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System of linear equations

  1. Apr 18, 2017 #1
    1. The problem statement, all variables and given/known data
    3.For which values of ##\lambda## does the following system of equations also have non trivial solutions

    hggfhgfhfg.jpg

    2. Relevant equations


    3. The attempt at a solution
    What I tried doing first is to put all variables on the same side and got
    ##
    v+y-\lambda*x=0\\
    x+z-\lambda*y=0\\
    y+u-\lambda*z=0\\
    z+v-\lambda*u=0\\
    u+x-\lambda*v=0
    ##
    and when I wrote the coefficient into the matrix i got
    ##
    \begin{bmatrix}

    -\lambda& 1 &0&0&1\\
    1&-\lambda&1&0&0\\
    0&1&-\lambda&1&0\\
    0&0&1&-\lambda&1\\
    1&0&0&1&-\lambda\\
    \end{bmatrix}
    ##

    here I noticed that all the columns sum to the same number ##2-\lambda## there I summed everything into the first row and got
    ##
    \begin{bmatrix}

    2-\lambda & 2-\lambda&2-\lambda&2-\lambda&2-\lambda\\
    1&-\lambda&1&0&0\\
    0&1&-\lambda&1&0\\
    0&0&1&-\lambda&1\\
    1&0&0&1&-\lambda\\
    \end{bmatrix}
    ##
    here I looked into 2 different possibilities if a) ##\lambda=2## and b) ##\lambda\neq2##.
    However a) is pretty simple and it's mostly b) that I'm having trouble with.
    Here I thought if ##\lambda\neq2## then I can devide the first row by ##2-\lambda##
    When I did this my matrix looked like this
    ##
    \begin{bmatrix}

    1 & 1&1&1&1\\
    1&-\lambda&1&0&0\\
    0&1&-\lambda&1&0\\
    0&0&1&-\lambda&1\\
    1&0&0&1&-\lambda\\
    \end{bmatrix}
    ##
    Then I subtracted the first row from the second and last one and got
    ##
    \begin{bmatrix}
    1 & 1&1&1&1\\
    0&-\lambda-1&0&-1&-1\\
    0&1&-\lambda&1&0\\
    0&0&1&-\lambda&1\\
    0&-1&-1&0&-\lambda-1\\
    \end{bmatrix}
    ##
    then I just rearranged some rows so that it would be easier for me to read
    ##
    \begin{bmatrix}
    1 & 1&1&1&1\\
    0&1&-\lambda&1&0\\
    0&-1&-1&0&-\lambda-1\\
    0&-\lambda-1&0&-1&-1\\
    0&0&1&-\lambda&1\\
    \end{bmatrix}
    ##
    then I added the second row to the third and forth one and switched the third and forth row
    ##
    \begin{bmatrix}
    1 & 1&1&1&1\\
    0&1&-\lambda&1&0\\
    0&-\lambda&-\lambda&0&-1\\
    0&0&-1-\lambda&1&-\lambda-1\\
    0&0&1&-\lambda&1\\
    \end{bmatrix}
    ##
    Lastly I added the last row to the forth one and switched them
    ##
    \begin{bmatrix}
    1 & 1&1&1&1\\
    0&1&-\lambda&1&0\\
    0&-\lambda&-\lambda&0&-1\\
    0&0&1&-\lambda&1\\
    0&0&-\lambda&1-\lambda&-\lambda\\

    \end{bmatrix}
    ##
    Here is where I get stuck. I don't know how to continue from here on out. Maybe I made a mistake somewhere in my addition however I went through it at least a few times and I was not able to find it:
    Any help / tips are greatly appreciated
    Thanks
     

    Attached Files:

  2. jcsd
  3. Apr 18, 2017 #2

    Mark44

    Staff: Mentor

    After your first step (when all the entries in row 1 were 1), you used the first entry in row 1 to eliminate all the entries below it. Continue this process by using the 2nd entry in row 2 (the pivot) to eliminate all entries above and below it. Continue this process until you have a diagonal matrix or until the matrix is as reduced as possible.
     
  4. Apr 18, 2017 #3
    Thank you for the reply.
    I though about doing that however does that mean that I have to multiply some rows with ##\lambda## to get the other ones to cancel out ?
    If so do I also need to check what happens when ##\lambda=0## ?
     
  5. Apr 19, 2017 #4

    Mark44

    Staff: Mentor

    Yes to the first question. For the second, you could go back to your first matrix and replace ##\lambda## with 0, and see what you get from that.
     
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