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# Gaussian Elimination help Watch

1. Hi, need help with this question.

Using Gaussian elimination solve this matrix

(A|b) =

1 1 -2 1
2 -1 2 1
c 0 1 0

its the bottom line with 'c' that i don't know how to reduce.
can someone tell me what i need to do with it?

thanks
2. Well, c is just a number, so how would you turn the c into a 1 so that you can reduce it? (And under what conditions are you allowed to perform this operation?)
3. im told to find (x,y,z) in terms of c
so i dont have to turn it into anything
4. i just need to reduce it in terms of C, but i dont know how to do that :/
5. (Original post by Smaug123)
Well, c is just a number, so how would you turn the c into a 1 so that you can reduce it? (And under what conditions are you allowed to perform this operation?)
^
|
6. (Original post by lar di da)
^
|
You know you can perform elementary row operations without changing the answer. Can you think of an elementary row operation that will turn (c,0,1|0) into (1, <some other numbers>)? (It might help to list the elementary row operations.)
Once you've got a 1 in that position, you can repeat in the usual inductive way to complete the Gaussian elimination; whenever you end up with something involving c in the "active slot" (there's probably an official term for that), you can perform a similar step to turn it into a 1.
7. Personally, I'd use a different method. Noting the y and z coefficients of the first two rows, I would add those two rows together first (in terms of elementary row operations, replacing the first or second row with the sum 3 0 0 | 2). From there, you can just multiply that row by some obvious factors and subtract it into the other rows to get to RREF.

In other words:
Spoiler:
Show

1 1 -2 | 1
2 -1 2 | 1
c 0 1 | 0

1 1 -2 | 1
3 0 0 | 2 #R2<=R1+R2
c 0 1 | 0

1 1 -2 | 1
1 0 0 | 2/3 #R2<=(1/3)R2
c 0 1 | 0

0 1 -2 | 1/3 #R1<=R1-R2
1 0 0 | 2/3
0 0 1 | -2c/3 #R3<=R3-(c)R2

1 0 0 | 2/3
0 1 -2 | 1/3
0 0 1 | -2c/3

Anyways, you just have to treat 'c' like any old random number. If you can't do that, then perhaps you need more practice with RREF?
8. (Original post by aznkid66)
Personally, I'd use a different method. Noting the y and z coefficients of the first two rows, I would add those two rows together first (in terms of elementary row operations, replacing the first or second row with the sum 3 0 0 | 2).
Ah, much neater I admit that I tend to find the hardest possible way first, and stick to it…
9. (Original post by aznkid66)
Personally, I'd use a different method. Noting the y and z coefficients of the first two rows, I would add those two rows together first (in terms of elementary row operations, replacing the first or second row with the sum 3 0 0 | 2). From there, you can just multiply that row by some obvious factors and subtract it into the other rows to get to RREF.

In other words:
Spoiler:
Show

1 1 -2 | 1
2 -1 2 | 1
c 0 1 | 0

1 1 -2 | 1
3 0 0 | 2 #R2<=R1+R2
c 0 1 | 0

1 1 -2 | 1
1 0 0 | 2/3 #R2<=(1/3)R2
c 0 1 | 0

0 1 -2 | 1/3 #R1<=R1-R2
1 0 0 | 2/3
0 0 1 | -2c/3 #R3<=R3-(c)R2

1 0 0 | 2/3
0 1 -2 | 1/3
0 0 1 | -2c/3

Anyways, you just have to treat 'c' like any old random number. If you can't do that, then perhaps you need more practice with RREF?
(Original post by Smaug123)
Ah, much neater I admit that I tend to find the hardest possible way first, and stick to it…

thanks both of you. i think i do need a bit more practice with RREF though when i did work it out and how ive been taught it slightly different you. ive been taught to go to REF and then RREF. Yours seems to skip straight to RREF?
10. (Original post by lar di da)
thanks both of you. i think i do need a bit more practice with RREF though when i did work it out and how ive been taught it slightly different you. ive been taught to go to REF and then RREF. Yours seems to skip straight to RREF?
I go to REF and then to RREF; in this particular instance, you can use a trick (which aznkid66 spotted) to get you closer to RREF much faster.

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