January 28th, 2012, 07:08 PM  #1 
Member Joined: Jan 2012 Posts: 31 Thanks: 0  Trivial solution only
Hi I reduced a matrix to this and the book says it has only the trivial solution. For variables For variables I got: Then I assign the free variables arbitrary values r and s, so such that: and But this isn't what the book has. It just says that it has the trivial solution only, which I can't easily see by inspection of this solution I have. Are they just saying that are equal to zero since their rows are all zero rows? David. 
January 28th, 2012, 09:18 PM  #2  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,984 Thanks: 995  Re: Trivial solution only Quote:
 
January 28th, 2012, 09:24 PM  #3 
Member Joined: Jan 2012 Posts: 31 Thanks: 0  Re: Trivial solution only
This one \begin{bmatrix} 0 & 1 & 3 & 2 & 0\\ 2 & 1 & 4 & 3 & 0\\ 2 & 3 & 2 & 1 & 0\\ 4 & 3 & 5 & 4 & 0 \end{bmatrix} Thanks. 
January 28th, 2012, 09:41 PM  #4 
Member Joined: Jan 2012 Posts: 31 Thanks: 0  Re: Trivial solution only
Ok I'm teaching latex more than getting my problem solved. Here it is again. 
January 28th, 2012, 09:44 PM  #5 
Member Joined: Nov 2011 Posts: 40 Thanks: 0  Re: Trivial solution only
What does the question actually ask? I double checked and your reduction is right. In your steps it looks like you have gone on to find the null space of the matrix. Did it ask to find the determinant of the matrix? Cause that has only the trivial solution since the matrix is singular. Shouldn't there also be a fifth variable x5 since is a 4x5 matrix? 
January 28th, 2012, 09:45 PM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,984 Thanks: 995  Re: Trivial solution only Top row: b + 3c  2d = 0 Next 3 rows: 2a + b  4c + 3d = 0 2a +3b + 2c  d =0 4a3b + 5c 4d = 0 ================= Add 'em up: OK? 
January 28th, 2012, 09:51 PM  #7 
Member Joined: Jan 2012 Posts: 31 Thanks: 0  Re: Trivial solution only
Yeah four variables.

January 28th, 2012, 09:52 PM  #8 
Member Joined: Jan 2012 Posts: 31 Thanks: 0  Re: Trivial solution only
The question said "Solve the homogeneous system of linear equations by any method."

January 28th, 2012, 10:02 PM  #9 
Member Joined: Jan 2012 Posts: 31 Thanks: 0  Stop!!
My bad, I was looking at the wrong answer in that back of the book. Terribly sorry. Having a bad day. I still haven't got the right answer yet though. Let me investigate more Here is the answer: They obviously used s and t as the arbitrary variables. Now I will go back and see if I can get mine to look like this. Sorry, welcome to complete this with me if you like. 
January 28th, 2012, 10:04 PM  #10 
Member Joined: Jan 2012 Posts: 31 Thanks: 0  Re: Trivial solution only
Yes that works out now. Sorry about that. 

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