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Math Help - Prove that det(A^2) = 1

  1. #1
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    Prove that det(A^2) = 1

    If A is row equivalent to I and I can be obtained from A by using only row operations of type I (interchange rows) and III (adding a multiple of one row to another), prove det(A^2) = 1.

    My attempt:

    If A is row equivalent to I, then A is the product of some elementary matrices. As a result, det(EA) = det(E)det(A) where E is some elementary matrix.

    *Assume det(A) is multiplied by the determinants of elementary matrices of type I. When this happens, that will change the value of det(A) by -1 since det(E) = -1 and det(EA) = det(E)det(A)

    *Similarly, assume det(A) is multiplied by the determinants of elementary matrices of type III. When this happens, det(A) is unchanged because det(E) = 1 and det(EA)=det(E)det(A)

    We can also consider the case with both row operations I and III, which again, will only change the det(A) by multiplying it by -1 or leaving it unchanged.

    A is row equivalent to I, so

    A = (E1)(E2) *** (Ek)(I)

    This implies that

    det(A) = det[(E1)(E2) *** (Ek)(I)]
    det(A) = det(E1)det[(E2) *** (Ek)(I)]
    ...
    det(A) = det(E1)det(E2) *** det(Ek)det(I)

    The determinant of the identity matrix I is 1 since the identity matrix is triangular, and the product of its diagonal elements is 1. Therefore, multiplying it by the determinants of elementary matrices I and III mean only two possible values of det(A) exist: 1 and -1. Squaring that in det(A^2) mean that det(A^2) = 1.

    Proof done.

    Note: The proofs for why det(E) = 1 or -1 where E is type I or III, respectively, are already in my textbook, so I can simply cite the proofs there. I inserted an asterik (*) at the start of those 2 statements just to make things clear for you as viewers.
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  2. #2
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    This argument is correct; well done! At the end you can say \text{det}(A^2)=[\text{det}(A)]^2 and so the minus sign doesn't matter.
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  3. #3
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    Thanks for the comment, I'll include that at the end of my proof.
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