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Regarding Determinants

  1. Jul 21, 2007 #1
    For an n x n matrix A, what is the relationship between det(A) and det(-A)?

    I tried it with a 1x1 matrix, and det (-A) = - det (A)
    I tried it with a 2x2 matrix, and det(A) = det(-A)
    I tried it with a 3x3 matrix, and the results were the same as that with a 1x1.

    This leads me to believe that for all odd n's, det(-A) = - det(A) and that for all even n's the two are the same.

    Is this the case? And if so, how would I show that in a more mathematical manner that just intuition?
  2. jcsd
  3. Jul 21, 2007 #2


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    Let A be an nxn matrix. What happens if we multiply a row/column of A with a scalar [itex]\lambda[/itex]? What happens if we do this to all the rows/columns? How does it affect the first relation?
  4. Jul 21, 2007 #3
    Why are we multiplying each individual row/column by a scalar?

    When we multiply the matrix by a scalar, doesn't this automatically distribute the scalar to every single entry of the matrix?

    I guess what I'm confused at is this:

    Suppose we let the square matrix B be obtained from matrix A by multiplying the matrix A by the scalar k. Then det(B) = k*det(A). Then if we let k = -1, as in my original question, then won't det(-A) = det(A) for all n?

    But obviously this isn't the case. What am I doing wrong?
  5. Jul 21, 2007 #4


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    Not so. You're thinking of the case when B is obtained by multiplying a single row of A by k.
  6. Jul 21, 2007 #5
    So since multiplying each row of a scalar by the constant gives us det(B) = k*det(A)...

    If we multiply each row of a n x n matrix by the same constant, i.e. n times, then the formula would be for the case of k = -1:

    det(B) = [(-1)^n]*det(A)

    Which would explain the difference for odd n, and the same for even n.

    Is this correct?
  7. Jul 21, 2007 #6


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    Yes, that's correct.

    There's an another way to see this if you know that det(A*B) = det(A)*det(B): If we let I be the nxn identity matrix, then det(-1*I) = (-1)^n, so that det(-A) = det(-1*I * A) = det(-1*I) det(A) = (-1)^n det(A). Of course this can be generalized to any k.
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