 And now we come to the very exciting bit of complex item values. So I have x prime of t equals 6x minus y, y prime of t equals 5x plus 4y. I say that's a linear system of the differential equations. I need to solve for that. If I let x equal this column vector xy, that means I'm going to have the following. x prime is going to equal 6 and negative 1. 5 and 4 is my matrix of coefficients times x. If that's xy, if I put x and y there, I multiply this out. I'm going to be exactly there. So if I just look at a minus lambda i, that is going to equal 6 minus lambda and negative 1 of 5 and of 4 minus... Why do I put y there? Let's do that. 6 minus lambda, negative 1, 5 and 4 minus lambda. If I get the determinant of this, a minus lambda i, and I've got a set that equal to 0, that's going to be 6 minus lambda and 4 minus lambda minus negative 5 which is plus 5 is going to equal 0. So I'm going to have a lambda squared. I'm going to have negative 4 lambda and negative 6 lambda which is negative 10 lambda. And I'm going to have 6 as well as 24 plus 5 is 29. It's going to equal 0. Now I have to use my quadratic equation to solve this. I'm going to say lambda equals negative b which is negative 10. So that's 10 plus minus the square root of 100 minus 4 times 29 and it is going to be over 2 times a which is this 2. So lambda is going to equal 10 divided by 2 is 5 plus minus 100 divided by 24 minus 9 that's 116, that's the square root of 16 which is 4i, 4 divided by 2 is 2, so it's 2i. So I have lambda sub 1 equals 5 plus 2i and lambda sub 2 equals 5 minus 2i. So I have complex eigenvalues here. Let's start with lambda sub 1. So that is going to be 6 minus 5 which is just 1. 1 minus 2i, we're going to have the negative 1a, we're going to have the 5a and then 5 minus 4 minus 5 minus 2i that's negative 1 minus 2i. So that is what I'm going to get for a minus lambda sub 1i. That's what I'm going to get for this. I have to multiply this by k so that I can just make some space in this case of 1 and case of 2 and that's going to equal 0. So what am I going to have? 1 minus 2i case of 1 minus case of 2 equals 0 in other words 1 and then you can do the same with the 5 and you'll just see if you multiply there will be something which you can multiply this first one with you'll just get the second one. These are just actually constant multiples of each other as always. So I'm going to have k2 equals 1 minus 2i case of 1. So if I let k sub 1 equal 1, my case of 1, eigenvector there is going to be 1 and 1 minus 2i, 1 minus 2i. Now you can work out case of 2 but you can do that but I'm going to show you if these are real, if the matrix of coefficients here is real values and you have complex numbers in your eigenvectors, these two are just going to be, the k1 and k2 is going to look exactly the same other than the fact that these will have to have, now remember that's plus 0i. So this was going to be 1 minus 0i plus 1 and 1 plus 2i. So what we're going to have is that k2, this is going to be the complex conjugate of k1. It's always going to be under the circumstances that I've mentioned. So that positive for the beta there, remember that's alpha and beta, alpha and beta for the complex number. So that positive becomes a negative, that negative 2 becomes a positive 2. So you're always going to have that and you can see here as well with lambda, remember that lambda2 is just going to be the complex conjugate of lambda1, lambda1 and as much as I had 5 plus 2i and I'm going to have 5 negative 2i for lambda2. So that's always going to work out. So you have these two values so you can have and you'll note that they are not, they're not constant multiples of each other. This one you'll use for x sub 1, this one you'll use for x sub 2 so it's going to be c sub 1 and then x sub 1 is just going to be this. 1 and 1 minus 2i and it's going to be e to the power that was 5 plus 2i t and plus c sub 2 which was going to be 1 and 1 plus 2i e to the power of 5 minus 2i t. I hope you can see that. So that was going to be my final, let me just take you away so you can see that that was going to be my final set, general set of solutions there. Now we'll expand a bit on this in the next example.