 Now it's good for us to get some examples done, here we have a matrix A being a 3 by 3 matrix so we at least can have 3 eigenvalues, some of them which might be repeat, some of them which might be complex numbers, let's have a look, let's first of all, there's so many ways you can go about this but let's call our matrix C to something new I'm doing minus lambda I and that is going to be this 3 minus lambda, we have negative 1 negative 1, we have 1 1 minus lambda and negative 1 and we have 1 minus 1 and 1 minus lambda, that's our new matrix C, now we want the determinant of matrix C to equal 0, so the determinant of matrix C, well good thing to do these kind of examples because there's so many things you can review, how do you get the determinant of a 3 by 3 matrix, let's decide on a row or column to go along, it looks to me like column 1 would be the easiest because we have a positive 1 and positive 1 there, just remember that this turns out to be positive, negative positive, co-matrix there and so what are we going to have, we're going to have 3 minus lambda and then we're going to have these two multiplied, minus those two multiplied, so it looks like 1 minus lambda squared minus negative, minus negative 1, negative 1 is plus 1 so that is there and this becomes a negative, now I take out this row and this column, so I'm left with this, it's a negative 1 multiplied by the 1 minus lambda and from that, remember this row, this column is gone, so again minus negative 1 equals positive 1, so minus 1 there, this is now positive, so that remains and this row goes, this column goes, so it's those two, minus those two, that looks like a 1, that's a negative 1, but it's minus negative 1 so that becomes plus, this one, 1 minus, 1 minus lambda and that has got to equal 0, that's got to equal 0 and you can well imagine how easy it is to make a mistake somewhere along the line, if I haven't already done so, we'll see, so this remains 3 minus lambda, this is going to become 1 minus 2 lambda plus lambda squared minus 1 and what do we have here, a negative, so here we're going to have lambda minus 1 minus 1 and here we're going to have plus 2 minus lambda equals to 0, so those two will cancel out, so I have 3 minus lambda, now you can do these many fewer steps that I've just done here, lambda squared minus 2 lambda and then we're going to have negative 2 so that becomes a positive 2 minus lambda, another positive 2 minus lambda, that's got to equal 0, let's do this quickly, so that's going to be a 3 lambda squared, it's going to be a negative 6 lambda, that is going to be a positive lambda, where am I going? See, I told you, negative lambda cubed, positive 2 lambda and here we're going to have a positive 4 and a minus 2 lambda equals to 0 and that leaves us with a negative lambda cubed, so that one goes and left with a positive 3 and let's see, it looks like I lost the lambda some way, this is going to do the step, I'll tell you, so easy, and this is typical of me anyway, 3 lambda squared minus 6 lambda, we've got this negative lambda cubed, negative positive 2 lambda squared, there was my mistake, so there's positive 2 plus 2 is 4, negative 2 lambda equals to 0, so I have negative lambda cubed still for that one and 3 plus 2 is 5 plus 5 lambda squared, negative 6, and then there's that negative 6 lambda, this is 3, negative 6 lambda so that makes it negative 8 lambda, for those two and I'm left with a, for those two and I'm left with a positive 4 equals to 0, so in the end I can multiply 3 up by negative so I'm left with that minus that plus that minus that and we've got to factor that out, so let's do it, this is leave, this is leave ourselves, this is leave ourselves that so we've got to factor that out and let's try lambda minus 1, let's start there, and we're going to write that as lambda 3 minus 5 lambda squared plus 8 lambda minus 4, let's try that one out, lambda squared, it gives me lambda 3 minus lambda squared, that gives me a negative 4 lambda squared plus 8 lambda, that is minus 4 lambda there, which gives me negative 4 lambda squared plus, is that right? Have I got the right one? I'm going to pause the video to skin this board because I'm going to make sure that this is not incorrect, okay apologies for that, I couldn't find any mistake anyway, so let's carry on, there's 4, so that leaves me 0, that leaves me 4 lambda, I bring down a negative 4 which just looks like a positive 4 there, that looks good, 4 lambda minus 4 which is nothing left, so I can factor that out as well, so it seems I'm going to have a lambda minus 1 and a lambda, if I factor that out, that's lambda minus 2 squared and that equals 0, so I can see I have repeated eigenvalues there, repeated eigenvalues, so my choice of lambda minus 1 is the first one worked out quite well, okay so I have lambda sub 1, let's do a bit more, I have lambda sub 1 equals 1 and I have lambda sub 2 equals lambda sub 3 equals 2, let's start with that because I now have to find out whether I have just a single eigenvector or whether I have multiple eigenvectors, so let's do that here, so if lambda equals 2, so a minus, so a minus this lambda 2 or lambda 3 I, that is going to equal 3 minus 2 is 1, I'm going to have negative 1, negative 1, I'm going to have 1 and then 1 minus 2 is negative 1, negative 1, I'm going to have 1, negative 1, 1 minus 2 is negative 1, so that and I've got to multiply that, so k1, k2 and k3, then you can see that's all going to be the same, whether it's k sub 1 equals k sub 2 plus k sub 3 if I do that multiplication, so it's always going to be like that, okay and I can see two fundamental ways to write this, if I let this equal to 1 and this equal to 0, I'm going to get a k sub 1 which is going to equal 1, 1, 0 but I can also let this be 0 and this be 1, so I'm immediately going to have two distinct eigenvectors and so that's going to be 1, 0 or 1, so at least those two, so it's not repeat eigenvectors, so those two would be distinct, so let's just do the lambda equal to 1, oh that's horrible, so the a minus lambda i is now going to be 3 minus 1 which is 2, negative 1, negative 1, 1, 1 minus 1 is 0, negative 1 and 1 and negative 1 and 1 minus 1 is another 0, I'm going to multiply this by k1, k sub 2, k sub 3, now in actual fact you've got to call this case a 4, 5, 6, doesn't matter, we all know we're going to end up, so 2 times k sub 1, 2 times k sub 1 is going to equal minus 1, minus 1, minus 1, 3 minus 1, 2, 2 times k sub 1 is going to equal k sub 2 plus k sub 3 and oh I'm going to have k sub 1 minus k sub 3, k sub 1 equals k sub 3 and I'm also going to have k sub 1 equals minus k sub 2, bring it to the other side equals k sub 2, so k sub 1 equals k sub 2 equals k sub 3, they're all the same and there's only one way where that is going to work, so that's going to be 1, 1, 1, so that's my last eigenvector, I cannot write my single solution, remember it's going to be c sub 1, k sub 1 for me was 1, 1, 0, e to the power that was 2t plus c sub 2, 1, 0, 1, e to the power 2t, they were the same plus c sub 3, you have 1, 1, 1, e to the power t with lambda, the last lambda being 1, so nice kind of exercise is these because you have to go through so many things, bring back old basic mathematics, you have long division of polynomials to factor out your third-order polynomial there, we have repeat eigenvalues, fortunately that gave us two distinct eigenvectors, so we didn't have to go into our complex equation to get our set of solutions, which means in the end we have quite a simple answer.