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    Prove:

    1) if A, B are square and AB=I_n, then det B=1/det A
    2) if A, B, S are square and B=(S^-1)AS, then det B=det A
    3) If A is square and (A^T)A=I_n, then det A= +/- 1
    4) if A is n x n and A^T= -A, then det A= (-1)^n det A. In particular, if n is odd, then det A=0

    © BrainMass Inc. brainmass.com December 24, 2021, 10:05 pm ad1c9bdddf
    https://brainmass.com/math/linear-algebra/proofs-determinants-441081

    SOLUTION This solution is FREE courtesy of BrainMass!

    Proof:
    (1) Since , then we have . Thus we must have

    (2) Since , then we have

    Therefore, we get .
    (3) Since , then . Thus we must have
    (4) Since , then we have

    So we have . Especially when is odd, we have

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 10:05 pm ad1c9bdddf>
    https://brainmass.com/math/linear-algebra/proofs-determinants-441081

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