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April 13th, 2010, 04:14 AM   #1
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Simplification

Hi,

I have the (*)

x_1 + 3x_2 + 3x_3 + 2x_4 = 1
2x_1 + 6x_2 + 9x_3 + 5x_4 = 1
-x_1 -3x_2 + 3x_3 = k

I have solved this using Gaussian Elimination down to the simplification of:

x_1 + 3x_2 + 3x_3 + 2x_4 = -1 (i)
3x_2 + x_4 = -1 (ii)

(x_2, x_4) are free variables denoted: A and B respectively.

(ii) is simplified to; x_3 = ((-1 - B) / 3)

Now, I'm having some trouble simplifying (i)

I have gone,

x_1 + 3A +3((-1 - B) / 3) + 2B = -1

x_1 = 1 - 3A - 1 - B - 2B
= 2 - 3A - B + 2B

How can I simplify this further ?

Thanks!

wulfgarpro.
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April 13th, 2010, 10:20 AM   #2
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Quote:
Originally Posted by wulfgarpro
x_1 + 3x_2 + 3x_3 + 2x_4 = 1
2x_1 + 6x_2 + 9x_3 + 5x_4 = 1
-x_1 -3x_2 + 3x_3 = k

I have solved this using Gaussian Elimination down to the simplification of:

x_1 + 3x_2 + 3x_3 + 2x_4 = -1 (i)
3x_2 + x_4 = -1 (ii)
What happened to the "k" in the third (original) equation?
stapel is offline  
April 13th, 2010, 06:13 PM   #3
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Re: Simplification

Well,

If k = -3 there are infinite possible solutions; namely,

x_1 + 3x_2 + 3x_3 + 2x_4 = -1 (i)
3x_2 + x_4 = -1 (ii)
0 - 0 + 0 = 0 (iii)

This is because in row echelon form, k is k+3. If k != -3 there are no solutions because the system is inconsistent. So I just dropped it in this case; I'm trying to solve the system when k = -3.

wulfgarpro.
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April 13th, 2010, 07:31 PM   #4
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You get x_3 = (-1 - B)/3 and x_1 = 2 - 3A - B. Since A and B can have any values, you can't simply further.
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April 13th, 2010, 10:49 PM   #5
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Re: Simplification

Hm,

I get x_1 = -3A - B

If possible, could you show me how you got your solution in a stepwise fashion ?

wulfgarpro.
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April 14th, 2010, 01:13 AM   #6
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Re: Simplification

I got it:

x_1 + 3A + 3((-1 - B) / 3) + 2B = 1

x_1 + 3A -1 - B + 2B = 1

x_1 + 3A - B + 2B = 2

2 - 3A - B = x_1

wulfgarpro.
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