You'll note that we found a lot of information by looking at each of the four terms in the conjugate matrix one by one. You might wonder why not the system matrix. And there is generally more power in the conjugate matrix when you want to, for example, put the system in focus because the conjugate matrix talks about conjugate plans. It knows about object and image distances. However, the system matrix is a useful thing, partly because it explicitly does not deal with the object and image distances and really just talks about the glass, the system itself. And so here we'll go through one example of that. And show explicitly how the two relate because sometimes that relationship is useful. So let's imagine we have a black box optical system defined by a system matrix M. And remember that is bounded by two refraction matrices and deals with all transfers in between, leaving off only the transfer matrices on the other end. So therefore, the relationship between that system matrix and the conjugate matrix, which does include the object and image distances, is simply the two extra transfers from 0 to the end of the system and from the last surface, glass surface K to surface K+1. So let's do that. Let's take and just assume we've been given or we've calculated M. And those are the four terms of M, A, B, C, D. And let's multiply the two transfer matrices in both sides. Notice, I've written the 0 transfer matrix with a distance minus d1. I could have done it with a positive d0 prime. But this is kind of convenient, it's in keeping with how we think about, for example, d1 systems where we measure backwards from the glass to the object. So it's equivalent but just seem like a better way to do it. So little bit of algebra, and we end up with a matrix N now, conjugate matrix, in terms of the system matrix M. So why would you care? We know something about the matrix M, in particular, if we were to set this one term here N12 equal to 0. That enforces the conjugate condition, it puts the system in focus. So let's do that, let's just take that term and set it equal to 0 and solve for dk prime, the image distance. So what we have now is a condition that depends on d1, the object distance, and the four terms of the system matrix M. So the point is, is now if you give me a system matrix N, I can write this down and it gives me a relationship between object and image distances. And these are the kinds of tools we need when we're designing optical systems because we need to put constraints on them like if they're in focus, we know where the image is. Just as a check, we can put in the A, B, C, D matrix for a single thin lens. Pause for a minute and think exactly what that matrix is, because it's quite simple. And if we do that, I put the various terms in here, rearrange slightly and out pops the Gaussian thin lines equation which is just what we'd expect. So this is the equivalent of the Gaussian thin lens equation, but now for a thick lens or a system of lenses described by a two by two element matrix.