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Determinant 2

  1. Nov 8, 2009 #1
    If I have a matrix [tex]A[/tex], and I use [tex]n[/tex] different row operations of this form: [tex]a_kR_i + R_j \rightarrow R_i[/tex] to construct a new matrix [tex]B[/tex], what is the determinant of [tex]A[/tex] in terms of [tex]B[/tex]?



    Solved!

    [tex]|A|=|B|\prod^n_{k=1}\frac{1}{a_k}[/tex]
     
    Last edited: Nov 9, 2009
  2. jcsd
  3. Nov 8, 2009 #2
    Not entirely. Write B out in terms of the rows of A, then use the multilinearity of the determinant over the rows of the matrix.
     
  4. Nov 8, 2009 #3
    [tex]B_i = a_kA_i + A_j[/tex]



    I'm not sure what you mean when you say:
     
  5. Nov 8, 2009 #4
    If we write A as a list of rows: A1,...,Am, where the Ai is the ith row of the matrix A, we know that det(A1, ..., r*Ai+Aj, ..., Am) = r*det(A1, ..., Ai, ..., Am) + det(A1, ..., Aj, ..., Am) for all scalars r and each Ak. That is, the determinant is a linear operator with respect to each row; it is multilinear.
     
  6. Nov 9, 2009 #5
    Wow, after a week of looking for it, I found what I was doing wrong, and it turns out it was just a stupid mistake.

    The actual formula to the problem in my first post should be:

    [tex]|A|=|B|\prod^n_{k=1}\frac{1}{a_k}[/tex]

    I assume that's what you were trying to hint at slider142?
     
  7. Nov 9, 2009 #6
    Yep. Each scalar can be pulled out as a factor due to linearity, while the second determinant in the sum vanishes, so you end up with the product of each scalar multiplied by the original determinant.
     
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