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Matrix one-parameter solution help

  1. Dec 24, 2006 #1
    hello people
    i was wondering if anyone could help me with this question:

    For what values of a and b does the system

    ax +bz = 2
    ax +ay +4z = 4
    ay + 2z = b

    have: (i) no solution
    (ii) a unique solution
    (iii) a one-parameter solution
    (iv) a two-parameter solution?

    this is what i did, i wrote it in a matrix form and i did gussian elimination, and across the diagonal i got a, a, and 1
    i multiplied the digits on he main diagonal and i got a^2, i made that equal to 0... then i made this post
    (basically i'm just doing random things with the numbers and i'm so stuck i don't know what to do! :frown: )

    any tips would be really appreciated
    cheers in advance!

    (tomorrow is christmas :biggrin: )
     
  2. jcsd
  3. Dec 24, 2006 #2

    arildno

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    Subtract the first from the second equation, and get:
    ay+(4-b)z=2 (2*)

    Subtract (2*) from your third equation, and get:
    (b-2)z=(b-2)

    Thus, you have the following system of equations to work with:
    ax+bz=2
    ay+(4-b)z=2
    (b-2)z=(b-2)
     
  4. Dec 24, 2006 #3
    Thus, you have the following system of equations to work with:
    ax+bz=2
    ay+(4-b)z=2
    (b-2)z=(b-2) (3*)

    then i divided 3* by (b-2)

    and i got
    ax+bz=2
    ay+(4-b)z=2
    z=1

    to work with...should i not do that last step?

    and (i think you know this but) to get a no solution or a unique solution etc. we should use the determinant of the matrix which is the multiplication of the numbers on the main diagonal...i bet i'm talking rubbish...then we make the multiplication of the numbers on the main daigonal equal to zero, then i get stuck for real!

    i just don't know what to do! lol
     
  5. Dec 24, 2006 #4

    matt grime

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    You get 0,1 or infinitely many solutions if there are, well, 0, 1 or infinitely many solutions. So have a look at the solution set and let's see if when we can have 0,1 or infinitely many solutions.

    We know (b-2)z=b-2. Now, you divided by b-2. That's fine *when it's allowed*. When is it allowed?
     
  6. Dec 24, 2006 #5
    When is it allowed?
    'clueless'
     
  7. Dec 24, 2006 #6

    cristo

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    Suppose we have the equation ax=4. Then this gives the solution x=4/a iff a is not equal to zero. Does this help?
     
  8. Dec 24, 2006 #7

    arildno

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    sara:
    It is perfectly true that if the determinant of a quadratic matrix is non-zero, then we'll have a unique solution of the associated system of equations.

    Thus, the "no solution" or "infinite solution" cases will have the value of the determinant equal to zero.
    But do you see the problem here?

    The problem is that if we just set the determinant equal to zero, gaining a relation between a and b, we haven't really DISTINGUISHED between the "no solution"/"infinite number of solutions" cases!!
    But to make that distinction is precisely one of your tasks..
    Thus, just setting the determinant equal to zero is insufficient to answer your particular problem.
     
  9. Dec 24, 2006 #8
    thanx all for your help!!
    i really really appreciated it, i think i know what to do now...you know i went on this other site to ask a question and by the time i got a response i had worked out the answer my self!
    cheers!
     
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