b(s)-Ia(s)=y2(川1()+V()]-I() 1.If L, (s)=BJ1(sy=IBkJY(s V,(s)+V(s) IBiki represent the strengths of current sources in branchi as a function of the element current in branch k. For one such source the matrix is zero;except for the element j,k.(the j, k-and the general branch have the same orientation, Pik->+) Let r(s)=BkYe(s)=BikJk(s It is a matrix of zeros except the jk-th element, whose value is BikJk(s) a(s)=Y(s川b(s)+V( 6(s)=[y2(s)+Y(s)Vb6(s)+V,(s)]_I,(s)I (s) I (s) b d • • − Y (s)[V (s) V (s)] I (s) e b s s • • • • = + − [ ]Y (s)[V (s) V (s)] j k e b s • • • =  + [ ]  jk represent the strengths of current sources in branch j as a function of the element current in branch k. For one such source the matrix is zero except for the element j,k. (the j,k-and the general branch have the same orientation, → +)  jk Let Y (s) [ ]Y (s) [ y (s)] =  j k e =  j k kk • •  It is a matrix of zeros except the jk-th element, whose value is y (s).  jk kk I (s) Y (s)[V (s) V (s)] d b s • • •  •  = + I (s) [Y (s) Y (s)][V (s) V (s)] I (s) b e b s s • • • •  • • = + + − 1. If I (s) [ ]I (s) d j k e • • = 