对于给定的n阶矩阵A,如果存在n阶矩阵B,使得AB=BA,则称B与A可交换.对于下列矩阵A,试分别求出所有与A可交换的矩阵.
A=
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设与A可交换的矩阵X= (x11 x12 x13 x21 x22 x23 x31 x32 x33), 则 AX= (0 1 0 0 0 1 0 0 0) (x11 x12 x13 x21 x22 x23 x31 x32 x33) =( x21 x22 x23 x31 x32 x33) 0 0 0), XA= (x11 x12 x13 x21 x22 x23 x31 x32 x33), (0 1 0 0 0 1 0 0 0) = (0 x11 x12 0 x21 x22 0 x31 x33) 因为X与A可交换, 则AX=XA,有 (x21 x22 x23 x31 x32 x33 0 0 0) = (0 x11 x12 0 x21 x22 0 x31 x32) 即x11=x22=x33,x12=x23,x21=0,x31=0,x32=0, 令x11=x22=x33=α,x12=x23=b,x13=c, 从而所有与A可交换的矩阵形如 (α b c 0 α b 0 0 α) ,其中α,b,c为任意常数.