解 01000 (A-2)X=X2,20101 (3) 0-2000 1-2000 于是得 40202 P=(,P2,B3,P4) P=X(3) =(X,X(2),X1,X2) 10 010 000 PAP 200 2 0201 210             → =             − 0 0 0 4 0 2 0 2 0 2 0 0 0 2 0 1 0 1 0 1 0 0 0 2 1 (3) X1 . 2 2 2 1 2 1 1             = − P AP ( 2 ) , (2) X X1 A− I = 解 . 0 2 0 1 2 0 0 0 0 1 0 0 1 0 2 0 1             − = (3) 1 2 3 4  = P X 3 1 (1) (2) (3) (1) 1 1 1 2 ( , , , ) ( , , , ) P P P P P X X X X = = 于是得