Undergrad This linear system has a solution? (Gauss' elimination)

Click For Summary
SUMMARY

The discussion centers on solving linear systems using Gauss' elimination method. Participants emphasize the importance of meticulous calculations, noting that mistakes often arise from skipping steps. A linear system calculator was referenced, indicating that a specific system is inconsistent due to a determinant of zero, confirming that no unique solution exists. The conversation highlights the necessity of re-evaluating calculations to ensure accuracy in solving linear equations.

PREREQUISITES
  • Understanding of Gauss' elimination method
  • Familiarity with determinants in linear algebra
  • Basic knowledge of linear systems and their properties
  • Experience with matrix operations
NEXT STEPS
  • Learn how to calculate determinants of matrices
  • Study the implications of a zero determinant in linear systems
  • Explore advanced techniques in Gauss' elimination
  • Utilize online linear system calculators for verification of solutions
USEFUL FOR

Students, educators, and anyone involved in solving linear algebra problems, particularly those using Gauss' elimination for linear systems.

requied
Messages
98
Reaction score
3
TL;DR
I have the following linear equation
-3x1 +x2-2x3 =-7
5x1 + 3x2-4x3 = 2
x1 + 2x2-3x3 = -1

And I wonder it has a solution, because I found the third line as:
0 0 0 = -49/6
I don't know the terms so I'm sorry if the informations at summary above is unclear. But I add a detailed photo of my calculations below. I use Gauss' Elimination laws.
www.jpeg
 
Physics news on Phys.org
I didn't check your work, but I urge you to do it again. I taught this for years and always made mistakes on every calculation unless I did every step in complete detail. no shortcuts at all.
 
mathwonk said:
I didn't check your work, but I urge you to do it again. I taught this for years and always made mistakes on every calculation unless I did every step in complete detail. no shortcuts at all.
I think that the calculations shown in photo is enough to someone who already know a little solving linear systems.
mathwonk said:
but I urge you to do it again.
Here you refer to solve again the question?
 
"I think that the calculations shown in photo is enough to someone who already know a little solving linear systems."

I think you are missing the point, i don't have a problem, you do.
 
mathwonk said:
"I think that the calculations shown in photo is enough to someone who already know a little solving linear systems."

I think you are missing the point, i don't have a problem, you do.
Sir, what kind of problem are you speaking of? I don't get it what I actually should do ?
 
Forgive me, I am only suggesting if you are in doubt as to whether a gaussian elimination is correct, you should do it again, and see if you get the same answer.
 
mathwonk said:
Forgive me, I am only suggesting if you are in doubt as to whether a gaussian elimination is correct, you should do it again, and see if you get the same answer.
I get it but I had wondered this case which I add below. Isn't it time wasting making an effort to solve whole question again
AAA.png
 
"Isn't it time wasting making an effort"? indeed whose time do you speak of?
 
  • #10
mathwonk said:
"Isn't it time wasting making an effort"? indeed whose time do you speak of?
Man it's 4 am here :D I'm doing a homework for 5 hours. So my time...
 
  • #11
requied said:
Man it's 4 am here :D I'm doing a homework for 5 hours. So my time...
I don't want to be misunderstood, I am wasting my time, not you.
 
  • #12
pardon me for being flippant. i share your experience at trying to get gaussian calculations correct as i said. but the sad truth is there is no other way but to do them over and over until all errors are eliminated. and if you are up that late doing hw, perhaps you could learn to do it earlier next time?
 
  • Like
Likes requied
  • #13
mathwonk said:
if you are up that late doing hw, perhaps you could learn to do it earlier next time?
You are right at this point, thanks to you for all motivations. All the best :)
 
  • #14
Calculate the determinant (I get 0). There is no solution! That's the problem!
 
  • Like
Likes Delta2 and requied
  • #15
mathman said:
There is no solution!
So the actual answer is this, right?
 
  • #16
Why don't you compute the determinant of the original matrix to see if it is 0 too? Not a guarantee if it is, but shows a problem if it is not.
 
  • #17
1590356360440.png

The determinant of the original matrix is 0. I don't know what happens if a determinant of a matrix is 0.
 
  • #18
requied said:
View attachment 263424
The determinant of the original matrix is 0. I don't know what happens if a determinant of a matrix is 0.
One of the consequences is that there is a vector b for which Ax=b has no solution. And the matrix is not invertible.
 
  • #19
The determinant is 0 means that the equations are not linearly independent, so you can't get a solution. Getting (0,0,0) for the left side after Gauss elimination is equivalent to a 0 determinant.
 
  • #20
A zero determinant means that there is no unique solution. There could be many solutions or there may be none.
In this case, there is no solution. Your calculation leading to an obvious false equation is proof of that (although I haven't checked your work). An easier calculation process would be to add 3*row3 to row1 and subtract 5*row3 from row2. That will give you new row1 and row2 which are clearly contradictory: ##7x_2-11x_3=-10## and ##-7x_2+11x_3=7##.
 
Last edited:
  • Like
Likes S.G. Janssens

Similar threads

  • · Replies 26 ·
Replies
26
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 23 ·
Replies
23
Views
2K