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Linear system of equations

  1. Nov 27, 2016 #1
    1. The problem statement, all variables and given/known data
    system of equations is as follows
    x+y=1
    2x+y-z=1
    3x+y-2z=1
    ##\begin{cases} x+y=1 |*(-1)\\2x+y-z=1\\3x+y-2z=1 \end{cases}##

    2. Relevant equations
    Gaussian elimination method technique

    3. The attempt at a solution

    ##\begin{cases} x+y=1 \\ x-z=0 |*(-2)\\ 2x-2z=0 \end{cases}##


    <=>

    ##\begin{cases} x+y=1 |*(-1)\\ x-z=0 \\ 0=0 \end{cases}##

    <=>

    ##\begin{cases} x+y=1 \\ -y-z=-1 \\ x=z \end{cases}##

    <=>
    at this stage I think you are supposed to plug in x=z into some equation in the system
    ##\begin{cases} x+y=1 \\ -y-z=-1 \\ x=z \end{cases}##

    ##\begin{cases} x+y=1 <=> x=1-y \\ -y-z=-1 \\ x=z \end{cases}##

    ##-y-x=-1##
    <=> x=-y+1

    from those two equations it can be seen that those equations at least are identical equations for the same line.

    So I think ultimately based on that geometry there should be infinite solutions (is that the correct way to do it and solve it?)
     
  2. jcsd
  3. Nov 27, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes. Your equations are not independent, so instead of three you have only two equations and the solution has one degree of freedom.
     
  4. Nov 27, 2016 #3
    Hi @late347:

    I confess I am unfamiliar with the notation "| * (-1)". However, there is a quick way to discover there there is a unique solution or an infinity of solutions. If you take the determinant of the matrix of coefficients, if the value of the determinant is zero, then there are an infinite number of solutions. If the value is not zero, then there is a unique solution. Here the matrix is:
    1 1 0
    2 1 -1
    3 1 -2
    and the value is zero, so there are an infinite number of solutions.

    To find out the dimensional nature of this infinity requires some additional analysis.

    Hope this is helpful.

    Regards,
    Buzz
     
  5. Nov 27, 2016 #4

    Our teacher only really taught us to manipulate the equations themselves instead of making the matrix such as you did.

    I think my notation just means | * (-2) means that both sides of the equation were multiplied by (-2).

    And the newly created equations in the further phases of the calculation were created by summing equations together
     
  6. Nov 27, 2016 #5

    BvU

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    Clear to me it was.
     
  7. Nov 27, 2016 #6
    Hi late:

    Thanks for the explanation.

    Regards,
    Buzz
     
  8. Nov 28, 2016 #7

    Mark44

    Staff: Mentor

    The geometry here is that the three equations represent three planes that intersect in a line. Each point on this line is a solution to the system of equations. One point on this line is (0, 1, 0). It can be shown (and you probably haven't seen this yet) that every point on this line is of the form (x, y, z) = r(1, -1, 1) + (0, 1, 0), where r is any real number.
     
  9. Nov 29, 2016 #8

    That may very well be... I think the more simple case of two-variable system of equations was easy to justify to myself.
     
  10. Nov 29, 2016 #9

    Mark44

    Staff: Mentor

    Sure, that's fine. all I was trying to do was to look at the problem from a geometric perspective. Since you mentioned Gaussian elimination, it's likely that you will soon be doing these problems using matrices, and working with vectors. For me, it's helpful to consider the geometry of these problems, not just the algebraic manipulations.
     
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