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Math Help - [SOLVED] Solving a matrix

  1. #1
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    [SOLVED] Solving a matrix

    Augmented matrix in row echelon form:

    x1|x2|x3|b
    1 -3 7 1
    0 1 4 0
    0 0 0 1

    x1= 1 + 3x2 - 7x3
    x2= - 4x3
    x3= 1

    Is this the solution set?


    Matrix #2

    1 0 8 -5 6
    0 1 4 -9 3
    0 0 1 1 2

    x1 = 6 -8x3 + 5x4
    x2 = 3 - 4x3 + 9x4
    x3 = 2 -x4
    x4 = ?

    I have trouble with this one especially.

    Edit: Okay I think I got matrix #2. x4 is a free variable and we can assign it the arbitrary value "t". In this case the solution set is:

    x1 = 6 -8x3 + 5t
    x2 = 3 - 4x3 + 9t
    x3 = 2 - t
    x4 = t

    Yes?
    Last edited by thekrown; February 12th 2010 at 10:26 AM.
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  2. #2
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    If that first matrix is written correctly, it does not say x3 = 1. It says 0=1.

    As for the second matrix, the system of equations can be parameterized with x4, and x4 can be used to describe ALL other leading variables; yet you have x3 as a seemingly "free variable" in the equations.
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  3. #3
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    What do you mean by 0=1 for the first matrix?

    For second matrix, so if I assign an arbitrary value to x4 I can then get these solutions:

    x1= 6 - 8(2 - t) + 5t
    x2= 3 - 4(2 - t) + 9t
    x3= 2 - t
    x4= t
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  4. #4
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    In the first matrix you have, in row three, you have managed to say that 0 = 1, which shows that there is no solution to your matrix. You however are saying x3=1, which is not the case (unless you have written the problem wrong).
    0 0 0| 1 literally means "0x+0y+0z=1. Of course this is impossible. I would recheck your work to make sure you did your row reductions properly.

    In the second matrix, making x4 equal to t is kind of redundant. I mean if it makes it easier to keep track of x4, sure, but theres no need to assign a different variable to x4. As for your solutions, yes that would be a solution. Do they want it in this form, or are you actually meant to describe the solution set?
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  5. #5
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    Matrix 1: Yep, I guess it is a trick question then if there are no solutions and they are asking us to find the set.

    Matrix 2: They always want us to assign arbitrary values but I understand what you mean.

    Well, if matrix 1 has no solution (and I can understand this now) and my solution set for matrix 2 is correct then I have no further questions. Thank you.
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