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Matrices, I can't seem to cancel out any more elements

  1. Sep 13, 2005 #1
    Hello everyone, every row operation I make now is taking out the 0's i just made, any ideas on what I can do from here?
    Here is my work:
    Click

    Also I was confused on how this works, my professor did an example:
    3x+2y+z = 8;
    x = a
    y = b
    z = 8 - 3a - 2b
    I get this part but then he does the following:

    Here

    Where is he getting those values? The first one looks like he let a = 0 and b = 0, then ur left with just 8, but i'm not sure.
    Thanks.
     
    Last edited: Sep 13, 2005
  2. jcsd
  3. Sep 13, 2005 #2
    You have 3 equations and 5 unknowns, so you will have 2 free variables. You can't reduce it any more than that.
     
  4. Sep 13, 2005 #3
    hm...What do i do from there then? How can you solve the system?
     
  5. Sep 13, 2005 #4
    He's just writing it as a sum of vectors multiplied by some coefficient. For example,

    3+x 3 1 0
    x^2 = 0 *1 + 0 *x + 1 *x^2
    X^2-4x+7 7 -4 1

    or if you write it out as 3 separate equations, which it actually is,

    3 + x = 3*1 + 1*x + 0*x^2
    x^2 = 0*1 + 0*x + 1*x^2
    x^2-4x+7 = 7*1 + -4*x + 1*x^2
     
  6. Sep 13, 2005 #5
    Why does it look like all he is doing is letting a and b equal 0 in the first one, which will make 0 0 8 then he's letting a = 1, and b = 0, and finding the values then, letting a = 0 and b = 1, and then finding the values?
     
  7. Sep 13, 2005 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Because that is what he is doing! There are 3 equations in 5 unknowns. You can solve for 3 of the unknowns (x,y,z) in terms of the other 2 but those 2 (a and b) can be anything. To write a general formula your teacher is saying, "First,suppose a= 1, b= 0. The x,y,z= something written as a vector v1. Now, suppose a= 0, b=1. Okay, then x,y,z= something else written as a vector v2. " The general formula is then av1+ bv2. That's a linear combination that obviously gives the correct answer when a=1, b=0 and when a=0, b=1 and that's enough to show it is the general solution to this linear problem.

    (Oh, the bit about letting a= 0, b= 0 is because this is not a "homogeneous" problem.)
     
  8. Sep 13, 2005 #7
    Well if that is infact whats going on, then how is this possible?>
    3x+2y+z = 8;
    x = a
    y = b
    z = 8 - 3a - 2b

    Okay he let a = 0, b = 0, so z = 8
    a = 1, b = 0, he has z = -3
    z = 8 -3(1) - 2(0) = 5?
    then he has
    a = 0; b = 1; z = -2
    z = 8 -3(0) -2(1) = 6?
    but he has -2.
     
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