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April 5th, 2013, 08:38 AM   #1
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Non-trivial solution of a homogeneous system

Hello, I'm stuck with this question. Can anyone help me?

Find all values of k for which this homogeneous system has non-trivial solutions:

[kx + 5y + 3z = 0
[5x + y - z = 0
[kx + 2y + z = 0

I made the matrix, but I don't really know which Gauss-elimination method I should use to get the result.
Thanks already!
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April 5th, 2013, 12:45 PM   #2
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Re: Non-trivial solution of a homogeneous system

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April 5th, 2013, 12:48 PM   #3
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The determinant of the coefficients matrix is 5(1 - k), and you need it to be zero, so k = 1.

Alternatively, multiply the first equation by 3 and add the second equation to it, then compare the result with the third equation.
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April 5th, 2013, 12:59 PM   #4
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Re: Non-trivial solution of a homogeneous system

$\color{#003399}{\bf{skipjack}}$, as a moderator you should put your post ahead of mine because I think it will help the OP to read your solution first since it's short and sweet.


Last edited by skipjack; June 20th, 2016 at 08:39 AM.
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April 5th, 2013, 10:40 PM   #5
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Re: Non-trivial solution of a homogeneous system

O-K I managed to reorder posts with $\color{#003399}{\bf{skipjack}}$'s suggestion in PM. I still want to post my work even though it's long-winded as I believe it to use valuable ideas from Linalgebra.

Quote:
Originally Posted by agentredlum
We already know it has the trivial solution x = y = z = 0. If it is to have at least 1 non-trivial solution, then it must have an infinite number of solutions since the only possibilities for systems of linear equations are...

1) No solutions

2) One unique solution

3) Infinite number of solutions

This means producing a last row of zeros, in order to do this, we must make column 1 a linear combination of columns 2 and 3 of your matrix post.

a[5, 1, 2] + b[3, -1, 1] = [k, 5, k] (I'm using row vectors here for typing ease)

This gives at once

5a + 3b = k (1)

a - b = 5 (2)

2a + b = k (3)

(1) - (3) gives

3a + 2b = 0 (4)

a - b = 5 (2)


and we get a = 2 and b = -3 and back substitution confirms k = 1 in equations (1) AND (3)

to check work, replace k with 1 then rref

[attachment=1:2e2b4j6p]MSP13871e6gd0b37dd990dh00004cc6aa5i8bgeia5h.GIF[/attachment:2e2b4j6p]
[attachment=0:2e2b4j6p]MSP13921e6gd0b37dd990dh000021h5f1h01bcfihfe.GIF[/attachment:2e2b4j6p]

Notice a last row all zeros meaning infinite non-trivial solutions.

You can get wolfram to confirm k = 1 is the only solution by clicking link below

http://www.wolframalpha.com/input/?i=so ... +%2C+k%29+


I make no claim this is the best way to do this problem, and I expect other MMF members will have better solutions. My method is a little flawed because I have to make a complicated argument (omitted here) as to why k = 1 is the only solution.


Last edited by skipjack; June 20th, 2016 at 08:42 AM.
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June 19th, 2016, 10:20 PM   #6
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I also have a same problem
for -1.73a+d=0
-1.73b+d=0
-1.73c+d=0
a+b+c-1.73d=0
How can I GET VALUES OF a,b,c,d
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June 20th, 2016, 08:57 AM   #7
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Is 1.73 an exact value or could it be exactly √3?
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June 20th, 2016, 09:27 AM   #8
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Trivially, $a=b=c$, so it's now a system of two variables
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June 20th, 2016, 05:51 PM   #9
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Form the coefficient matrix and then solve for the values, but it might be a bit tricky for a 4*4 matrix
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June 20th, 2016, 10:09 PM   #10
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Quote:
Originally Posted by skipjack View Post
Is 1.73 an exact value or could it be exactly √3?
it can be root 3 also..........
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