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TomLikesPhysics

  • 3 years ago

I need help with a matrix / Gauß.

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  1. TomLikesPhysics
    • 3 years ago
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    I have no clue how one can get from the left matrix to the right matrix. To be more specific - I have no clue what I have to do to get the top line in right matrix.

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  2. TomLikesPhysics
    • 3 years ago
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    I totally get how I can get the middle and bottom line in the right matrix, but the one on top just seems wrong (the ratio).

  3. TomLikesPhysics
    • 3 years ago
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    I just would like to know if the right matrix is right or if there must be a mistake.

  4. phi
    • 3 years ago
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    Yes, it looks like a typo. You could get rid of the fractions If you multiply the 2nd row (after it is 0 1 2 | 0 ) by 8/5 and add to the top row you can get 1 2 3 | 0 for the top row but that does not match their row.

  5. TomLikesPhysics
    • 3 years ago
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    I don´t know if this is important I am supposed to find the eigen-vector but still I think I can not do some trick to get that row, right?

  6. amistre64
    • 3 years ago
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    1 2/5 -1/5 0 18/5 36/5 ; 5/18 0 36/5 72/5 ; 5/36 1 2/5 -1/5 0 1 2 0 1 2 ; this row is a multiple of the other 1 2/5 -1/5 ; *5 0 1 2 0 0 0 5 2 -1 0 1 2 0 0 0 i think it cant be turned

  7. amistre64
    • 3 years ago
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    now if direction is what is important and not size, <5,0,0> = <1,0,0> but i cant verify that to be a good move

  8. phi
    • 3 years ago
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    I'm sure it is a typo (somewheres). But exactly what are you trying to do here?

  9. TomLikesPhysics
    • 3 years ago
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    Who are you talking to, phi? Me or amistre64? If you meant me. I was just trying to figure out if there is some way to rewrite the top line as it shown in the picture of if it is a typo. I thought it also might be a special case of operation that one can do because I am supposed to find the eigen-vectors with this matrix so perhaps there is someway to get that top line because you can do stuff with this eigen-stuff your normally can not do with a matrix.

  10. phi
    • 3 years ago
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    I meant Tom. The reason I asked is that row operations do not preserve the eigenvalues (it would be nice if they did, because the eigenvalues sit on the diagonal of a triangular matrix) however, once you get an eigenvalue, you subtract it from the diagonal of the matrix and do gaussian elimination to find its corresponding eigenvector.

  11. TomLikesPhysics
    • 3 years ago
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    That is what is happening here. The eigenvalue was 0 and I wanted to solve this matrix. The solution is the one above but because of the typo I was not sure I couldn´t follow it or if it simply was wrong. So, you are saying we can not multiply by five?

  12. phi
    • 3 years ago
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    I tend to reduce the matrix all the way to reduced row echelon form starting with the original matrix on the left I get 1 0 -1 0 1 2 0 0 0 now I read off the answer negate (-1 2) and append a 1 to get (1 -2 1) as the eigenvector

  13. phi
    • 3 years ago
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    I got this from Strang's course http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-7-solving-ax-0-pivot-variables-special-solutions/ and maybe the following lecture

  14. TomLikesPhysics
    • 3 years ago
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    Ok. Thanks a lot for your help phi. I saved my day. :)

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