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Linear Algebra

  1. Mar 18, 2004 #1
    This seems like a very simple problem, yet I can't seem to come up with the correct solution.

    Determine all the values of c for which the system has nontrivial solutions, and then find all the solutions.

    x + 2y + cz = 0
    3x - y = 0
    -2x + y + z = 0

    OK, the system has nontrivial solutions if the determinant = 0.
    So I set the determinant = 0 and solve for c.

    Using cofactor expansion, I have:
    c(3 + 2) + (-1 - 6) ==> c = 7/5.

    I plug this into the system and using elementary row operations I still only come up with only the trivial solution x = 0, y = 0,
    z = 0. I can't go through the steps here. It would be way too tedious, but I've gone over it several times and come up with the same result. What gives? I was promised that if the det = 0 the system has either 0 solutions or infinitely many solutions. Since x,y,z = 0 is a solution there must be infinitely many solutions.

    Where have I gone wrong?

  2. jcsd
  3. Mar 18, 2004 #2
    You're doing a good job of ensuring that the system of equations has only trivial solutions by setting the determinant equal to zero (which is why those are all you're getting), but it doesn't help much if you're looking for non-trivial solutions.

  4. Mar 18, 2004 #3
    So I need to do the elementary row operations first and reduce to row-echelon form? Once I'm there I set c blah, blah, blah = 0 to eliminate the last equation thereby ensuring infinitely many solutions to the system? This doesn't seem to make use of anything new we learned that the homework problems should refer to. It's a section determinants, but it doesn't seem that determinants are very helpful here.
  5. Mar 18, 2004 #4
    What're you trying to do in the problem? Are you trying to find non-trivial solutions or trivial ones?

    If you're looking for trivial solutions (i.e. x,y,z = 0), then you've done it.

    If you're looking for non-trivial solutions, you're not going to set the determinant to 0.

  6. Mar 18, 2004 #5
    But doesn't a det = 0 just mean that the system will have either no solutions or infinitely many solutions? And in the case of a homogeneous system, wouldn't it guarantee that you have an infinite number of solutions?

  7. Mar 18, 2004 #6
  8. Mar 18, 2004 #7
    OK, I checked Mathworld/Cramer's Rule and found the following where D is the determinant of a system of equations:

    If d = 0 and D = 0, the system has no unique solution.

    Wouldn't this mean that if D = 0 and the system of equations is homogeneous then it must have infinitely many solutions?
  9. Mar 18, 2004 #8


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    cookie monster must have had his eyes crossed when looking at this!
    discoverer02 is completely correct and cookie monster (much as it hurts me to say it!) is, in this case, completely wrong.

    A system of linear equations has a unique solution if and only if the determinant of the coefficient matrix is non-zero. Since a homogenous system of equations always has the trivial solution, the only way there can be non-trivial solutions is if that not a unique solution: in other words, if the determinant is0 just as discoverer02 said.

    The only thing wrong with discoverer02's solution is that he calculated the determinant incorrectly!

    The determiant is |1 2 c|
    |3 -1 0|
    |-2 1 1|

    Expanding on the third column, that is
    |3 -1| |1 2|
    c|-2 1|+ 1|3 -1|= c((3)(1)-(-1)(-2))+ (1(-1)-3(2))
    = c(3-2)+ (-1-6)
    (not "c(3+2)"!)
    = c- 7= 0 so c= 7, not 7/5.

    In particular, if c= 7, then x= 1, y= 3, z= -1 satisfies

    x+ 2y+ 7z= 1+ 2(3)+ 7(-1)= 7- 7= 0
    3x- y = 3- 3 = 0
    -2x+ y+ z= -2+ 3 - 1 = 1- 1= 0.
  10. Mar 18, 2004 #9
    Thanks very much for the help.

    I looked at this at least a dozen times and didn't catch that.

    Last edited: Mar 18, 2004
  11. Mar 18, 2004 #10
    Maybe I should stop sticking my nose in where it has no business...

  12. Mar 19, 2004 #11
    Actually cookiemonster, our brief dialogue really got me thinking about this problem and these types of problems, so whether you think so or not, you were a great help.

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