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Math Help - det(A)=-det(B)

  1. #1
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    det(A)=-det(B)

    We have matrix A in the order of (n x n) , and we have matrix B that can be derived from A by adding to the row j the multiplication of row i with a scalar.
    Then det(B)=-det(A)
    I tried to prove it with the multilinear property but I am stuck... please help. thank you
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  2. #2
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    Re: det(A)=-det(B)

    http://www.maths.nuigalway.ie/~rquin...section2-5.pdf

    has everything you could want to know about row operations and determinants
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  3. #3
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    Re: det(A)=-det(B)

    Again, what you are trying to prove IS NOT TRUE. What IS true is that det(A) = det(B). For convenience we will assume j < i (if i < j, apply the matrix that switches rows i and j first, see note below).

    Let P be the matrix:

    P = \begin{bmatrix}1&\cdots&0&\cdots&0&\cdots&0\\ \vdots&\ddots&\vdots&{}&\vdots&{}&\vdots\\0&\cdots  &1&\cdots&r&\cdots&0\\ \vdots&{}&\vdots&\ddots&\vdots&{}&\vdots\\0&\cdots  &0&\cdots&1&\cdots&0\\ \vdots&{}&\vdots&{}&\vdots&\ddots&\vdots\\0&\cdots  &0&\cdots&0&\cdots&1 \end{bmatrix}

    then PA has the effect of adding r times row i to row j.

    It follows that det(B) = det(PA) = det(P)det(A). Expanding by successive minors along the first column shows that det(P) = 1, so:

    det(B) = det(A).

    Or, regarding det as a multilinear function of the rows of a matrix, we have:

    det(B) = det(B1,...,Bj,...,Bi,...,Bn) = det(A1,...,Aj,...rAj+Ai,...,An)

    = det(A1,...,Aj,...rAj,...,An) + det(A1,...,Aj,...,Ai,...,An)

    = r(det(A1,...,Aj,...Aj,...,An) + det(A)

    = r(0) + det(A) = det(A) (since the first term is a determinant of a matrix with 2 equal rows).

    Note (from above):

    If i < j, then switching the rows multiplies the determinant by -1 (the determinant is an ALTERNATING multilinear function). Then adding r times row j (which *was* row i) to row i ( which *was* row j, before) does not change the determinant, as above.

    Now we have our original row i times r plus our original row j currently in the i-th row, and our original row i in the j-th position. Switch them again, which again multiplies the determinant times -1.

    This gives us our original row i times r plus our original row j in the j-th position, as desired, and puts our original row i back in the i-th position. Thus:

    det(B) = (-1)(det(A))(-1) = det(A).

    (Or: you could simply use the matrix:

    Q = \begin{bmatrix}1&\cdots&0&\cdots&0&\cdots&0\\ \vdots&\ddots&\vdots&{}&\vdots&{}&\vdots\\0&\cdots  &1&\cdots&0&\cdots&0\\ \vdots&{}&\vdots&\ddots&\vdots&{}&\vdots\\0&\cdots  &r&\cdots&1&\cdots&0\\ \vdots&{}&\vdots&{}&\vdots&\ddots&\vdots\\0&\cdots  &0&\cdots&0&\cdots&1 \end{bmatrix}

    instead of P).
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