设B = b1 b2b3 b4若 AB=BA, 则有b1+b3 b2+b4 b3 b4=b1 b2+b1b3 b4+b3所以有b1+b3 = b1b2+b4 = b2+b1b4 = b4+b3解得: b3=0, b1=b4所以,所有与A可交换的矩阵为a b0 a满意请采纳 有问题请消息我或追问