Results 1 to 8 of 8

Math Help - Find Lambda for the linear system

  1. #1
    Newbie
    Joined
    Apr 2011
    Posts
    10

    Find Lambda for the linear system

    Hi guys I am trying to figure out the following question can anyone help:
    Find \lambda for the Linear System:
    \lambdax+y+z=1
    x+ \lambday+z=1
    x+y+ \lambdaz=1

    For infinite solutions.
    A unique solution.

    And it's not asked but no solutions I wouldn't mind knowing either.

    I understand how to do these with numbers
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by userod View Post
    Hi guys I am trying to figure out the following question can anyone help:
    Find \lambda for the Linear System:
    \lambdax+y+z=1
    x+ \lambday+z=1
    x+y+ \lambdaz=1

    For infinite solutions.
    A unique solution.

    And it's not asked but no solutions I wouldn't mind knowing either.

    I understand how to do these with numbers
    by inspection is should be obvious that is lambda=1 all three equations are the same so you will get an infinite number of solutions, but there could be other values as well.

    Now use equation 3 to eliminate the x variable from equations one and two to get

    -\lambda E_3+E_1=(1-\lambda)y+(1-\lambda^2)z=0 =E_3


    -E_3+E_2=(\lambda-1)y+(1-\lambda)z=0=E_4

    Now use these two equations to eliminate y Can you finish from here?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2011
    Posts
    10
    Great thanks very much - when I add E3 and E4 I get a quadratic equation in lambda which gives lambda =1 or lambda = -2. So ruling out lambda=1 which gives an infinite number of solutions then lambda = -2. Would I be right? Thanks again
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by userod View Post
    Great thanks very much - when I add E3 and E4 I get a quadratic equation in lambda which gives lambda =1 or lambda = -2. So ruling out lambda=1 which gives an infinite number of solutions then lambda = -2. Would I be right? Thanks again
    We need to be vary careful right now.

    Both \lambda=1\quad \lambda =-2 both give an infinite number of solutions.

    Back in my first post I recommended that you muliply one of the equations by

    \lambda right?

    What value can't lambda be? Asked another way when we multiply an equation by a number can we use any number or are there number(s) we can use?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2011
    Posts
    10
    I think we can use any number except 0? I'm not sure I'm kind of lost
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by userod View Post
    I think we can use any number except 0? I'm not sure I'm kind of lost
    You are correct. What happens when you put  \lambda =0 into the original system does it have a solution then?
    If so is it unique?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Apr 2011
    Posts
    10
    Ah yes lambda=0 seems to give a unique solution .5,.5,.5.

    But it was just trial and error to give this? The simultaneous equation part didn't seem to help this atall?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,325
    Thanks
    1296
    If you are lost, it might be best to go back and think about what you are doing. You originally said, "Find lambda for the linear system", and then gave a system of equations involving x, y, z and parameter lambda. Find lambda so that what happens? I suspect you want to find lambda so that the system has a unique solution. Or, the contrary- find lambda so that there is more than one solution. Why would that be of interest? I can't imagine why you would say "The simultaneous equation part didn't seem to help this atall?" Without the system of equations, you would only be saying "Find lambda" without any conditions on it. And that would make no sense at all!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: February 27th 2011, 11:11 AM
  2. Find solutions of linear system
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 19th 2010, 01:58 PM
  3. Replies: 2
    Last Post: December 5th 2009, 01:30 AM
  4. Another linear system
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 29th 2009, 05:14 AM
  5. linear system
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 28th 2009, 08:17 AM

Search Tags


/mathhelpforum @mathhelpforum