So we have this and now let's look at what the contributions are to our stiffness matrix and our force vector. What I'm going to do now to do that in order to get there is to simply carry out the process of assembly. Right. Of these two equations the last two equations on the slide of that and that, okay? Recognizing, of course, that we have this map, right? So this map between local and global degrees of freedom, okay. So what we do is we carry out now the process of finite element assembly, okay, recognizing that our vectors such as c1e, c2e, right, is equal to, this is actually identically equal to, our entry c e plus one. Right. And next, and basically the same thing for the [inaudible] for the D degrees of freedom as well. Okay. So in element e, d one e, d two e or d sum e and d sub e plus one. Okay. These are local degrees of freedom and these are global degrees of freedom. Okay? So when we put all of this together what we will get is the following, right? We get here the matrix c, sorry the roll back to c two, c three, all the way up to c nel multipying EA over he. And I didn't stated it explicitly but you note that I have already fallen to using our special case where all the element lengths are the same. Okay? All right. So we have that and now the matrix that we will have here will have the following fall. We'll have minus one here, two, minus one minus one, two minus one. Okay. So this c nel corresponds to C one nel, right, in terms of local node numbering and what we see is that this process essentially ends with a one here and a, sorry, there's a one on the diagonal here. There is a one on the diagonal here and a minus one here, okay? This is multiplied by a column vector which is full, right. There is nothing lost in this column vector for d degrees of freedom, comes all the way down to d nel plus one, okay? This is equal to, I'm going to write it down here because I just need some more room. We have for the forcing vector, we have c two, c three, all the way up to c nel, okay? Again using the special case of all the element lengths being the same. We have fAhe over two, okay? And now the contribution that we get here is of a sort where all the entries are two. Okay, as far as the sizes are concerned we note that this now has size nel minus one, right? Clearly if something is going from two up to nel, right, this now has length nel minus one, okay, whereas this vector has length nel plus one. Accordingly, the dimensions of this matrix, right, are nel minus one rows and nel plus one columns. Right. This is also nel minus one and therefore so is this nel minus one. All right. And here now that we are, we are getting used to thinking about these kinds of ideas, we make the observation that d one is now equal to u not, right, using the [inaudible] delta property, right. Whereas d nel plus one which is the trial solution degree of freedom corresponding to the last node which is the second Dirichlet boundary condition that x equals l. This is u given and both these quantities are known. Okay. And also note that now that we have this form recall that the other elements in this column vector which I'm going to mark out in a step that is recommended only for clarity and not something that you should normally do in a Matrix, right. We have over here two, we have zeroes, right,. all the way up here. Well, it's a good thing I turned back to look at this matrix because I realized that I had forgotten that in this very last element here the one I made an error, right? There is a contribution coming to that element or to that entry from the nel minus one element so that entry there properly is two, okay? If you just go again, if you would go back again and you would go back and look at our contribution to the eventual stiffeners matrix from our Dirichlet [inaudible] problem you would see that there is a two there. Allright, I was simply forgetting the fact that there is a, an element before it, right? Okay, so those are traps for the unwary and for some reason I seem to introduce the one here. That's a zero, okay. All right, so essentially the first column and the last column have a minus one in one position right there and zeroes and the very last column has a minus one in the very last position and zeros about it. Okay. We recognize the fact that in the multiplication on the left hand side of this matrix vector product that column vector multiplies u zero whereas this column vector multiplies u g. Okay. And since u not and u g are known we can move them to the right hand side. Okay. We'll do that. And so what we're left with then is c two up to c nel minus one, multiplying EA over h e. There's a matrix here which has the form two minus one minus one two minus 1 going all the way down to two here. And now we would have a minus one here and the minus one here and just for completeness let me put a two there as well. This multiplies now. Now what we are left here is a vector starting with d two and coming all the way down as far as d nel. Okay. The first and last entries from the trial solution degrees of freedom vector have been lost. Okay. But they haven't been lost forever. They've just moved to the right hand side and on the right hand side we have our C vector consist, starting with C two going as far as C nel minus one. This multiplies. First fA over he divided by two, it multiples the force vector now with entries two in every postion. Okay. Additionally it has EA over he, which is the contribution from the left hand side, right, correct? That the contributions that arise from the fact that our first and last nodes have Dirichlet boundary conditions specified on them. Okay. So you get EA over he and I realize I have given myself very little room there so let me just move it to the next line, plus EA over he multiplying first a column vector with u not in the first position and zeroes every where else. And this is the contribution from the Dirichlet boundary condition at the left end that X equals L. Where we have the trial solution value specified to be equal to u not. And then we also have EA over he multiplying a column vector which is zeros everywhere except in the very last position where it is Ug. Okay. We close it here. Note that this parenthesis, this right parenthesis closes that left parenthesis. Okay. Well, now as before we have, you know, everything is the same as before. We have C transpose. Right. This is our stiffness matrix K, which has dimensions nle minus one cross nle, this one nle minus one cross nle minus one. Okay. And this is our reduced D vector for this problem. Okay. These are the unknown, no real trial solution values. Okay. Once again here we have our C transpose vector and everything that multiplies it from that left parenthesis to the right parenthesis forms our f vector. Okay. So we have, once again we have C transpose K d equals C transpose F for all vector C now belonging to the smaller domains in space are nel minus one. Okay. And then what this implies, of course, as before K d equals f. Okay. Our final finite element equations to be solved. What I'm going to do now is just write out our final forcing vector in the [inaudible] equations. And not, it's contributions. Right. F is the force vector in this case. Right. Where it is equal to f A h e over two with entries two everywhere. Right. And this is the contribution from the forcing function, right, our function f. All right. Plus EA over he multiplying u zero, a column vector with u zero in the first entry and zeros elsewhere. Right. This one is the contribution from the Diritchlet forcing, right, at x equals zero. Okay. Of course if u not equal to zero this thing would just drop out. Okay. Plus EA over he multiplying a column vector with zeros everywhere except in the very last position which is ug which now is the Diritchlet data at x equals f. Okay. So here you have a situation where clearly having only the Diritchlet boundary conditions is actually driving the problem. Right. There are no [inaudible] data in this problems. Purely Diritchlet. Okay. So this would be a good place to stop this segment.