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Cramer's rule

  1. Apr 15, 2005 #1
    It won't work and I don't see what I'm doing wrong.

    Find the solutions of the following system of linear equations:

    x + 3y - z = 1
    2x + y + 2z = 3
    5x + z = 2

    I put these into the form Ax = b, where

    A = (1 3 -1)
    (2 1 2 )
    (5 0 1 )

    x = (x)
    (y)
    (z)

    b = (1)
    (3)
    (2)

    I worked out det A = 20.

    Cramer's rule says the solutions are given by:

    x = (1/det A) | 1 3 -1 | => x = 1/10
    | 3 1 2 |
    | 2 0 1 |

    y = (1/det A) | 1 1 -1 | => y = 9/10
    | 2 3 2 |
    | 5 2 1 |

    z = (1/det A) | 1 3 1 | => z = 11/10
    | 2 1 3 |
    | 5 0 2 |

    These solutions are wrong, where have I gone wrong?? Grr.

    When I work out the answers algebraically, I get x = 1/5, y = 3/5 and z = 1. These are correct.
     
  2. jcsd
  3. Apr 15, 2005 #2
    det A is 30 not 20

    marlon
     
  4. Apr 16, 2005 #3
    If you are not limited to using Cramer's Rule, I always find that just computing the cofactors of the matrix is much easier... and also a good way to check if your Cramer's Rule method is correct.
     
  5. Apr 17, 2005 #4
    ARGH, thank you.

    I didn't understand cofactors :(.
     
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