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 November 23rd, 2008, 03:39 PM #1 Newbie   Joined: Nov 2008 Posts: 4 Thanks: 0 Homework question The directions state: Find the solution space of the following systems of linear homogeneous equations: x-y+z-w=0 2x+y-z+2w=0 2y+3z+w=0 Is this the same as finding the solutions, because I did that and got (-z, -3z, 3z, z). But I don't know what a solution space is. Help. Thanks you so much.
 November 24th, 2008, 04:33 PM #2 Newbie   Joined: Nov 2008 Posts: 4 Thanks: 0 Re: Homework question To answer my own question: the solution space is actually the subspace you get. For instance, this problem yields a one dimensional subspace.
 December 2nd, 2008, 07:45 PM #3 Newbie   Joined: Nov 2008 Posts: 4 Thanks: 0 Re: Homework question I don't think that solution set is right. The vector -1,-3,3,1 does not give you 0 when used as weights in the original equation. It is a system of 4 unknowns in 3 rows there has to be a free variable somewhere. The solution (-1,-3,3,1) does not satisfy ax=0. That is if you still care about this problem :P, unless im missing something..
 December 7th, 2008, 08:04 AM #4 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 Re: Homework question Oh, wow, I completely missed this problem... Anyway, yes, the vector (-1,-3,3,1) forms a basis for the vector space. It isn't the solution itself--- z is the free variable. So, undermine, you are missing something
 April 15th, 2015, 09:35 PM #5 Newbie   Joined: Apr 2015 From: usa Posts: 1 Thanks: 0 Thanks! Looking forward to it. I think I've spend more time watching your show this at work than actually working! Fantastic stuff!
April 16th, 2015, 05:17 PM   #6
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Quote:
 Originally Posted by orionankh The directions state: Find the solution space of the following systems of linear homogeneous equations: x-y+z-w=0 2x+y-z+2w=0 2y+3z+w=0 Is this the same as finding the solutions, because I did that and got (-z, -3z, 3z, z).
This makes no sense because you haven't said what variables each of those places means. Normally, I would assume (x, y, z, w) since that is the order of the variables in the equations. But x= -z, y= -3z, z= 3z, w= z makes no sense.
Quote:
 But I don't know what a solution space is. Help. Thanks you so much.

April 16th, 2015, 06:04 PM   #7
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Quote:
 Originally Posted by Country Boy This makes no sense because you haven't said what variables each of those places means. Normally, I would assume (x, y, z, w) since that is the order of the variables in the equations. But x= -z, y= -3z, z= 3z, w= z makes no sense.
You are aware that this thread is seven years old? It was triggered by a troll.

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