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HM am i not allowed to use this method? Finding the determinant vs row reduce

  1. Oct 28, 2005 #1
    Hello everyone, I would rather find the cofactors and find the determinant than row reducing this, but is it possible, its not square! But our teacher is acting like its possible, so it must be! here is the equations:
    x+y+z = 4
    3y-z = 1

    so i got:
    1 1 1 4
    2 -1 4 9
    0 3 -z 1

    a 3x4! but is there anyway for me to solve the following system other then row reduction or augmenting it with the idenity matrix? Thanks!
  2. jcsd
  3. Oct 28, 2005 #2


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    Homework Helper

    Sure there is. Use Cramer's Rule, where's a link:

  4. Oct 28, 2005 #3


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    Science Advisor

    It's not square because you are using the "augmented" matrix.

    The matrix representing the coefficients is square- its 3 by 3. The augmented matrix is not square because it has the right hand side of the equations added as a 4th column.
  5. Oct 29, 2005 #4
    So if i don't augment it, will i still get the correct value if i use cofactor expansion? what do i with the other set of vectors t hough? the vectors that are = [4 9 1]^T
  6. Oct 29, 2005 #5


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    Have you understood how Cramer's rule works?
    You always work with the coefficient matrix. To find the n-th unkown, you replace the n-th column in the coefficient matrix ("A") by the column of the constants ("B") and you take its determinant and divide it by det(A). You do this for each unkown. Realise that this only works for non-singular (so regular) matrices A, since det(A) can't be 0.
  7. Oct 29, 2005 #6
    Ohh, I had no idea thats what cramer's rule really ment. Thank you! I'll see if I can figure these out and get the right values. When you say its gota be singular, what would be a case when i couldn't apply the determinant?
    Last edited: Oct 29, 2005
  8. Oct 30, 2005 #7
    awesome, that works, but i just found out that row reduction is the fastest, unless its going to be a big one!
  9. Oct 30, 2005 #8


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    Gold Member

    Cramer's rule is generally not of computational interest... however, it is good to know because it can be pretty handy for proving theorems.
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