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

In summary, the conversation discusses solving linear systems using Gaussian elimination and the importance of doing every step in complete detail to avoid mistakes. The speaker also suggests using a linear system calculator to check for consistency. The conversation concludes with the realization that the determinant of the original matrix is 0, indicating that there is no unique solution to the system. The importance of linear independence and an easier calculation method is also mentioned.
  • #1
98
3
TL;DR Summary
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
  • #2
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.
 
  • #3
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?
 
  • #4
"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.
 
  • #5
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 ?
 
  • #6
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.
 
  • #7
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
 
  • #9
"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

Suggested for: This linear system has a solution? (Gauss' elimination)

Replies
2
Views
165
Replies
26
Views
4K
Replies
5
Views
1K
Replies
3
Views
847
Replies
11
Views
1K
Replies
10
Views
782
Back
Top