设n阶非零实数矩阵A满足A的伴随矩阵等于A的转置,试证A的行列式等于一,且A为正交矩阵

首先,当n > 1,关于伴随矩阵的秩,有如下结果:
若r(A) = n,则r(A*) = n;
若r(A) = n-1,则r(A*) = 1;
若r(A) < n-1,则r(A*) = 0.
证明:当r(A) = n,有A可逆,|A| ≠ 0.
于是由A*A = |A|·E可得A* = |A|·A^(-1)也可逆.
当r(A) = n-1,A有非零的n-1阶子式,故A* ≠ 0,r(A*) ≥ 1.
又A*A = |A|·E = 0,故r(A*)+r(A) ≤ r(A*A)+n = n,即得r(A*) = 1.
当r(A) < n-1,A的n-1阶子式全为0,故A* = 0,r(A*) = n.
回到原题,由条件A* = A'得r(A*) = r(A') = r(A).
当n > 2,根据前述结论,只有r(A) = n,故|A| ≠ 0.
对A*A = |A|·E取行列式得|A*|·|A| = |A|^n.
于是有|A|^2 = |A'|·|A| = |A*|·|A| = |A|^n,解得|A| = 1 (|A|为非零实数).
进而得A'A = A*A = E,即A为正交矩阵.
n = 1,2时是有反例的,例如A = 2E.谢谢亲。。。。