0 000 5-30301 0 -350-10-3 000040 (A-41)= 101 (A-4) 0 -13010-1 000000 00000 1280-80-4 120408 (A-41)3 80408 00000 -80004 00000 求(A-41)3x=0(A-41)2x=0设x=[515253545 (4-4)x=0其通解为5,+5+5 0 0 000010 0 0 (4-41)2x=0其通解为引10 0 可取P3=[000010 P2=(4-4)23=[-1-1010- P1=(4-41)P2=[0040002 1 0 1 1 0 1 2 0 0 1 1 1 1 0 1 0 1 ( 4 ) 0 1 0 1 1 0 0 0 0 0 0 0 1 0 0 0 1 1 A I   − − −   − −     − − − − =   − −         − − 2 5 3 0 3 0 1 3 5 0 1 0 3 0 0 0 0 4 0 ( 4 ) 1 3 0 1 0 1 0 0 0 0 0 0 3 1 0 1 0 1 A I   −   − − −     − =   − −         − − 3 12 8 0 8 0 4 8 12 0 4 0 8 0 0 0 0 0 0 ( 4 ) 4 8 0 0 0 4 0 0 0 0 0 0 8 4 0 4 0 0 A I   − − −   −     − =   −         − 求 3 ( 4 ) 0 A I x − = 2 ( 4 ) 0 A I x − = 设  1 2 3 4 5 6  T x =       3 ( 4 ) 0 A I x − = 其通解为 1 3 5 1 0 0 1 0 0 0 1 0 1 0 0 0 0 1 1 0 0                                  + + −                         2 ( 4 ) 0 A I x − = 其通解为 1 3 1 0 1 0 0 1 1 0 0 0 1 0                       + −                 可取 P33＝0 0 0 0 1 0 T 32 33 ( 4 ) 1 1 0 1 0 1   T P A I P = − = − − − 31 32 ( 4 ) 0 0 4 0 0 0   T P A I P = − =