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    I'm intrigued to know how this can be found. Would anyone care to enlighten me or provide a link to a method?

    Cheers :rolleyes:

    And by hand. I know you can do it with a calculator.
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    Row reduction.

    set up the matrix you want to invert (A)|(I) and work to get (I)|(A^-1)

    (There are other methods)
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    http://en.wikipedia.org/wiki/Gauss%E...an_elimination (the same as Nota's suggestion, but with some explanation)
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    Ah man 3 by 3 was enough for me! (Hey that rhymes!)
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    tbh you can use the same method that use used for 3x3 extended to 4x4 but that takes forevarrr.
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    (Original post by Totally Tom)
    tbh you can use the same method that use used for 3x3 extended to 4x4 but that takes forevarrr.
    Lol that was what I was thinking.

    Does Gauss-Jordan elimination still work?
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    The determinant of a square matrix,  A is always given by:

     A^{-1} = \frac{\text{adj}(A)}{\det(A)}
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    (Original post by Jamie Innes)
    Lol that was what I was thinking.

    Does Gauss-Jordan elimination still work?
    ?

    err yes.
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    Same as with a 3x3.
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    (Original post by bigmo7)
    Ah man 3 by 3 was enough for me! (Hey that rhymes!)
    Seconded. Although you do it the same way.
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    (Original post by Phugoid)
    The determinant of a square matrix,  A is always given by:

     A^{-1} = \frac{\text{adj}(A)}{\det(A)}
    You mean "inverse of a square matrix".
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    (Original post by DFranklin)
    You mean "inverse of a square matrix".
    Woopsies. Typo indeed.
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    (Original post by Phugoid)
    The determinant of a square matrix,  A is always given by:

     A^{-1} = \frac{\text{adj}(A)}{\det(A)}
    Sorry, what's the adj(A)?
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    The adjugate matrix. Finding it is about as difficult as finding the inverse matrix though, so I don't think that's very helpful...
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    (Original post by Zhen Lin)
    The adjugate matrix. Finding it is about as difficult as finding the inverse matrix though, so I don't think that's very helpful...
    Agreed. In fact, I've never seen someone advocate using the adjugate matrix unless they were going to then use the recursive formulation of the determinant to calculate each term. For large n, that's about as bad a plan as you can get: if memory serves you need roughly (n+1)! calculations to find the inverse that way.
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    Yeah, rref is the best way to do it already
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    http://www.intmath.com/Matrices-dete...rse-matrix.php have a look at this it shows how you can find the inverse of 4x4 5x5 and 6x6, finding 4x4 is basically the same as 3x3.
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    (Original post by lnx)
    http://www.intmath.com/Matrices-dete...rse-matrix.php have a look at this it shows how you can find the inverse of 4x4 5x5 and 6x6, finding 4x4 is basically the same as 3x3.
    I can't see the method they finally use for the 4x4 matrix, but if it involves determinants or adjoint/adjugate matrices, it's almost certainly grossly inefficient.
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    (Original post by DFranklin)
    I can't see the method they finally use for the 4x4 matrix, but if it involves determinants or adjoint/adjugate matrices, it's almost certainly grossly inefficient.
    Why?
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    See post #15. Basically, to find each element of the adjugate, you need to find a 3x3 determinant. So that's 16 3x3 determinants to calculate. Each 3x3 determinant requires 12 multiplies and 5 adds, so you need 192 multiplies and 80 additions.
    If you want a 5x5 it gets worse. To find each element of the adjugate you need to find a 4x4 determinant. That's 25 4x4 determinants. Each 4x4 determinant requires finding 4 3x3 determinants, so takes at least 48 multiplies and 20 adds. So you need well over a thousand multiplies to find the adjugate.
    6x6, you need something like 9000 multiplies.

    In contrast, Gauss-Jordan elimination will solve a 6x6 matrix in a couple of hundred multiplies. 4x4 matrix will be comfortably under a hundred ops.
 
 
 
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