Finding the inverse of matrices


by SomeRandomGuy
Tags: inverse, matrices
cronxeh
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#19
Mar16-05, 09:35 AM
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I tried coming up with general number of operations you'd perform with either method to determine which is faster, and I just dont have time right now..

matt, which one is faster - augmented matrix or adjoint/determinant method? Ive counted about 47 operations on adj/det method, and I'm guessing there is more in augmented method
ComputerGeek
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#20
Mar17-05, 09:03 PM
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Quote Quote by Hurkyl
One thing that's sure to work (though it might not be the easiest) is to compute the adjoint (I think that's the word used) -- the matrix whose (i,j)-th entry is the (i,j)-th subdeterminant. The identity A * adj A = (det A) I will yield the inverse.

Another approach is to use the chinese remainder theorem. Work out the inverse mod 13 and mod 2, then combine them to get the mod 26 answer.

All that said, you probably made an arithmetic error -- I don't think that last row is correct. (Because if that 22 is right, all of the terms on the RHS should be even)
the second option might be best because then every element on mod 2 and mod 13 will be a unit and you can find an inverse for every number. mod 26 is harder because not every element will have an inverse and you may need to do some careful row switches.
ComputerGeek
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#21
Mar17-05, 09:06 PM
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Quote Quote by Hurkyl
Yep, but then again, the matrix is invertible if and only if the determinant is invertible.
don't you mean not equal to zero?
Data
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#22
Mar17-05, 11:59 PM
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Not equal to zero and invertible mean the same thing for elements of a field~
matt grime
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#23
Mar18-05, 04:23 AM
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But we aren't dealing with a field. Invertible is indeed the correct term.
Hurkyl
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#24
Mar18-05, 05:49 AM
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For example, an integer matrix has an integer inverse if and only if its determinant is 1 or -1 --- because those are the only invertible integers!
ComputerGeek
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#25
Mar18-05, 12:11 PM
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ah... that makes sense.


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