  1 -17/6 | -47/6 ]
[ 0  0  1 |  44/17 ]

Add 17/6 times the third row to the second row (R2 = R2 + (17/6)R3):

[ 1  0  0 | -19/34 ]
[ 0  1  0 | -47/6 + (17/6)*(44/17) ]
[ 0  0  1 |  44/17 ]

Simplifying the second row:

[ 1  0  0 | -19/34 ]
[ 0  1  0 | (-799 + 748)/102 ]
[ 0  0  1 |  44/17 ]

[ 1  0  0 | -19/34 ]
[ 0  1  0 | -51/102 ]
[ 0  0  1 |  44/17 ]

[ 1  0  0 | -19/34 ]
[ 0  1  0 | -1/2 ]
[ 0  0  1 |  44/17 ]

Therefore, the reduced row-echelon form of the augmented matrix is:

[ 1  0  0 | -19/34 ]
[ 0  1  0 | -1/2 ]
[ 0  0  1 |  44/17 ]

Reading the Solution

From the reduced row-echelon form, we can directly read the solution:

x = -19/34
y = -1/2
z = 44/17

So, the unique solution to the system of equations is x = -19/34, y = -1/2, and z = 44/17.

Verification

To ensure our solution is correct, let's substitute these values back into the original equations:

Equation 1: x + y - 5z = -14

(-19/34) + (-1/2) - 5(44/17) = -19/34 - 17/34 - 440/17 = (-19 - 17 - 880)/34 = -916/34 = -14

Equation 2: 3x - 3y + 2z = 5

3(-19/34) - 3(-1/2) + 2(44/17) = -57/34 + 3/2 + 88/17 = (-57 + 51 + 176)/34 = 170/34 = 5

Equation 3: x + 3y - 5z = -24

(-19/34) + 3(-1/2) - 5(44/17) = -19/34 - 3/2 - 440/17 = (-19 - 51 - 880)/34 = -950/34 = -24

Since the values satisfy all three original equations, our solution is correct.

Conclusion

We've successfully used the Gauss-Jordan method to solve the given system of equations. The solution we found is unique: x = -19/34, y = -1/2, and z = 44/17. The Gauss-Jordan method provides a systematic way to solve systems of linear equations. By performing elementary row operations, we transformed the augmented matrix into reduced row-echelon form, making the solution straightforward to read.

Further Reading: For a deeper understanding of linear algebra and the Gauss-Jordan method, check out resources like Khan Academy's Linear Algebra section. There, you'll find comprehensive lessons, practice problems, and videos to enhance your skills.