Help with Gauss-Jordan Elimination

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In summary, the student is trying to solve a system of three equations using the Gauss-Jordan elimination, but is having difficulty understanding the process.
  • #1
clickclick
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Hello everyone, this is my first time posting here, I found the website looking for help. For my finite math homework one of the questions ask to solve a system through the Gauss-Jordan Elimination; here's what I have so far.

Homework Statement



3x + y -2z = 2
x - 2y +z = 3
2x - y -3z = 3

Homework Equations



So when I put it in the matrix I get this:

[ 3 1 -2 2 ]
[ 1 -2 1 3 ]
[ 2 -1 -3 3 ]

The Attempt at a Solution



My problem is that I have not really mastered the Gauss-Jordan elimination. So I don't know the exact rules as to how I can proceed with this. I don't know if there are any limitations as to how much I can multiply, substract, add rows to get to:

[ 1 0 0 ? ]
[ 0 1 0 ? ]
[ 0 0 1 ? ]

Thank you,
 
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  • #2
You can multiply/divide a row by any number and add/subtract any multiple of a row to a different row.
 
  • #3
Here is what I got so far:

[ 3 1 -2 2 ]
[ 1 -2 1 3 ] -1
[ 2 -1 -3 3 ] -2

[ 3 1 -2 2 ]
[ 0 -3 0 2 ]
[ 0 -3 -5 1 ] + row 2

[ 3 1 -2 2 ] + 2
[ 0 -3 0 2 ]
[ 0 0 -5 3 ]

[ 5 3 0 4 ] + r2
[ 0 -3 0 2 ]
[ 0 0 -5 3 ]

[ 5 3 0 4 ] + r2
[ 0 -3 0 2 ]
[ 0 0 -5 3 ]

[ 5 0 0 6 ] /5
[ 0 -3 0 2 ] /-3
[ 0 0 -5 3 ] /-5

[ 1 0 0 1.2 ]
[ 0 1 0 -.666... ]
[ 0 0 1 -.6 ]

I don't know what I am doing wrong.
 
Last edited:
  • #4
clickclick said:
Here is what I got so far:

[ 3 1 -2 2 ]
[ 1 -2 1 3 ] -1
[ 2 -1 -3 3 ] -2

[ 3 1 -2 2 ]
[ 0 -3 0 2 ]
[ 0 -3 -5 1 ] + row 2
You're not allowed to do that. You can't simply subtract all the entries of a row by some constant. You can only change the numbers in each row by either:
1. Adding some multiple of another row to it or
2. Multiplying that row by a constant.
 
  • #5
Using Gauss-Jordan to solve linear equations is no different to if you were going to do it without matricies.

If you were solving this problem without matricies would you simply subtract or add random constants to coefficients?? No, necause that would not make sense. What you would do is to add multiples of one equation to another equation, or multiply equations by some constants (as Defennder said).

Hopefully you see why you can't simply add constants to coefficients now.

-Spoon
 

Related to Help with Gauss-Jordan Elimination

What is Gauss-Jordan Elimination?

Gauss-Jordan Elimination is a method used in linear algebra to solve systems of linear equations. It involves transforming a matrix into its reduced row echelon form using elementary row operations, such as swapping rows, multiplying rows by a constant, or adding a multiple of one row to another.

When should I use Gauss-Jordan Elimination?

Gauss-Jordan Elimination is useful when you need to solve a system of linear equations with multiple variables. It can also be used to find the inverse of a square matrix, which is helpful in solving many engineering and scientific problems.

What are the steps involved in Gauss-Jordan Elimination?

The steps for Gauss-Jordan Elimination are as follows:1. Write the system of equations in matrix form.2. Use elementary row operations to transform the matrix into its reduced row echelon form.3. Interpret the transformed matrix to obtain the solution to the system of equations, if one exists.

What are the benefits of using Gauss-Jordan Elimination?

Gauss-Jordan Elimination is a straightforward and systematic method for solving systems of linear equations. It can also be used to find the inverse of a matrix, which has many applications in mathematics and engineering.

Are there any limitations to using Gauss-Jordan Elimination?

While Gauss-Jordan Elimination is a powerful tool for solving systems of linear equations, it can become computationally expensive for large matrices. Additionally, if the matrix is ill-conditioned, the solutions obtained may not be accurate. In these cases, it may be more efficient to use other methods, such as Gaussian elimination or LU decomposition.

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