求所有与A可交换的矩阵,只能用待定系数法么?比如说A=1 1 0 0 1 1...答:设 B = [ b11, b12, b13][ b21, b22, b23][ b31, b32, b33]且 AB=BA, 即 AB-BA=0.而 AB-BA = [ b21, b22 - b11, b23 - b12][ b31, b32 - b21, b33 - b22][ 0, -b31, -b32]所以 b11=b22=b33, b12=b23, b21=b31=b32=0 故 B = a b c 0 ...