0100)(0100 20102010 X 0-2000000 000 4020)(0000 0 解(A-2D)X=X1 P=X(),P=X() 01001 20100 →)Ⅹ (2) 0-200-2 0100 X=0+C1和+C2X 0102 取C1=0cC2=2,即x(2)=1 P=X(2)9             =             −  =             →             − 1 0 0 0 , 0 2 0 1 0 0 0 0 0 0 0 0 2 0 1 0 0 1 0 0 4 0 2 0 0 2 0 0 2 0 1 0 0 1 0 0 (1) 2 (1) X1 X (1) 2 2 (1) 1 1 (2) 0 (1) 1 0 0 1 0 , 0 0 1 0 4 0 2 0 0 0 2 0 0 2 2 0 1 0 0 0 1 0 0 1 ( 2 ) X X C X C X A I X X + +             =             → =             − − 解 − =           = = = 2 0 1 0 0, 2, (2) 取C1 C2 即 X1 (1) (1) 1 1 4 2  = = P X P X , (2)  = P X 2 1