Question

矩阵A=|2 1 0| 矩阵B满足ABA*=2BA*+E A*是A伴随矩阵 E为单位矩阵 求矩阵B |1 2 0| |0

Answer 1

|A| = 3.

由 ABA*=2BA*+E

等式两边右乘A得 ABA*A=2BA*A+A.

因为 A*A=|A|E=3E

所以 3AB = 6B+A

所以 (3A-6E)B = A

所以 B=(3A-6E)^-1A

3A-6E=

0 3 0

3 0 0

0 0 -3

(3A-6E)^-1 =

0 1/3 0

1/3 0 0

0 0 -1/3

B=(3A-6E)^-1A=

1/3 2/3 0

2/3 1/3 0

0 0 -1/3