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Define coefficients for system of equations with 3 variables

  1. Sep 9, 2009 #1
    1. The problem statement, all variables and given/known data
    I have the following system og equations:

    [itex]
    \left\{\begin{array}{l}
    y + 2z = 1 \\
    x + 4y + 3z = C \\
    x + y + Bz = B
    \end{array}\right.
    [/itex]

    a) Create the augmented matrix and reduce to row echelon form (I've already done this)
    b) For which values of A and B is the system inconsistent?
    c) For which values of A and B does the system have only one solution?
    d) Determine the complete solution to the system for all values of A and B.

    2. Relevant equations
    The system in row echelon form:
    [itex]
    \begin{pmatrix}
    1 & 4 & 3 & C \\
    0 & 1 & 2 & 1 \\
    0 & 0 & 1 & \frac{B - C + 3}{B + 3}
    \end{pmatrix}
    [/itex]


    3. The attempt at a solution
    Earlier when solving this kind of problems with systems with 2 variables, I solved it geometrically as lines being parallel, intersecting etc. Of course this could be solved geometrically too as planes, but as I can't imagine how to do it geometrically with >3 variables, I thought I've missed something.

    If someone could point me in a direction, then I'll figure out the rest myself.

    Thanks.
     
  2. jcsd
  3. Sep 9, 2009 #2
    Ok, now go back of the system of equations. You got:

    [tex]\left\{\begin{matrix}
    x+4y+3z=C\\
    y+2z=1\\
    z=\frac{B-C+3}{B+3}
    \end{matrix}\right.[/tex]

    What if B=-3 ?
     
  4. Sep 9, 2009 #3
    As that's not possible, I'll guess the system would be inconsistent?
     
  5. Sep 9, 2009 #4
    Yup, you're right.

    Now find x,y,z solutions and again tell me for which is inconsistent and for which values its consistent. :smile:
     
  6. Sep 9, 2009 #5
    I'm not sure what you want... Do you want me to back substitute to get expressions for y and x too, and then define values for B and C, which results in division by 0?
     
  7. Sep 9, 2009 #6
    Yes, you need to do back substitution.

    But where is A ? I can't see any A in the problem.
     
  8. Sep 9, 2009 #7
    There isn't any A... Only B and C... Don't know why
     
  9. Sep 9, 2009 #8
    Maybe C=A ? Anyway, do the back substitution and find the values of x,y,z in terms of B,C.
     
  10. Sep 11, 2009 #9
    I don't see the relevance as x and y would only incorporate multiples of z, which would be the same fraction and still be inconsistent for B = -3 and thereby be consistent for B not equal to -3?
     
  11. Sep 12, 2009 #10
    Yes, that's completely correct.
     
  12. Sep 13, 2009 #11
    After 5 hours of searching. I finally found an example I could understand a general solution in.

    [itex]
    z=\frac{B-C+3}{B+3} \iff z(B+3) - B + C - 3 = 0
    [/itex]

    With the equation above, it's easy to pick the values for B and C.

    b) No-solutions: B=-3 and C!=0
    c) One-solution: B!=-3
    d) Complete solution: B=-3 and C=0 ... (-4+5z, 1-2z, z)

    Is this completely wrong?
     
  13. Sep 13, 2009 #12
    That's correct again.

    For d) I would rather say infinite many solutions. :smile:
     
  14. Sep 13, 2009 #13
    Just using the problems own words... As it's written by my professor and I like high grades, that's what I useually do :P

    But thanks a lot... The feedback was just what I needed and much appreciated.
     
  15. Sep 13, 2009 #14
    You're welcome!:approve:
     
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