All right. Let's now write this out entirely in terms of matrices, getting rid of the summation over elements. All right. Okay. In order to do that, what we get is our big vector, c2, c3, c4, c5, c6, so on until we come to the last elements, c2nel -1, c2nel, c2nel + 1. Okay? Now, in order to write this, allow me to assume that every element, or allow me to consider the case where every element has the same length. Okay? So if h e = h for all e. Okay? That's the case I'm considering. It just makes it a little more, makes what I have to write a little less cumbersome. Right? In particular, what it does is that it allows us to write here 2EA over he. Okay? Sorry, over h, like we just agreed. Right? Now, we are going to write out our big matrix. Giving myself plenty of room. Okay. For the very first element, we have contributions only from c2, c3. Right? And if you go back and look at what we had on the previous slide, I need to just pull it up for a second. Right? Okay, look at the contributions with c2 and c3 on the left-hand side here. Okay? We get here -4 over 3, 8 over 3, -4 over 3. And here, we get 1 over 6, -4 over 3, 7 over 6. Okay? We move on to the next element. Element two has contributions from c3, c4, c5, right? Or alternatively, from global nodes 3, 4, 5. Global degrees of freedom 3, 4, 5. Okay? It is going to get slotted into these positions here. Okay? Right. That's where the contribution from c3, c4, c5 will come. All right? Now, the contributions on the side of columns will come in here, right, which will come all the way down here. It can come all the way up to this column and to the next two. Okay? So what that means is that when we come here, we add on 7 over 6, right, from the one-one component of the stiffness matrix of element two. Because I need a little more room here, let me move these arrows across the middle. Okay? So you get 7 over 6, and then, the rest of this difference matrix of element two just fills itself here. Minus four-thirds, 1 over 6. Right? We get here -4 over 3, 8 over 3, -4 over 3. And here we get 1 over 6, we get -4 over 3. And here we get 7 over 6. Okay? The important thing is that is where the contribution from the common node or the common degree of freedom of elements one and elements two comes apart, and that is from that degree of freedom. Okay? The process continues, of course, until we arrive at our very last element. Okay? And for it, we observed that it's going to form this little block in the very bottom right of our matrix. Also, observe that as we go onto element three and so on, we are going to add some more terms here. Right? That's where terms are going to get added, right? That's where components are going to get added into this difference matrix. When we come down here to our very last element in the end, we see that it will give us a contribution 7 over 6. But that contribution itself will be added onto a contribution that was already put in there from element nel-1. Okay? So we'll get 7/6 here, -4 over 3, 1 over 6, -4 over 3, 8 over 3, -4 over 3. And here we get 1 over 6, -4 over 3, and 7 over 6. Okay? The d vector here is full, right? It starts out at d1, d2, d3, d4, d5. And d5 is the last trial solution degree of freedom for element two. Okay? We carry on until we come here to d2nel-1, d2nel, and d2nel+1, and all of this then get's closed. Okay? So that is our stiffness matrix. Right? Okay. Those are the contributions from the left-hand side of the waveform. All of this now equal to, on the right-hand side, again, our c vector just as before, c2, c3, c4, c5. We go on until we come to c2nel-1, c2nel, and c2nel+1. Right? Now, there are two types of terms which need to go in here, right? The contributions from the forcing function, as well as the contributions from the traction. Okay? So, I will open a parenthesis here. Since I've allowed us to assume that we have a fixed element length, we don't have to use he, we use h instead. Okay? Right. Which is uniform for all the elements. It's the same element length for each element. So, we get a contribution fah over 2, and now we get our vector from the forcing function. For element one, because those are the only degrees of freedom from the waiting function, we go back and look the way we wrote out our vector from the forcing function. We get four tones. Okay? We get 1 over 3 from element one. We come to element two, and those are the contributions. Okay? From the waiting function degrees of freedom. All right? And if we go back and look at how we wrote the sum of all elements of the contributions from the forcing function, we see that we do get indeed a one-third added here, we get four-thirds from the c4 degree of freedom, and we get a 1 over 3, but this one over 3 will then be joined by another one-third from element three. Right? And this process continues until we come all the way down to the contributions from the last element. Okay? The contributions from the last element will have the first node from the last element. It will give us a contribution of one code where the first degree freedom will, but it will be added to an existing contribution from element nel-1. Second degree of freedom gives us four-thirds, and the very last degree of freedom gives us one-third. Okay? This vector then contains the contributions from the forcing function. For the traction, remember, if you go back and look at the way we wrote out the traction vector contribution down here, it's the very last term on the slide. The only contribution comes c2nel+1. Okay? Since we have the entire c vector sitting here, what that tells us is that in order to write the traction contribution as a vector, also, we fill it with 0s all the way down, except for the very last component, which is tA. All right? And we need to close the parenthesis. Okay. Now, if you go back and look at the matrix that we have on the left-hand side here, we sum over, we add up the terms that needed to be added, likewise, we add up the terms that needed to be added here, we end up with a matrix vector problem. Okay? One thing I want you to note, and this is what we are going to pick up when we start the next segment, is that the length of this vector is, how many components of that vector? 2 times nel, whereas that vector has 2 times nel+1. Consequently, this matrix that we have here is 2nel times 2nel+1. Right? This vector, again, here, is 2nel. Each of those vectors here, right? Each of those vectors is okay, of the same length, right? Because they are forming a sort of dot product with the c vector. All right? So we're going to use this when we return to the next segment.