Post by: squance on January 06, 2009, 09:03:48 pm
Use Gaussian elimination to reduce the augmented matrix which represents the linear system
3x + 2y - z = -15
x + y -4z = -30
3x + y + 3z = 11
3x + 3y - 5z = -41
to row echelon form
__________
So I placed all these coefficients into a matrix
According to my notes, we have to make the top left element a leading entry (i.e. make the leading entry non-zero). The leading entry is already zero so I just use the matrix I already have. Then I have to hold the first row constant and add multiples of the first row to all the other rows so that all the entries from row 2 and down are zero in the first column.
To get the new matrix, I did:
Second row - 1/3 x first row
Third row - 1 x first row
Fourth row - 1 x first row
As a result of this, I got
The next step is to reorder rows 2 and so on so that the next leading entry is in row 2. In this case, there is no need to reorder the rows since \frac{1}{3} is already the entry in row 2.. Row 2 is now kept constant and I have to make all the other entries from row 3 and down in the column with the second leading entry zero.
I did:
Third row + 3 x second row
Fourth row + 0 x second row
Then I kept row 3 constant and repeated the step above to the last row by doing fourth row - \frac{4}{7} x third row.
My final answer:
And i was just wondering if that was the right answer. This is the first time I have encountered something as hard as this and im not sure if Im doing it right. (I have to fill in the gaps in my lecture notes by myself :( )
I also would like to know that if we were given a set of linear equations (for example, like the one given in this question), is it okay to reorder them in such that maybe the cofficients of the second equation are in the first row of the matrix when doing this type of question? Would reordering the equations make a difference in the final answer?
Any responses would be appreciated :)
Post by: brendan on January 06, 2009, 09:16:18 pm
Assuming my arithmetic is right, the augmented matrix in row echelon form is:
From that you can use back substitution to find the unique solution to the linear system.
Post by: squance on January 06, 2009, 09:23:25 pm
Thanks for pointing that out :)
I'll try to get the right answer now.
Post by: squance on January 06, 2009, 09:40:13 pm
But I finally got the right answer!!! YES! (after several hours of pain.... :( )
But if anyone has the answers to the questions at the bottom of my first post, that would be great
Post by: brendan on January 06, 2009, 09:44:31 pm
Sorry for double posting.
But I finally got the right answer!!! YES! (after several hours of pain.... :( )
But if anyone has the answers to the questions at the bottom of my first post, that would be great
is it what i got
Post by: kj_ on January 06, 2009, 09:48:45 pm
Sorry for double posting.
But I finally got the right answer!!! YES! (after several hours of pain.... :( )
But if anyone has the answers to the questions at the bottom of my first post, that would be great
EDIT: I re-read your initial question, and if you mean you want to place the x + y - 4z = 30 as row 1, and 3x + 2y - z = -15 as row 2, then no, it wouldn't make a difference. You would get the same answer.
(This is actually just a row-swap, which is legit in a matrix)
Post by: squance on January 06, 2009, 09:52:55 pm
And the unique solutions I got are x = -4, y = 2 and z = 7 :)
Post by: kj_ on January 06, 2009, 10:02:24 pm
Hmm...I'm not sure. I think the row echelon form you are talking about is the "reduced echelon form" because in my case I don't have to have any leading 1's...
And the unique solutions I got are x = -4, y = 2 and z = 7 :)
What I've been taught, generally row echelon form has all leading 1's, whereas reduced row echelon form has all leading 1's and 0's otherwise only, barring the last column. (which is of course your solution to the coeff :P)
Post by: squance on January 06, 2009, 10:07:04 pm
Thanks for that. I think I'll go and ask my lecturer next time at uni because he hasn't made it clear for me.
Post by: Mao on January 06, 2009, 10:14:17 pm
Hmm...I'm not sure. I think the row echelon form you are talking about is the "reduced echelon form" because in my case I don't have to have any leading 1's...
And the unique solutions I got are x = -4, y = 2 and z = 7 :)
What I've been taught, generally row echelon form has all leading 1's, whereas reduced row echelon form has all leading 1's and 0's otherwise only, barring the last column. (which is of course your solution to the coeff :P)
that's more of a 'preferred' thing, and it looks a lot nicer.
Post by: brendan on January 06, 2009, 10:33:33 pm
Post by: squance on January 06, 2009, 10:46:39 pm
Now I have to reduce that matrix that was in row echelon form to the form kj_ was talking about.
Im stuck here.
Edit:
1 hour later and finally I got the answer...
Dumb lecture notes are useless !!! :(
But Im really glad I finally got the answer.