 So let's look at yet another problem. I have this 3 by 3 matrix, and you'll see it's a very special type of matrix. If you remember matrix algebra you'll see on the diagonal we have all 2s and everything 0 under that. Okay, you'll remember those matrices. Anyway, let's look at a minus lambda i. That is going to be 2 minus lambda 1, 6. And we're going to have a 0 of 2 minus lambda 5 and a 0 is 0 and 2 minus lambda. That's a matrix and we've got to get the determinant of that, but we're going to be clever about it. We're going to get it along either column 1 or row 3 because we have 2 zeros in there. Remember it will be positive, negative, positive, negative, positive. So it's going to be that one. So the determinant is going to be 2 minus lambda. And we're going to multiply that, so we take it along this row, close that off. So it's going to be this minus 0, which is now going to be 2 minus lambda squared. And that's going to equal 0. And then, behold, what do I have here? It's 2 minus lambda cubed equals 0. This is my m, so this is multiplicity 3. So I now have an x sub 1, x sub 2, x sub 3. But I first need to find out for this eigenvalue of mine now, my eigenvalue being... Now, just have a look at this. Now, you've just got to be careful with this negative here. It says 2 minus lambda, in other words, lambda equals 2, not negative 2. So lambda sub 1 equals 2. So with this I need to find out how many eigenvectors I can get from that. So if I have a minus 2i times k is going to equal 0. There's the short way of doing it. There's the long way of doing it. Let's do the longer way. Maybe you can also do the augmented matrix. So I'm going to have 0, 1, 6, 0, 0, 5, 0, 0, 0, 0. And I've got to multiply that by k sub 1, k sub 2, k sub 3. And then just got to equal the 0 column matrix. So what are we left with? We are left with k sub 2 plus 6 times k sub 3 there. And then we're left with 5 times k sub 3 there. And we're just left with 0 there. So those two have got to equal each other. We can clearly see that k sub 3 at least. So if we just look at k, k sub 3 has got to be 0 there. And if I let that 0, that means k sub 2 has also got to be 0. But that leaves me in case of 1 to be anything at once because I've got to multiply it by 0 anyway. We're not going to choose 0 because then we have that 0 eigenvector, which means nothing, so the next thing we'll put in is a 1. So I can put in a 2 there. A 3 doesn't matter, but that would just be constant multiples of this eigenvector. So what I can see here is that I can only get 1 eigenvector for this eigenvalue. And if I do that, remember now I've got to go move on to say that a minus 2i times p has got to be equal to k. That would be my next step. Remember if I could get more than one eigenvector, I could have just gotten, I just could have used all of them, but I can only get a single eigenvector. Case of 1 can be anything because it's multiplied by 0. But that will just be constant multiples of this, so that won't help me. So let's do this one now. And now it's going to equal 1, 0, 0. And that now becomes p sub 1, p sub 2, and p sub 3. So again I'm going to be left with p sub 2 plus 6 times p sub 3. Is that and p sub 3 is going to equal that and lastly 0 equals 0. So p sub 3, if I just look at p, p sub 3 is just 0. p sub 3 equals to 0. And if p sub 3 is 0, that means p sub 2 must be 1. I'm left with p sub 1 being anything I want. I'm always going to start with 0. I didn't want to start with 0 there, otherwise it's all the same. But yeah I can start with 0 because I'm still left with this. The next one I have to do is multiplicity 3. So I'll have to go to q and that is going to equal p. So let's do that one. Now this is going to equal 0, 1, 0. And again, so it's q, so it's 1, q sub 2, and q sub 3. And so again q sub 2 plus 6 times q sub 3. There I'll just have 5 times, I hope I put the 5 in before, and 0. So now I see q sub 3, q sub 3, 5 times q sub 3 equals 1. So q sub 3 is going to be equal to 1 over 5. And if that's 1 over 5, so that becomes 6 over 5, so that means q sub 2 is negative 6 over 5. And again q sub 1 can be anything, but I'm going to start with 0. So there we go, I have all of these now. And now I've just got to remember how to construct my general solution set. My general solution set. So let's look at that. For x sub 1, I'm just going to have k, what was my k, remember it was at 0, 1, 0, 0. So that's just going to be k, so that's 1, 0, 0, e to the power, what was my, it was 2t, 2t. x sub 2 was going to be k, that's 1, 0, 0, t e to the power 2t plus p was 0, 1, 0, e to the power 2t. And x sub 3 was going to be 1, 0, 0, t squared over 2 e to the power 2t plus p was 0, 1, 0, t e to the power 2t and q was 0, negative 6 over 5, 1 over 5 e to the power 2t. So there we go. Because I could only get one eigenvector for my eigenvalue, that's what I have there. And then for the general solution set, it will be x equals c sub 1, x sub 1 plus c sub 2, x sub 2 plus c sub 3, x sub 3, x sub 1, x sub 2 and x sub 3 are not constant multiples of each other. In other words, they each get their own, so that's going to be c sub 1 times that plus c sub 2 times that whole one plus c sub 3 plus that, c sub 3 times that whole one. Okay, so this would be the general solution. That's quite easy to do.