Let's just see the concrete example corresponding to the case 2-2, right? So let's consider the following example, okay? We'd like to solve the following system x prime is equal to the coefficient matrix is minus 1 and 2 and 1. And the 0, -1 and 3, and 0, 0, -1. Okay, so this is the coefficient matrix A, right? Okay, so the characteristic equation, corresponding characteristic equation, determinant of A minus lambda i3, and that is equal to 0. Okay, computing the determinant, you can easily see that this is minus of lambda plus 1 to the 3. So and that means there is only one Eigen value lambda 1 equal to lambda 2 is equal to lambda 3, and that is equal to -1, okay? Okay, so this is the -1 is the only Eigen value. Trivially there's a multiplicity 3, right, okay? Eigen value with, okay, Eigen value with multiplicity 3, okay? Try to compute to the corresponding Eigen vector. Okay, say we'd like to solve that the A minus lambda. So A plus the 1 times of i3, right and the K. Okay, and that is equal to 0, 0, 0, and 0 and 0. Right, then okay, if we're to solve this one then, okay, there's a simultaneous system then there is only one. Okay, there is only one linearly independent Eigen vector say K1 that is equal to 1 and 0 and 0, right? This is the only one, okay? Only one, the corresponding Eigen vector, okay? The theory we have introduced before, right, says that in this case there are three linear independent, the solutions of this system of equation. Okay, corresponding to this single Eigen value -1 with multiplicity 3 of the following form, right, okay? First the solution x1. This is equal just k1 e to the -t. Well, what is the K1? K1 is already given here, right, that's the Eigen vector, okay? Eigen vector corresponding to Eigen value lambdas of 1, okay? The second solution will be, okay, is of the form, right? This vector K1 times t times e to the t, okay? And plus some unknown vector e to the minus t where the unknown vector P is a solution to the following system. Okay, A plus i3 times P, that is equal to K of 1, okay? That's a simultaneous equation required for the unknown vector P. Okay, so the solution will be this is equal to, okay, that K1 times t squared over 2 times e to the t plus p times of e to d. Not t but here, the -t, right? I'm sorry, e to -t and plus a known vector Q times e to the -t. And e to the -t where the Q is the solution of the following system of equations. A plus i3 and Q, that is equal to P, okay? That's something we have made through the claim, right, in the last lecture, right? First solution is given by the K1 times e to the -t, where K1 is 1, 0, 0. And the second solution is of this form where the unknown P satisfying this relation. So the solution is given by that form where the unknown solution, the unknown vector Q satisfied the relation. Okay, this is the system of equations, right, it's easy to solve. Okay, it's easy to solve and the solution is not unique, right, because the determinant of this matrix is 0, right, okay? So anyway, what we can say is I will leave the, some are tedious computations to you. Okay, solving these systems, we already get K down there. Okay, and furthermore, okay, by suitable choice of the arbitrary constants involved, right? We get the solution possible solution for P. We may get the vector P is equal to 0, 1/2 and 0. And the solution, possible solution for Q will be 0, -1/12 and 1/6, right, okay? So that with those the K1, okay, what was the K1? Let's remind the K1 is equal to 1, 0, 0. Okay, so using those three vectors, we have three independent solutions x1, x2, x3, okay, of the given problem. Okay, of the given problem and the general solution will be, right? So that therefore, the general solution is arbitrary linear combination, right, of those three. Okay, c1x1 plus c2x2 plus c3x3, okay? This is a general solution, okay? [MUSIC]