Simple gauss-jordan elimination

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In summary, the conversation discussed the difficulty of solving systems of linear equations and the use of Gauss-Jordan elimination. The first system had a solution of C=-1, B=2, A=1, while the second had a solution of x=2 and y=-5. The conversation also highlighted the importance of understanding the algorithm and having a goal in mind when performing row reductions. It was also noted that not all matrices can be reduced to have every column as a pivot column, resulting in an infinite number of solutions.
  • #1
The_ArtofScience
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Homework Statement


I'm having trouble with these very strange (and deceptive-looking) systems of linear equations that I would like to know more about. Here they are:

(1) A+C=0
B-2C=4
-A+B+C=0

(2) 3x+y= 1
x-3y=17


The Attempt at a Solution



(1) Using algebra the solutions to the first are C=-1, B=2, A=1. When I did the Gauss method it failed. I've tried thinking of ways to make it work but it just won't come to me.

(2) Algebra gives x=2 and y=-5

This is how my Gauss method turned out:

a) 1/3 R1
b) R1 -1/9 R2 =R1 giving 8/9, 0, -14/9
c) 9/8R1 giving 1, 0, -7/4
d) -1/3R2 giving -1/3, 1, -17/3
e) R2 +1/3R1

but I get -7/4 and -25/4 (?)
 
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  • #2
1. Gaussian elimination should work just fine. You need to post your work here so that we'll know your mistakes (though it can be tedious to type out all your steps).

b) R1 -1/9 R2 =R1 giving 8/9, 0, -14/9
Here is your mistake. The 2nd row, 2nd entry is -3. You should multiply by 1/9 and not -1/9 to reduce the above row to zero.
 
  • #3
Defennder said:
1. Gaussian elimination should work just fine. You need to post your work here so that we'll know your mistakes (though it can be tedious to type out all your steps).

Here is your mistake. The 2nd row, 2nd entry is -3. You should multiply by 1/9 and not -1/9 to reduce the above row to zero.

Thanks Defender for tracing back to my steps. I got the soln

I couldn't get the first one to work. So here's the work

(1) R3+R1 = R3 giving 0, 1, 2
(2) 1/2 R3
(3) R1 -R3 =R1 getting 1, -1/2, 0

this is where I got stuck, can't figure out what to do because I keep going back and forth over and over again...
 
  • #4
You need to understand how to perform the algorithm in a fashion which terminates meaningfully, because it seems you're performing row reductions without having any goal in mind.

(1.) Looks fine to me. What should do next is to reduce the 3rd row using the second row. Remember that you want to add a multiple of the 2nd row to the third row to make sure that the 2nd entry of the third row becomes zero (remember that the purpose of Gaussian-jordan elimination is to express every column as a pivot column if possible) so that the second row would be the only row which has a non-zero 2nd entry.

Once this is done, use row-reduction again to make sure that the third row has only one entry and this entry is at the last column. Observe that the end result is a matrix with 1's arranged in a staircase pattern. That is the purpose of Gauss Jordan elimination. Swap rows if necessary to achieve that pattern to get the end result.
 
  • #5
Ok the answer seems very clear to me. Your help is amazing!
 
  • #6
No prob, one last thing I should add:
It may not always be possible to reduce all the matrices you encounter into one in which every column is a pivot column. In this case, you should note that there are an infinite number of solutions to the system of equations, which you can express explicitly by reading them off the reduced-row echlon form and assigning each variable an arbitrary parameter like s or t.
 

FAQ: Simple gauss-jordan elimination

What is Simple Gauss-Jordan Elimination?

Simple Gauss-Jordan Elimination is a mathematical method used to solve systems of linear equations by transforming the equations into a simpler form and eliminating variables until a solution is reached.

How does Simple Gauss-Jordan Elimination work?

Simple Gauss-Jordan Elimination works by converting a system of linear equations into an augmented matrix, and then using elementary row operations to reduce the matrix into a form known as "row-echelon form". This process is repeated until the matrix is in "reduced row-echelon form", which represents the solutions to the system of equations.

What are elementary row operations?

Elementary row operations are specific manipulations that can be performed on a matrix to transform it without changing the solutions to the system of equations. These operations include swapping two rows, multiplying a row by a non-zero constant, and adding a multiple of one row to another row.

What are the advantages of using Simple Gauss-Jordan Elimination?

Simple Gauss-Jordan Elimination is a straightforward and systematic method for solving systems of linear equations. It allows for the quick and efficient computation of solutions, and can be easily programmed into a computer algorithm. Additionally, the method provides a clear understanding of the steps involved in solving a system of equations.

Are there any limitations to Simple Gauss-Jordan Elimination?

Simple Gauss-Jordan Elimination may not work for all systems of linear equations. If the matrix does not reduce to the desired form, or if there are infinitely many solutions or no solutions, then the method cannot be used. Additionally, the method can become computationally intensive for larger matrices, making it less efficient.

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