A2=H1A=00-3 0-33 (2)求H2,作A3=H242=R 2=sgn(a2)∑ e(2)定Sgn(O)=1) 2l1=0,2=a2+a2=3,l3=a3)=-3,=(0,3,-3) 4           = = 0 3 3 0 0 3 3 3 3 2 1 - - - - A H A (2)求 H2 ，作 A3 = H2 A2 = R 1 sign ( )( ) 3 ( sign (0) 1); 2 2 (2) 2 (2) 2 2 2 2 =  = = = 约定 i  a ai  ( ) 2 0, 3, 3, (0,3, 3) ; 2 2 3 3 2 (2) 1 2 2 2 T u = u = a + = u = a = − u = − 