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Gauss-Jordan Elimination Method Help

  • Thread starter heavyarms
  • Start date
  • #1
11
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Gauss-Jordan Elimination Method Help!!!

Hi, I'm having a problem solving the following using the Gauss-Jordan Elimination Method. I am wondering if someone can help me...
2x + 5y - z = -3
x - y + 4z = 20
3x + 2y - z = 3
i.e.
Code:
[ [2   5  -1  -3]
[1  -1   4  20]
[3   2  -1    3]]
I've tried solving it by... R1<-->R2, R2+(-1)R1, R3+(-3)R1, but I'm stuck with...
Code:
[ [1  -1   4    20]
[0   7   9   -43]
[0   5  -13 -57]]
I'm stuck because my textbook says we're not suppose to use fractions to solve it. I know I can do R3+(-5/7)R2, but that'd be using fractions. So I'm wondering if someone can lend me a hand and show me the correct way to complete this question! Thanks!
 

Answers and Replies

  • #2
For your first step of using matrixes,

R1[ 2 5 -1 | -3 ]
R2[ 1 -1 4 | 20 ]
R3[ 3 2 -1 | 3 ]

Second Step:

R1 [ 2 5 -1 | -3 ]
R1-2R2 [ 0 7 -9 | -43 ]
3R1-2R3[ 0 11 -1 | -15 ]

Third Step:

[ 2 5 -1 | -3 ]
[ 0 7 -9 | -43 ]
11R2 - 7R3[ 0 0 -92 | -385 ]
 
  • #3
11
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forevergone said:
[ 2 5 -1 | -3 ]
Do I not have to isolate a 1 in the top left hand corner?
i.e. [ 1 X X | X ] for the top row?

forevergone said:
3R1-2R3 || 11R2 - 7R3
Also isn't it not allowed to changed two rows at a time?
 
Last edited:
  • #4
heavyarms said:
Do I not have to isolate a 1 in the top left hand corner?
i.e. [ 1 X X | X ] for the top row?
Also isn't it not allowed to changed two rows at a time?
No. The way I learned to do Gauss-Jordan Elimination was to leave the 1st row alone. For the second and third row, you make the first terms zero and apply it to the rest of the numbers in that row. In the second step, you make the second number zero from the third row by subtracting it from the second row.
 
  • #5
JasonRox
Homework Helper
Gold Member
2,314
3
forevergone said:
No. The way I learned to do Gauss-Jordan Elimination was to leave the 1st row alone. For the second and third row, you make the first terms zero and apply it to the rest of the numbers in that row. In the second step, you make the second number zero from the third row by subtracting it from the second row.
Do whatever is easiest, which is avoid fractions for the most part. Then just switch around the rows at the end. Same thing.
 
  • #6
11
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JasonRox said:
Do whatever is easiest, which is avoid fractions for the most part. Then just switch around the rows at the end. Same thing.
So as long as I obtain an answer, it is still correct no matter the method I use? If this is the case then my textbook is really odd... it states that (1,1) has to be either +/- 1 and that multiplication changes can only occur in one row when adding.
 
  • #7
What is the full question that asks you to do for this set of matrixes?

Usually, matrixes are used to determine the intersection of 2/3 planes.
 
  • #8
11
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forevergone said:
What is the full question that asks you to do for this set of matrixes?

Usually, matrixes are used to determine the intersection of 2/3 planes.
The full question is as follows:

Solve the following system by the Gauss-Jordan elimination method.
2x + 5Y - Z = -3
X - Y + 4Z = 20
3X + 2Y - Z = 3
Indicate the row operation used in each step.
 
  • #9
heavyarms said:
The full question is as follows:
Solve the following system by the Gauss-Jordan elimination method.
2x + 5Y - Z = -3
X - Y + 4Z = 20
3X + 2Y - Z = 3
Indicate the row operation used in each step.
Gauss Jordan elimination is not necessary for obtaining the values of the three variables. Gaussian elimination is quicker, which is only making the triangle of zeros in the bottom left corner of the augmented matrix. If you have a TI-83 calculator, you can find the values of the three variables by pressing 2nd x^-1, and edit matrix A accordingly. You can then quit out of the program, enter matrices again, go to math, rref (reduced row echelon form) and then select matrix A
 
  • #10
11
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Leonardo Sidis said:
Gauss Jordan elimination is not necessary for obtaining the values of the three variables. Gaussian elimination is quicker, which is only making the triangle of zeros in the bottom left corner of the augmented matrix. If you have a TI-83 calculator, you can find the values of the three variables by pressing 2nd x^-1, and edit matrix A accordingly. You can then quit out of the program, enter matrices again, go to math, rref (reduced row echelon form) and then select matrix A
Thank you for your suggestion. I've just tried it on my TI-83 Plus SE and it gave me..
[[1 0 0 3]
[0 1 0 -1]
[0 0 1 4]]
I know that is the correct answer, but the textbook requires me to show all the steps. The problem is I am stuck midway to obtaining the answer. (See the first post in this thread). As such, I'd be very grateful if you can lead me in the right direction. Thanks.
 
  • #11
heavyarms said:
Thank you for your suggestion. I've just tried it on my TI-83 Plus SE and it gave me..
[[1 0 0 3]
[0 1 0 -1]
[0 0 1 4]]
I know that is the correct answer, but the textbook requires me to show all the steps. The problem is I am stuck midway to obtaining the answer. (See the first post in this thread). As such, I'd be very grateful if you can lead me in the right direction. Thanks.
Heavyarms, I'll try to do this at school and find how this works out. Your first row though, I'm not sure why but it should never ever change.
 
  • #12
513
0
I threw the matrix in Matlab and got the same results using G-J Elimination:
1 0 0 3
0 1 0 -1
0 0 1 4
 
  • #13
11
0
forevergone said:
Heavyarms, I'll try to do this at school and find how this works out. Your first row though, I'm not sure why but it should never ever change.
Thanks! Your help is very much appreciated.

BTW, Merry Christmas!!! :biggrin:
 
  • #14
360
0
Do a rref on your calculator. i am sure gauss wouldn t be mad at your.
 

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