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Math Help - Det(AB)=Det(A).Det(B)

  1. #1
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    Det(AB)=Det(A).Det(B)

    Can Someone prove this too?

    det(AB)=det(A).det(B)
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  2. #2
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    Re: Det(AB)=Det(A).Det(B)

    Thanks from davidciprut
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  3. #3
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    Re: Det(AB)=Det(A).Det(B)

    I have a question, I have uploaded a picture of the proof, and I understood everything, however there is something that isn't clear, (it's in Hebrew but I think you can understand what he did.) So in the last part he defined a new permutation sigma from 1 to k1, 2 to k2,..... n to kn, and then he changed the the expression and said that it's equal to detA. However the sign(sigma) (of permutation) is missing? Is there a mistake there? becauseif there is no sign maybe it will be equal to -det(A) at the end? Appreciate it if you can help me out. Thank you.
    Attached Thumbnails Attached Thumbnails Det(AB)=Det(A).Det(B)-okhakha.png  
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  4. #4
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    Re: Det(AB)=Det(A).Det(B)

    Couldn’t resist opportunity to promote Einstein summation convention to answer your question:

    Let A,B,C be matrices, and lAl be detA, etc.

    Theorem:
    If AB=C then lABl=lCl=lAllBl

    Little Preliminaries:
    1) lAl=eijka1ia2ja3k, definition of determinant
    2) eijkariasjatk=erstlAl, Permute rst starting with 123. Any transposition corresponds to a transposition of rows, which changes sign.

    Proof:
    AB=C, cij=aimbmj
    lABl=eijkc1ic2jc3k
    =eijk(a1rbri)(a2sbsj)(a3tbtk)
    =(eijkbribsjbtk)(a1ra2sa3t)
    =lBlersta1ra2sa3t, from 1) and 2)
    lABl=lBllAl=lAllBl=lCl

    One caveat, think of eijk… for larger matrices.

    note: repeated index is a dummy index: aibi =ajbj=a1b1+a2b2+a3b3+…
    also note bonus for determinant of matrix product.

    Better than wading through all the summation signs, isn't it?
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