 Good, let's get to an example. Now the letters are a bit small, the numbers are a bit small, so you'll perhaps have to view this in a higher resolution in YouTube. We have X prime there with this 3 by 3 matrix A and we see it there multiplied by X. So there's our matrix A and I also have put in there the matrix A minus lambda I. This is how we would usually do it, and we would get the determinant of that and we'd set that equal to 0. And you'll have to know how to do 3 by 3 matrix determination. Here I've used the first row and just remember that you have to change the signs as you move along. So the 1 minus lambda will be positive. The negative 2 in row 1 column 2 should be minus negative 2 and then a positive in front of the law last 2. So you'll have to know how to do the determinant and there are videos on my channel as well about linear algebra. What you would get to is the second to last line. After all, that's a very simple algebra. We'll have the negative outside, which we can just divide both sides by negative 1, so that'll disappear. So we have lambda plus 1 squared and lambda minus 5. So we have a repeated eigenvalue. Lambda sub 1 equals lambda sub 2 and that's negative 2 and then we have lambda sub 3 equals 5. So whichever way you want to look at it, we have this repeated eigenvalue and it'll be one of two possibilities. We're either going to get a single eigenvector for this or more than one eigenvector. And I showed you in the previous video what is going to happen under those two circumstances. So let's have a look. We'll have lambda sub 1, which is then also lambda sub 2, and we'll first get this k sub 1 eigenvector, which might also have a k sub 2. We don't know. And we'll just call it k sub 1, k sub 2, k sub 3. It's a column vector, an eigenvector. So I'm going to have a minus lambda, which is now negative 1i, multiplied by k. And what I've done here is just a short-hand way of doing this. Remember that the way that I do my matrix multiplication, I'm going to write the one matrix slightly to the upper right of the of the other one. And the left lower one would be the first one in the multiplication. The right top one will be the second one so that the sizes of the matrix, matrices coincide so that I can actually do this matrix multiplication. And what are you going to do then? Well, you're going to get a three by one matrix. And that three by one matrix has got to equal the zero vector, column vector. And that means you're going to have three unknowns and three equations. And you can solve a linear system like that just with Gaussian, Gaussian Jordan elimination. And that's all I have there. So I've skipped all those steps. They would have just gotten me a matrix of coefficients, which you'd see there, two negative two two, negative two, two, negative two, two. And we have the zero column vector on the right hand side. And I can just do Gaussian Jordan elimination. Once again, you'll have to watch videos on how to do linear algebra. And that's what you'll get to. You'll get these, the row one, negative one, one, which means one times case of one, plus a negative one case of two, plus a one case of three equals zero. And then a case of one, if I just simplify that case of one equals case of two minus, that should read case of three. Case of one equals case of two minus case of three. And if I look back very quickly, if I let case of two equal one and case of three equals zero, that'll leave me with a case of one of one. If I let case of two equal one and case of three equal one, I'm free to do that because the other two rows, which is zeros, was going to leave me a case of one of zero. And those are linearly independent of each other. They're not constant multiples of each other. So there's at least one, or there is more than one, eigenvector for this lambda sub one equals lambda sub two equals negative one, eigenvalue, this repeated eigenvalue. In other words, I have two eigenvectors there. And my set of solutions in the end is simply just going to be a c sub one times that first eigenvector times e to the power, now my lambda sub one was negative one, so it's going to be e to the power negative t and c sub two times that case of two times e to the power negative t. So nothing untoward there. If we look on the right hand side, though, the flow on this is just a bit awkward. So you have the case of one and the case of two up at the left top. Then we move across to the right. Let's have lambda sub three, which is now five. That's not a repeated eigenvector. So I'm going to have a minus five i and I'm going to multiply that by k sub three. And that's going to equal the zero vector, not just zero. Once again, I'm going to end up with three equations and three unknowns. Homogeneous in as much as it equals the zero column vector. And I can just write that as a matrix of coefficients together with this augmented matrix with the homogeneous part on the right hand side. Gas Jordan elimination again. And that's what I get one zero negative one zero zero one one zero and zero zero zero zero. So now we flow back to the left hand side. Apologies for that. So that means one times case of one plus zero case of two minus a case of three equals zero. In other words, case of one equals case of three. And in case of two must equal negative case of three. If I just take the case of three over to the other side, I get that from the second row. So I this is not to repeat it. I don't have to come up with repeats. I can just choose something very simple and I only need to choose one eigenvector for this. So if I let k sub one equal one, that means k sub three has got to equal one and k sub two is going to equal a negative one. So this is my eigenvector for the eigenvalue five. So if I swing over to the bottom again, I see that my case of three is that one negative one one. So my x sub three is going to be one negative one one e to the power of five t. So I can I can now write my final set of solutions, my c sub one. And there's my case of one eigenvector, my c sub two and my case of two eigenvector and my c sub three and my k sub three eigenvector. And you see the lambda is the negative one for the first one, negative one for the second one and five for the last one. So this was an example where we had repeated real eigenvalues and it turned out that there was more than one eigenvector. So I needn't make use of that complicated equation where we're going to use t to the power m minus one over m minus one factorial. I needn't use that because I've got more than one eigenvector here. And I can simply use this form of the solution set.