 So let's do another problem as far as repeated eigenvalues are concerned. So just a term, I might have used it before, but just to mention it properly, multiplicity. That is where we're going to get a repeated real eigenvalue, say to the power 2, lambda plus 3 squared, that will be multiplicity 2. And that multiplicity we call m. Remember we had that long equation to write x sub m. And that m, I think before, I might also have just written less than or equal to n, where n is the size of the matrix A, the square in matrix A. And I want to remind you, let's do a problem here with multiplicity 2. So we're going to have these two solutions, x sub 1, which is going to be the eigenvector k e to the power lambda sub 1 t. And x sub 2 is going to be k e to the power lambda sub 1 t plus p e to the power lambda sub 1 t. In other words, this comes from that long equation we remember, the way that you could write x sub n with the t to the power m minus 1 over m minus 1 factorial, etc. Just to remind you, then, k e is for this multiplicity 2. It's going to be an eigenvector k sub 1, k sub 2, and p is going to be a column vector as well, p sub 1, p sub 2. So there's our problem, x prime equals our square matrix here, 3 negative 18, 2, and negative 9x. As per usual, we're just going to look at what A minus lambda i is, because we need to get, first of all, our eigenvalues. So that is simply going to be 3 minus lambda and negative 18 and 2 and negative 9 minus lambda. We're going to get the determinant of this, and we're going to set that equal to 0. So the determinant of this, let's just do it. So it's going to be 3 minus lambda times negative 9 minus lambda, and negative 36, say, but minus negative 36 becomes positive 36, and that's going to be equal to 0. So let's see, we have negative 27, we have negative 3 lambda, we have a positive 9 lambda, and we have a positive lambda squared and a positive 36. So it's going to equal 0, so we have lambda squared plus 6 lambda plus 9 equals 0, and then lo and behold, that's lambda plus 3 squared. So the multiplicity of the repeated eigenvalue here is 2. 2 is m, m larger or smaller than n. n size is 2, it's a 2 by 2 square matrix. So indeed, we fall there with multiplicity 2, we know we're going to have these two solutions. We don't have a third, it's not a 3 by 3 matrix, so that's all we have. So we know lambda sub 1 equals lambda sub 2 has got to be negative 3. Now we're going to use that fact, first of all, for our x sub 1, we know that lambda minus negative 3, which will become positive 3i, so that's positive not lambda a. What am I doing? a, that's negative minus negative 3, which is positive 3i, times k has got to equal the 0 vector. This is vector multiplication, so that's not simply 0, that has got to be the 0 column vector. So what are we looking for? We're looking for k, our eigenvector. So how many eigenvectors can we make from this single eigenvalue? Remember I said you're either going to just make one or more than one and that's going to give you two different types of solutions. So what is a plus 3i? So 3 plus 3 is 6 and negative 18 and 2 and negative 9 plus 3 is negative 6. So that's our vector there. We've got to multiply it by k1 and k2. Let's just do it the long way. I think before I showed you how you can just do an augmented matrix to get those solutions, so it's going to be 6 times k sub 1 minus 18 times k sub 2 equals 0. So we have 6 times k sub 1 equals 18 times k sub 2 or k sub 1 equals 3 times k sub 2. So I can look at one example. My k eigenvector here would be 3, 1 because I've finite k sub 2 equal 1, that means k sub 3. It's got to be 1 and anything will just be constant multiples of that. So here I just have a single, I can only get a single eigenvector for this repeated eigenvalue. Therefore, I've got to go for that. Remember it was k sub m equals, what was it? k1m and t to the power m minus 1 over m minus 1 factorial e to the power minus 1t plus, then we're going to have k2m and that was going to be t to the power m minus 2 over m minus 2 factorial e to the power m minus 1t and if you're plugging those values you are going to get for m1 you're going to get that and for m2 you're going to get that and instead of writing k2m we're writing this as p and k1m we're writing as k. Let me just clear the board, I'm going to stop the recording just to clear the board and we'll carry on. Good, now we forgot to have the fact that a minus negative 3 so that will be 3i times p. The vector p is now going to equal the vector k. So I'm still left with that and now I'm just going to multiply by p1 and p2 so I'm still left with 6 negative 18 to negative 6 and now I'm going to multiply by p1 and p2 and what do I get and that's going to equal k and k is equal to 3i so I'm going to get 6 times p1 minus 18 p2 or 2 times p1 minus 6 times p2 that's exactly the same as equal vector k which is 3i so let's have a look, 6 times p1 minus 18 times p2 is equal to 3 and I can say p1 equals to by everything by 6 so that will be a half and plus 18 divided by 6 plus 3 times p2 so let's say p2 equals 0 so in other words I can write one as a half zero as a half zero so here I would have now my set of solutions now what I'm interested in I've got this now which will be 3i e to the power negative 3t that's my one set I'm going to put a c sub one there and I've got to put a c sub two here because I'm interested in a general set of solutions so x is going to be c sub one for that first solution which was 3i e to the power negative 3t plus c sub two then for my second set which is there which is going to be what it's k which is 3 and 1 t e to the power negative 3t and p which we got as a half and zero plus a half zero e to the power negative 3t okay so don't get us confused you will need for the general set solution set you need both of these x sub one and x sub two so there's my x sub one which is just a k in and there's the x sub two remember I had to make use of this equation because for my repeated eigenvalues my repeated eigenvalue of multiplicity two I could only get one vector I could only get one eigenvector therefore I had to go for this but this is not the only solution remember you've got to have x one you've got to have x one in there as well