# determinants

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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

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

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