Last time I introduced the notion of the linear independence and the linear independence of a finite set of solutions for the homogeneous system, right? Let's remind it shortly. We are considering that the linear homogeneous system X prime is equal to eight times of X, with the n components, right? From the X1 of T, X2 of T and so on, X of n, right? The coefficient metrics a it is the n by n matrix given by the the containing the entries a's of IJ of T, right? Okay? Let's remind them about the I have introduced in the last time, okay? The I said the set of finite lee many solutions, right? XJ, X equal to 1 to k, okay? Solutions, okay? This given the homogeneous system, okay? Then we say that this XJ is, okay? Are, okay? Linearly dependent. An I, okay? Solutions of the system on the interval I, okay? Linearly dependent on I, right? If there is the K constant, which I call by CJ, okay? Constant such that, okay? Not all zero, okay? Which means that the sum of the one to the K of cjk, this is strictly positive, right? Not oj, and the satisfying the linear combination, okay? CJ, XJ of T is identical equal to 0 and I, okay? Then we call this is the definition, right? Our definition of the notion of linear independence of define Italy many solutions of the homogeneous system, right? Okay? And we say that otherwise, okay? Otherwise we say that the system is, okay? They are linearly independent, right? Linearly independent on I, okay? What does that mean? Okay? We need the negation of this statement down over there, right? Okay? So it should mean that the following things, right? It should mean, okay? Linear independent on I, okay? If make the linear combination, CJ, XJ of T, okay? And require that that is equal to identical is zero, right? This is not a scale of zero, but this is a zero vector, okay? So let's denoted by this way, okay? Vector zero. And this is zero vector, and I, okay? If this statement implies implies all the CJs are equal to 0, right? So C1 equal to C2 equal to and so on. And the Ck must be equal to zero, right? That's the linear independence of that. Financially many solutions, right? Okay? Checking this condition is sometimes it can be rather tedious, instead of we have the following alternative, okay? That is the so called the scan test, okay? For the linear independence of the solutions, okay? What I mean is let me introduce the theory. This is a run scan, Test for linear, Independence, okay? Again, consider the same, the homogeneous system of equation, okay? And let now XJ 1 to the n, okay? B solutions, Of the homogeneous system, okay? Say X prime is equal to, The coefficient matrix A times X, right? Reminded that, okay? The unknown function has n components right? Unknown function X has this is the n component through the X1 of T, X2 of T and X of n of T, right? And we have the same number of solutions for this system, okay? We have n solutions, okay? For the system of equation with an unknowns, right? Then, okay? Let's denote the XJ. This is the column vector, right? This is the column vector let XJ checks J let's denoted by the entries of X of J is X1 of Jft, X2 of Jft. And also on X of n of Jft, right? This is n vector, right? Okay? Each one so that for OJ from 1 to the n, right? So that from this one, we can make the the environ matrix say force the column X1, okay? Second column X2 and so on. Last s column is X of n, right? This is equal to, right? Okay? So the entry wise this matrix is exactly X I j f T where I and J are moving from one to the n, right? This is the n biometrics, right? Okay? This is an end biometrics, okay? And biometrics of functions, okay? So that we can define its determinant, right? Okay? So we can think about the determinant of the matrix say X1 and X2 and X of n. In other words determinant of this enviro metrics given by the X of I of J and T, okay? We give a special lame on it, okay? This determinant to be recorded, we denoted by this is a capital WX1 and X2 and so on, X of n of T or in a short cut, okay? Short notation is a capital W of T, okay? And call it to be a determinant, okay? And I call this one to be the rumskim determinant, okay? Or simply just run scan, okay? We call it as a run scan or the run scan determinate, right? Now is the main claim of this run skin test, okay? Think about this homogeneous system of equations with an unknowns. Now we have an solution vectors, okay? And now here's the conclusion, right? These XJ's, okay? And solutions one to the n, they are linearly independent, okay? Independent on the interval I, okay? There are the solutions of this problem on the interval I, okay? The linear independent on I, okay? The claim is this run scale matrix, okay? Run skin determinant, or the run scan WT is never zero for any TNI, okay? They're equivalent, okay? So, okay? In order to check the end solutions of such a system to be linearly independent or not, okay? It's enough to look at the just the run skin of that. The n solutions, right? And whether that this is equal to zero or not, okay? One another, the striking result is in fact, you don't have to check the run skin is not equal to 0 for any T in I bar. It's enough to show that run skin of T0 is not equal to 0 for any, I'd better to say not any, but for some, okay? T0 and I, okay? Although the three statements are equivalent, right? Okay? The equivalent surplus. These two comes from the following fact, okay? That's another the interesting fact that we can see, okay? The last claim, the equivalence of those two is coming from the following effect, right? For any answer of solutions they're run scan. Let me say in this way, right? For any such a set since either brown skin is identical 0 on I or run skin is never 0 for any T in I, okay? For any and solutions of this homogeneous system always this is true, okay? Its corresponding one skin is either identically 0 on I or never 0 on I. So that okay? If this is not equal to zero for any single point, then it will be never zero for any other points too, right? And that is equivalent to that, okay? These n solutions are linearly independent, okay? This is a very the convenient and the useful, right? Useful test for the for checking the linear independence the of the set of solutions for and given the homogeneous system, okay? Let's see the simple example. Before the introducing the example, let me introduce the just one more terminology. Say, a set of solutions of the homogeneous problem. X prime is equal tax of n. Eight times the coefficient matrix A times X, right? Where X has the n components, right? Okay? Is a fundamental solution, okay? Is a, Fundamental solution, Sub of the problem X prime is equal to 8 times X on I, okay? I'm introducing this determinology, okay? Right? If this XJ is, okay? 1 to the n are linearly independent solutions of the problem X prime is equal to A of X, right? Okay? Then we call it as simply, okay? We call simply the fundamental set of solutions, okay? Or fundamental solutions, okay? Let me say fundamental set of solutions that will be better, okay? Fundamental, Set of of solutions, The problem X prime is equal to 8 times X, right? Okay? So let's look at the simple example, okay? The system I'm considering is the two unknowns and the two linear equations, okay? So I'm considering that, okay? X prime is equal to 2 by 2 coefficient matrix, that is the constant matrix I'm taking. 2, 1 -1 and 0, and times X definitely X is equal to just the two components. It has two components X1 of T and X2 of T, right? Okay? I'm claiming forced to that, okay? Then X over 1 which is 1 and -1 times E to the T. And X of 2 that is equal to 1, 3 and the E to the T + 4- 4 X T times E to the T, right? Okay? Both X1 and X 2, okay? Are solutions, okay? Of this given system, okay? Definitely because it has a constant coefficient so, okay? The coefficient matrix is a continuous everywhere on the real line, so that they are the solutions of that the problem on the whole real line R, okay? It's the easiest thing to check away, okay? You just plug this expression into that and to see that the system is satisfied. The same for the extra to plug this form into that equation and to see that the equality holds, right? So, the both our solutions, right? This is the problem of two unknowns. And now we have two solutions. So we can make the the two by two matrix formed by the X1 and X2, right? What's this matrix? This is the first component to E to the T, okay? And that's a minus E to the T. From here you get the E to the T + 4 X 2 X E to the T, that's the first one second. The one is three times of E to the t- four times the tlv to the T, right? That's the two by two matrix, right? Okay? Let's the computers run scan, okay? What is the means determinant of this? Two by two matrix is, right? Okay? And the determinant of this two by two matrix is the easy to compute, right? Okay? You just multiply this main diagonal, then subtract the product of the sub diagonal, right? So these times that will be 3 times E to 2T. Then- 4 times T times E 2 T, right? That's the from the main diagonal, right? Subtracted the product of these two. So, that means minus this is- C E to the T2 T. And this times that that is equal to minus four times T times E to the 2 T, right? That's the determinant, right? Okay? So, what? Okay? This negative 4 times T times E to the 2T. And here you have a negative and negative. That is a plus four times T times E to the 2 T. They canceled out, right? Okay? And finally, we'll get three times the 32 T plus the 32 T. So, totally, you have four times the V 22 T. And as you know, where exponential function is never zero. So this is strictly positive on the whole real line R, right? Okay? So what oversee a conclusion then then miss these two solutions. They are not just the solution of this system, but they are linearly independent solution, right? This is a system of two for the two unknowns. And you have a two linear independent solution so that you may call, right? Okay? This XJ 1 and 2. This is okay? As I introduced here, okay? This is a fundamental set of solution, right? Fundamental set of solutions. Let me say let's repeat it here, okay? XJ 1 to 2. This is okay? Fundamental set of solutions for okay? For the given problem, X prime is equal to eight times of X on the whole real line. Are okay? That is our conclusion, right? Okay? [MUSIC]