Need help with finding the flaw in the proof

Statement: For any n x n matrix A, if A is orthogonal, then det(A) = 1. Flawed Proof. First consider the case that A is 2 x 2. Let A = -b a ). Suppose that A is orthogonal. Then AT = A-. This means that AA = AA-1 = 1 = A-' A = ATA. Therefore, by taking the determinant and using that det (A) = det(AT) we get 1 = det ((6 1) ) = det(A) det ( A ") = det(A)?= (@) + 6)?. Since A is invertible det(A) = a + 63 > 0, and by taking square roots of the last equation, we get a2 + 62 = 1. Therefore, det(A) = a2 + b? = 1. For the general case when A is n x n, we can just take a b 0 O -b a 0 A = 0 1 0 0 with a b a in the top left corner and I's down the rest of the diagonal and the same argument works. This proves the statement