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 January 28th, 2012, 06:08 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 $x_{1}, x_{2}, x_{3} and x_{4}$ $\begin{bmatrix} 1 & 0 & \frac{-14}{4} & \frac{5}{2} & 0 \\ 0 & 1 & 3 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ For variables $x_{1}, x_{2}, x_{3} and x_{4}$ I got: $x_{1}= \frac{14}{4}x_{3} - \frac{5}{2}x_{4}$ $x_{2}= -3x_{3} + 2x_{4}$ Then I assign the free variables arbitrary values r and s, so $x_{3}= r$ $x_{4}= s$ such that: $x_{1}= \frac{14}{4}r - \frac{5}{4}s$ and $x_{2}= -3r + 2s$ 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 $x_{3} and x_{4}$ are equal to zero since their rows are all zero rows? David.
January 28th, 2012, 08:18 PM   #2
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Re: Trivial solution only

Quote:
 Originally Posted by Bucephalus I reduced a matrix to this .....
WHAT matrix did you reduce?

 January 28th, 2012, 08:24 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, 08:41 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. $\begin{bmatrix} 0 & 1 & 3 & -2 & 0\\ 2 & 1 & -4 & 3 & 0\\ 2 & 3 & 2 & -1 & 0\\ -4 & -3 & 5 & -4 & 0 \end{bmatrix}$
 January 28th, 2012, 08:44 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, 08:45 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,122 Thanks: 1003 Re: Trivial solution only $\begin{bmatrix} 0 & 1 & 3 & -2 & 0\\ 2 & 1 & -4 & 3 & 0\\ 2 & 3 & 2 & -1 & 0\\ -4 & -3 & 5 & -4 & 0 \end{bmatrix}$ Top row: b + 3c - 2d = 0 Next 3 rows: 2a + b - 4c + 3d = 0 2a +3b + 2c - d =0 -4a-3b + 5c -4d = 0 ================= Add 'em up: OK?
 January 28th, 2012, 08:51 PM #7 Member   Joined: Jan 2012 Posts: 31 Thanks: 0 Re: Trivial solution only Yeah four variables.
 January 28th, 2012, 08:52 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, 09:02 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: $x_{1} = 7s - 5t x_{2} = -6s + 4t x_{3} = 2s x_{4} = 2t$ 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, 09:04 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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