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Math Help - Question on determinants

  1. #1
    Newbie Doktor_Faustus's Avatar
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    Question on determinants

    Hi guys,

    This may have been answered already.

    Let A be an nxn matrix and c a non zero scalar. If matrix B is obtained from A by multiplying the elements of a row by c then det (B) = cdet (A).

    Show that det (cA) = c^n det (A).

    Would appreciate any help whatsoever. I've looked through my book on linear algebra ( by Howard Anton...pretty good book too ), but I can't find anything besides it being stated without proof.

    Cheers.
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  2. #2
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    Prove It's Avatar
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    You should know \det{(AB)} = \det{(A)}\det{(B)}.


    In your case, you are wanting to find \det{(cA)}.

    Note that cA = cI_nA, which is now the product of two matrices, cI_n and A.

    So \det{(cA)} = \det{(cI_nA)}

     = \det{(cI_n)}\det{(A)}.


    Since cI_n is a diagonal matrix, its determinant is the product of the diagonal entries. In this case they are all c, so their product is c^n. Thus \det{(cI_n)} = c^n, and finally we have...


    \det{(cA)} = c^n\det{(A)}, which completes the proof.
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  3. #3
    Newbie Doktor_Faustus's Avatar
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    Quote Originally Posted by Prove It View Post
    You should know \det{(AB)} = \det{(A)}\det{(B)}.


    In your case, you are wanting to find \det{(cA)}.

    Note that cA = cI_nA, which is now the product of two matrices, cI_n and A.

    So \det{(cA)} = \det{(cI_nA)}

     = \det{(cI_n)}\det{(A)}.


    Since cI_n is a diagonal matrix, its determinant is the product of the diagonal entries. In this case they are all c, so their product is c^n. Thus \det{(cI_n)} = c^n, and finally we have...


    \det{(cA)} = c^n\det{(A)}, which completes the proof.
    Thank you. Much appreciated.
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  4. #4
    MHF Contributor

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    Quote Originally Posted by Doktor_Faustus View Post
    Hi guys,

    This may have been answered already.

    Let A be an nxn matrix and c a non zero scalar. If matrix B is obtained from A by multiplying the elements of a row by c then det (B) = cdet (A).
    If you are given this then you can prove det(cA)= c^n(A) by repeating it n times. Let B1 be the matrix obtained by multiplying the elements of row 1 by c: det(B1)= cdet(A). Let B2 be the matrix obtained by multiplying every element of row 2 of B1 by c: det(B2)= cdet(B1)= c^2 det(A). Repeat as needed!
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  5. #5
    Newbie Doktor_Faustus's Avatar
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    Thanks guys.

    Apologies for not having the thread in the correct forum.
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