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chrisplusian

  • one year ago

Linear algebra question dealing with transpose

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  1. chrisplusian
    • one year ago
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    I can't figure out how to input matrices here but \[b = \left[\begin{matrix}1 \\ 1 \\ -1 \end{matrix}\right]\]

  2. chrisplusian
    • one year ago
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    i figured it out..... find \[b^{T}b\]

  3. chrisplusian
    • one year ago
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    what I got was a three by three \[\left[\begin{matrix}1 & 1 & -1\\ 1 & 1 & -1\\ -1 & -1 & 1\end{matrix}\right]\]

  4. chrisplusian
    • one year ago
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    just wanted to check and see if this is correct

  5. dan815
    • one year ago
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    it should say b*b^T if u got that

  6. dan815
    • one year ago
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    3 by 1 matrix * 1by 3 matrix = 3 by 3 matrix, ^--------^---- these inner dimensions have to be equal, that is a way to check if solution exists

  7. dan815
    • one year ago
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    rows by columns

  8. chrisplusian
    • one year ago
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    Sorry that is correct it should have said b*b^T

  9. chrisplusian
    • one year ago
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    would that be correct for b*b^T ?

  10. dan815
    • one year ago
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    yes its right

  11. chrisplusian
    • one year ago
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    thank you

  12. dan815
    • one year ago
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    okay but btw

  13. dan815
    • one year ago
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    if u do b^t * b you will get 3

  14. chrisplusian
    • one year ago
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    Yeah I actually did that problem first, for some reason I was second guessing myself on this part though

  15. dan815
    • one year ago
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    ah okay

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