Magister said:
But is the CKM matrix in the the B^0 bar-B^0 system the same as in the k^0 bar-k^0 system?
Yes!
I'll try to explain why your question doesn't make so much sense :)
When you consider an elementary interaction involving a quark q, another quark q', and a W, in the calculation of the amplitude you will have to multiply for g (the weak interaction coupling, which is *universal*, i.e. does not depend on the process), and a factor which depends on the quarks, let's call it Vqq'. This factor is the element in the q-th row and the q'-th column of the CKM matrix.
The K^0 is composed of a strange and a down quark (one is a quark and the other an antiquark, and which is which depends on whether it is K^0 or \bar K^0). So the coefficients of CKM involved will be Vus, Vcs,Vts and Vud, Vcd,Vtd (because the weak interaction can couple the "low" quarks only to "high" quarks, I mean that d,s,b can only couple to u,c,t and not to other d,s,b).
In the B^0 the quark composition is one bottom and one down quark, in the B^0_s the composition is one bottom and one strange. So the coefficients will be different coefficients in general (Vub, Vcb, etc.) but the matrix is always the same good CKM matrix.