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    If you are using this method on a 4x4 matrix and it works out so you get all zeros on the bottom row can you still solve it?

    i.e
    x1-x2-x3-x4 = y1
    x1-x2-x3-x4 = y1
    x1-x2-x3-x4 = y1


    Also, if you start using similtaneous equations is it wrong. Why?
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    Well, this happens when the matrix does not have full rank, so, either you have no solutions, or you have an infinite number; in the case where you have one row of zeros, you will have one free variable in your solution set.
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    Im still unsure whether there are any answers or not the qu is....

    z1 -4z2 +z4 = 12
    -3z2 +3z4 = 6
    z2 -2z3 +3z4 = 24
    2z1 -6z2 -4z3 +8z4 = 72

    Can anyone have I go at solving it?
    Thanks
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    (Original post by geo3)
    Im still unsure whether there are any answers or not the qu is....

    z1 -4z2 +z4 = 12
    -3z2 +3z4 = 6
    z2 -2z3 +3z4 = 24
    2z1 -6z2 -4z3 +8z4 = 72

    Can anyone have I go at solving it?
    Thanks
    You might like to play with this applet.

    http://www25.brinkster.com/denshade/...imination.html
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    ???
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    help???
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    (Original post by geo3)
    help???
    Did you not understand how to use the applet I gave you a link for.

    Enter 4 in the rows/columns box and then enter the coefficients with the answers in the last column. Then hit calculate. You get a complete row of zeroes at the bottom. The implications of this were explained by another poster.
 
 
 
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