Results 1 to 14 of 14

Math Help - Non-trivial solutions

  1. #1
    Member
    Joined
    Sep 2007
    From
    Calgary, Alberta
    Posts
    77

    Non-trivial solutions

    Find b such that the homogeneous system with augmented matrix below has non-trivial solutions:

    A = <br />
\begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;-2 & 4 & -18 & 0\;\\\;-4 & 0 & b & 0\;\\\;2 & 1 & 3 & 0\end{vmatrix}<br />

    So I know a system of equations is homogeneous if all the constant terms are zero, and I know to make a nontrivial solution I need at least one variable to have a nonzero value.

    My question is, what is b supposed to be? Is there a single number that it should end up being? Or can it be 1 or 2 or 3, etc.?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,380
    Thanks
    1473
    Awards
    1
    This is going to be a different approach.
    Please investigate what happens to your system if b = -12.
    After doing so, try to convince yourself that is the answer.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2007
    From
    Calgary, Alberta
    Posts
    77
    That gives me:

    A = <br />
\begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & 3 & 0\;\\\;0 & 1 & -3 & 0\;\\\;0 & 0 & 0 & 0\end{vmatrix}<br />

    So:

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

    Is that the answer? I'm not getting this.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,380
    Thanks
    1473
    Awards
    1
    That makes me question if you understand what is going on here?
    Is the triple \left( {3t, - 3t, - t} \right) a solution to your system if t =  - 1,1,2, - 2 or t is any number?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2007
    From
    Calgary, Alberta
    Posts
    77
    Quote Originally Posted by Plato View Post
    Is the triple \left( {3t, - 3t, - t} \right) a solution to your system if t =  - 1,1,2, - 2 or t is any number?
    Yes.

    I'm wondering where you got b = -12 though, that confuses me.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,380
    Thanks
    1473
    Awards
    1
    Quote Originally Posted by Thomas View Post
    I'm wondering where you got b = -12 though, that confuses me.
    EXACTLY THAT! That is what I asked you to find out in the begining.
    If you can understand where that comes from then you will understand this problem.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Sep 2007
    From
    Calgary, Alberta
    Posts
    77
    I tried solving the matrix while keeping b in the matrix.

    I got:

    A = <br />
\begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & \frac{-b}{4} & 0\;\\\;0 & 1 & \frac{-b - 36}{8} & 0\;\\\;0 & 0 & \frac{5b + 60}{8} & 0\end{vmatrix}<br />

    I noticed b = -12 would make the \frac{5b + 60}{8} = 0

    Am I getting any closer?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Sep 2007
    From
    Calgary, Alberta
    Posts
    77
    Bump.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,380
    Thanks
    1473
    Awards
    1
    Quote Originally Posted by Thomas View Post
    Bump.
    Bump, what is this bump?
    b=-12 makes the rows dependent. What does that imply?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Sep 2007
    From
    Calgary, Alberta
    Posts
    77
    Quote Originally Posted by Plato View Post
    Bump, what is this bump?
    b=-12 makes the rows dependent. What does that imply?
    I haven't heard the term dependent when doing matrices before, so I don't know.

    Could you possibly tell me how you came up with -12 so I understand?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Thomas View Post
    I got:

    A = <br />
\begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & \frac{-b}{4} & 0\;\\\;0 & 1 & \frac{-b - 36}{8} & 0\;\\\;0 & 0 & \frac{5b + 60}{8} & 0\end{vmatrix}<br />

    I noticed b = -12 would make the \frac{5b + 60}{8} = 0

    Am I getting any closer?
    Yes, you're getting very close. If b = -12 then the bottom row of that matrix becomes all zeros. The corresponding equation is 0 = 0 (which is automatically satisfied, of course), and we're left with only two genuine equations for three unknowns. So you should be able to find a nontrivial solution for them.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Sep 2007
    From
    Calgary, Alberta
    Posts
    77
    Okay, awesome.

    When they refer to "trivial" does this just mean the only answer to the equations is 0, and "nontrivial" means the answer can involve numbers other than zero?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Thomas View Post
    When they refer to "trivial" does this just mean the only answer to the equations is 0, and "nontrivial" means the answer can involve numbers other than zero?
    Yes. A homogeneous system of linear equations always has at least one solution, namely when all the variables are zero. That is called the "trivial" solution. Any other solution is "nontrivial" (even if some of the variables are zero, just so long as they are not all zero).
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Sep 2007
    From
    Calgary, Alberta
    Posts
    77
    Okay, thank you for clearing that up.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. What is a trivial cocone?
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: November 6th 2011, 02:50 PM
  2. Replies: 0
    Last Post: October 2nd 2011, 08:33 AM
  3. linear algebra proof with trivial solutions
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: August 6th 2010, 09:28 PM
  4. Need help with Trivial/Non trivial solution
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 6th 2010, 05:53 AM
  5. Trivial/Non Trivial Solutions
    Posted in the Advanced Algebra Forum
    Replies: 14
    Last Post: October 15th 2008, 05:17 PM

Search Tags


/mathhelpforum @mathhelpforum