 just to make things 100% clear, let's just see what this terminology is telling us, that x prime is this x prime, y prime, that's what it's saying, and x equals x and y, so what we are saying there is 1353, 1353, that matrix times this matrix x and y, that's what we are saying, so it's going to be x plus 3y and so it's that column, now this row is 5x plus 3y and that has got to equal that equals x prime, so this equals x prime, y prime, so what I'm saying is that x prime equals x plus 3y and y prime equals 5x plus 3y, so I've got two differential equations here but they form a system and I need to solve them both together, up till now or in previous lecture series on differential equations, we're just dealing with one equation, so that is what I'm trying to solve and our solution set is here, it says xy, that is x and y equals e to the power negative 2t and negative e to the power negative 2t, so x is that and y is that, okay let's just do that, let us just do that, can we get that? Well I'm suggesting here that x equals e to the power negative 2t, that's what I'm saying and if I get x prime just from that, just from that, it's negative 2 e to the power negative 2t, let's see if I can get that here because it's going to be equal here, so that is going to equal x and what did we say in this instance x equals e to the power negative 2t plus 3 times y and what is y? Well here we said y equals negative e to the power negative e to the power negative 2t and that equals e to the power negative 2t minus 3 times e to the power negative 2t and lo and behold that leaves me with negative 2 e to the power negative 2t, so indeed it is correct and you can try it with the other eigenvalue which was lambda sub 2 which was 6 and which will give you this solution set here, which will give you the solution here and both of them will work out to exactly the same values for all of these, for all of these will work out exactly the same