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HELP NEEDED - Matrices

  1. Feb 10, 2005 #1
    I know how to determine the determinant, and the inverse, and how to sovle a system with the inverse of a matrice, but I have no idea what these two questions are talking about:
    1. Find the values of [tex]k[/tex] for which the system
    [tex]kx+2y=1[/tex]
    [tex]3x+(k-1)y=1[/tex]
    does not have a unique solution. If [tex]k[/tex] does not have these values, find the unique solution. For each value of [tex]k[/tex] for which no unique solution exists, determine whether or not any solution of the system exists.

    what is this question asking me for? What is a unique solution?

    2. Find the values of [tex]k[/tex] for which the system
    [tex]x+2y+2z=1[/tex]
    [tex]2x+y+2z=4[/tex]
    [tex]3x+3y+kz=5[/tex]
    has a unique solution. Find this unique solution and solve the system for any values of [tex]k[/tex] for which the unique solution does not exist.

    again, wtf?
     
  2. jcsd
  3. Feb 10, 2005 #2

    dextercioby

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    What's the condition for a nonhomogenous system to have unique solution?

    Daniel.
     
  4. Feb 10, 2005 #3
    I do not know what a unique solution is, nor have I ever heard the term nonhomogenous.
     
  5. Feb 10, 2005 #4

    dextercioby

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    Then my advice is to study theory first and then get busy (trying to) solve(ing) problems...

    Daniel.
     
  6. Feb 10, 2005 #5
    this is not some independant study topic. This is from a practice booklet the teacher gave us, a question which the tacher recomnded we do. So, my current knowledge should suffice in answering this question.
    What I know: -determinants
    -inverses
    -simple matrix operations
    -solving systems with matrix inverses
    -there being no solution if a system forms a singular matrix
     
  7. Feb 10, 2005 #6

    dextercioby

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    The IV-th item of your "list" is the key.

    Yes,setting the determinant of coefficients to zero will ensure nonuniqueness of solutions.

    Daniel.
     
  8. Feb 10, 2005 #7
    For the 2nd question, if nonuniqueness equates to there being no solution, the that is easy, k = 4, but for a unique solution, could not that be anything but for in the field of k?
     
  9. Feb 10, 2005 #8

    dextercioby

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    I'm sorry,i really didn't understand your question.Could you please rephrase it using other words...?

    Daniel.
     
  10. Feb 10, 2005 #9
    I really dont know what a unique solution is. One of the answers it gives in the answers section is k cannot equal 4, which is the restriction should there be a solution to the system, if k = 4 detA = 0. So if k=4 means the system has no unique solutions, what does it mean if it does have unique solutions? How could you figure that out? Would not any value of k other than 4 bring about a system with unique solutions?
     
  11. Feb 10, 2005 #10

    dextercioby

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    Yes,any value different than 4 would make unique solution.

    Daniel.
     
  12. Feb 11, 2005 #11

    HallsofIvy

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    I am sorely tempted to ask "Do you know what a "dictionary" is"???
    (but I won't).

    "Solution" a value (in this case a vector or set of numbers) that satisfies the equation.

    "unique" only one.

    You don't actually need to know anything about matrices to answer these questions.

    kx+ 2y= 1
    3x+ (k-1)y= 1

    Try to solve the pair of equations. For what values of k can you NOT get a single (unique) solution? If you can get a single solution what is it?

    Yes, you could write this as a matrix equation. You can solve for a single (unique) result as long as you can take the inverse of the matrix- as long as the determinant is not 0. What value of k makes the determinant 0?
     
  13. Feb 11, 2005 #12

    xanthym

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    Also keep in mind that for a system of Linear Equations, there are only 3 possibilities:
    1) No Solutions
    2) 1 Solution (The "Unique" Solution)
    3) Infinite Number of Solutions

    ~~
     
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