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4x6 matrix

  1. Aug 18, 2011 #1
    I have a problem that involves 4 simulationious equations with 6 unknown variables.
    Been trying to solve it using elimination but getting stuck in a loop.
    Any advice on how to solve this? I know there will be multiple answers.
    Is elimination the best method?

    Can't upload the problem now cause I have writing this from a phone, so if you need the question I'll upload it as soon as I can.

    Thank you for any help.
     
  2. jcsd
  3. Aug 18, 2011 #2

    micromass

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    Yes, elimination seems to be the right way to go.
    So we'll try to help once you uploaded the problem and what you tried...
     
  4. Aug 18, 2011 #3
    Thank you sorry for the useless post haha I'll get it up as soon as I get home
     
  5. Aug 18, 2011 #4

    Ray Vickson

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    Such systems are met with millions of time per day when solving _linear programming_ problems. There, the concept of a *basic* solution arises; this is a solution in which four of the variables are solved for as functions of the other two, then setting those two to zero, assuming that the 4x4 submatrix of those 4 variables is nonsingular. If every choice of 4x4 matrix is allowed (i.e., all are nonsingular), the number of basic solutions is C(6,4) = 6*5/2 = 15. If some of the 4x4 matrices are singular, the corresponding basic solution is non-existent (by definition).

    While there could be as many as 15 different basic solutions there are infinitely many non-basic solutions, simply by assigning arbitrary non-zero values to the right-hand-side variables in each basic system.

    For more on basic solutions, see, eg.,
    http://www2.isye.gatech.edu/~spyros/LP/LP.html .

    RGV
     
  6. Aug 20, 2011 #5
    sorry for the ridiculously late reply. Finally got home so here is the question and my attempt at a solution.
     

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  7. Aug 20, 2011 #6

    HallsofIvy

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    Well the first thing I notice is that if you add twice the second equation to the first equation, you eliminate three unknowns at a stroke! [itex]x_3[/itex], [itex]x_4[/itex], and [itex]x_6[/itex] all cancel leaving [itex]4x_1+ 4x_2- 7x_5= -2[/itex]. Since you know you will want to solve for four of the unknowns in terms of the other two, I would choose the two to be [itex]x_1[/itex] and [itex]x_2[/itex] so that I already have [itex]x_5= (4/7)x_1+ (4/7)x_2+ 2/7[/itex].

    Replace [itex]x_5[/itex] in each of the equations by that and continue.
     
    Last edited: Aug 20, 2011
  8. Aug 21, 2011 #7
    Of course. Thank you, will continue working and let you guys know how I go.
     
  9. Aug 31, 2011 #8
    So I left this problem and I have come back to it.
    I have ended up with
    x1=-3-x2
    x3=-1+x6
    x4=3
    x5=-2

    I am stuck...is this the answer?
    I know 2 will be defined, 2 will be infinite and 2 will be defined by the infinite variables.
    I am just not sure how to get to that conclusion.
     
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