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 April 13th, 2010, 04:14 AM #1 Senior Member   Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0 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.
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?

 April 13th, 2010, 06:13 PM #3 Senior Member   Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0 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.
 April 13th, 2010, 07:31 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,835 Thanks: 2162 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.
 April 13th, 2010, 10:49 PM #5 Senior Member   Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0 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.
 April 14th, 2010, 01:13 AM #6 Senior Member   Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0 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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