Linear algebra - gaussian elimination

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Homework Help Overview

The discussion revolves around solving a system of linear equations using Gauss-Jordan elimination. The equations presented involve four variables: w, x, y, and z. Participants are exploring the implications of their calculations and the nature of the solutions obtained.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants describe their attempts at performing Gauss-Jordan elimination, including creating tableau forms and applying row operations. There are questions about the validity of the solutions, particularly regarding the presence of arbitrary solutions and the potential for unique solutions.

Discussion Status

Some participants express uncertainty about their calculations and seek clarification on whether the system can yield varying or infinite solutions. There is acknowledgment of mistakes in the process, and some participants are reconsidering their methods and the implications of their results.

Contextual Notes

Participants note constraints such as time limitations affecting their ability to present full solutions or detailed steps. There is also mention of a specific mistake in the calculations that may have impacted the results.

The_ArtofScience
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Homework Statement



Use Gauss-Jordan elimination to solve this sys of linear eqs

2w+3x-y+4z =0
3w-x+z =1
3w-4x+y-z=2



The Attempt at a Solution



I wrote 9 tablabeaus and ended up with three arbitrary solns. I just want to know if there anything wrong, here they are: 225a/121 = 39/605 550b/121 =-488/605, 148c/11=-126/55.

If there is a unique soln please let me know. Sorry I can't type everything out, don't have time. Gotta rush to/
 
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If you don't have time to show us steps, we don't have time to check your answers! :)))
 
rocomath said:
If you don't have time to show us steps, we don't have time to check your answers! :)))


2 3 -1 4 0 1/2R1=R1, R2-3R1=R2
3 -1 0 1 1
3 -4 1 -1 2

1, 3/2, -1/2, 2, 0 R3-3R1=R3
0, -11/2, 3/2, -5, 1
3, -4, 1, -1

1, 3/2, -1/2, 2, 0 2/5R3=R3, 2/-11R2
0, -11/2, 3/2, 5, 1
0, -17/2, 5/2, -5, 2

1, 3/2 -1/2, 2, 0 R3+ 17/5R2= R3
0, 1, -3/11, 10/11, 2/-11
0, -17/5, 1, -2, 4/5

1, 3/2, -1/2, 2, 0
0, 1, -3/11, 10/11, -2/11
0, 0,

...Ok I took a long look at my paper. I realize I made one mistake but I still can't figure out how to get back at 1 after that 0. The eq R3+ 17/5R2= R3 doesn't really work out. Its a bit messy to work back. Is there a better method?
 
1, 3/2, -1/2, 2, 0 2/5R3=R3, 2/-11R2
0, -11/2, 3/2, 5, 1
0, -17/2, 5/2, -5, 2

You made a mistake here, this should be -7.
 
I worked it out again and still got 3 different arbitrary solns. Can a sys of equations have varying or infinite number of arbitrary solns?

My method is as follows:

(1) 1/2R1
(2) -2/11R2
(3) R3+17/5R2=R3
(4) R1+ 55/44R3=R1
(5) R2 +15/4R3=R2
(6) 55/4R3
(7) R1- 2/15R3 =R1

Solving I get 2w=1/2 5x=31/11 15y=11 and no z. I just hope I never see this again in cal 1 :-/
 
Last edited:

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