With backgrounds that we have studied in section 1.1. Basic theory. Okay, we are now ready to discuss the methods of solving systems of linear first order differential equations with constant coefficients. Okay, so lets the consider the following problem. Okay, X. Prime is equal to eight times of X. Right. Where the X is unknown vector having any components? Right? This is a known vector and the A. Is the N by N matrix having components entries A. I. J. This is a constant constant environ matrix. Okay. Right. How to solve this system of equation. Okay, think about it then. Okay. It's quite nature to to assume that there is a solution. Right? So we are making guests for the top of the solution. Right? There is a solution Okay. Of tai X is equal to OK. Some unknown constant vector K times E. To the land of T. Okay. Where the the case of actor having end components K one, K two. And the case of n. Okay, okay, this is the question vector. And land a constant. Okay. We assume that there is such a solution. Okay, one thing we we need to note that is if case zero vector. Okay then actually is a zero vector. Right? If case zero vector then actually becomes a zero vector and it will be a solution of that homogeneous system of equations trivially. Okay, that is a so called the trivial solution. Right? Any homogeneous linear system of equations having zero vector as a trivial solution. Okay, so we are seeking for a solution of this type but nontrivial solution. Okay, for this type of solution to be a nontrivial solution, that means we need a non trivial vector K satisfying. Okay of this type, satisfying the differential equation. So okay, constant. Non zero vector. Okay. Non zero vector. Okay. In other words, this is not the zero vector we are looking for. Okay, this is the solution of the system. We can check it easily. Just plugging this one into that. Then what is X prime? X. Prime is equal to K times lambda times E. To the lambda T. Right. That's the left hand side, right hand side is eight times K times E. To the lambda T. Right, Okay, what I'm saying is, let me have a space here. Right. We are looking for the non zero vector K. Satisfying this equation. Right. So this is what X is equal to K times E. To the lambda. T is a solution. Okay. Is a nontrivial solution more precisely? Okay, nontrivial solution of. Okay, ex prime is equal to eight times of X. If. Okay, lambda in case satisfies this relation. Okay, because the exponential function is never zero. Actually always positive, strictly positive. So divided the both sides, spy into the lambda T. Then we get this one. So this is a solution. A nontrivial solution of that. K is a nontrivial back to satisfying this relation. In other words, A K O K is equal to lambda K. In other words a K minus lambda K. That is equal to zero. Okay, but because the cake is comin we'd better to write it as as a minus lambda times K. That is a zero vector. Right? That is a zero vector. Okay, Okay. Mhm. But you cannot subtract the scalar from the environ metrics unless n is equal to one. So that you'd better to write this one as a minus 11 times ice event. Right, Okay. Okay. So that we can make a subtraction of the n biometrics minus another environ matrix. Right, okay. What is isil went down there right, Where the is equal to and buy an identity matrix. Right, Let's summarize what we have the done so far. Okay, now we know the following things. Right? Actually is equal K times E to the lambda T were case another of the vector. Okay. Is a nontrivial solution of the problem of the given system? The homogeneous system, lambda and K satisfied this equation. Ok, look at this last equation. Okay, 10 minutes the word. Okay, okay. This is a simultaneous system of equations for the unknown vector. K. Right? We have k unknowns from K one and K two and K N. Okay, satisfying this linear system of equations. Okay. And we'd like to find the solution of this system of equation simultaneous system of equations where the case and non zero vector. Right, okay, this is simultaneous system of equations have a nontrivial solution. K. Right now we know that, concentrate on that one A minus lambda times myself and K. Has a nontrivial solution? Okay, solution vector say K. Okay. We know that if I don't leave. Okay, determinant. Herb a minus lambda times I Serbian and that is equal to zero. Right. Right. Can you remind us is such a factor from the linear? Right, okay. Mhm. What's the determinant? Right. Another notation for the determinant is a minus lambda survey in that is a 11 minus lambda A 12 and so on A one N. The second role will be a 21 and a 21 minus lambda and the ace of two N. The last rule is will be a N one, a N two. And so on a basis of n n minus lambda. Right, okay. That's the matrix enviro matrix a minus lambda times. Right. And we compute is determinant and the saturday, quarter zero. Right, Okay. If you compute the determinant of this environment a tricks then I think you will get the following polynomial equation and minus one to the end times lambda to the end plus some constant lambda to the n minus one plus and so on. Some suitable constant plus lambda and the constant tom. And that is equal to zero. Right, That's the determinant. Right, okay. Can you remind the name of this equation. Okay, that is the so called the characteristic equation of. Okay, okay, characteristic equation of the n biometrics. A right, okay, so now. Okay, let's have a space here. The last question we have, which is in fact the polynomial equation of degree and for the unknown lambda. Okay, so with the determinant of a minus lambda times I serve N is equal to zero. Okay, this equation is a so called characteristic equation of equation of the environ metrics. A Okay, okay, that is a polynomial equation in lambda. Okay, polynomial equation in lambda degree N. Okay. As you can see from this one. Okay, okay. And ruta saab this characteristic equation determinant of lambda by N is equal to zero. Right, okay. They are called the so called Eigen values of the determinant. A right, okay. Our Eigen various. Okay, I can values over there the environ matrix A right, okay. So what we have so far is the following thing. Right, okay. For x is equal to K times of E to the lambda. T is a solution of linear homogeneous system. X prime is equal to lambda of X. Okay, okay, we get the one necessary condition. Right, What lambda should be. Okay, lambda must satisfy this equation. In other words, lambda must be in Eigen value of the environ metrics. A Okay, that's a necessary condition. Right, okay. If this is the solution then we get lambda must be an Eigen value of the the coefficient matrix A right, okay. What can you say about the vector K then. Okay, for this I made a mistake. This is the equation is right, this is zero. And that equation okay, has a nontrivial solution. K. Lambda is an Eigen vector of Eigen value of the metrics A. How about the K. Then? Okay. If lambda is an Eigen value of A, that means okay, this equation should have at least one nontrivial solution for K. Okay, okay, so we have the name for that vector. Okay, so routers of this equation is to be called it as an Eigen values of A. And any and any responding. Okay, non zero Betta. Okay, satisfy a minus lambda survey in and the K is equal to zero. Okay, this is a zero vector. Right, Okay, it's cold. Called an Eigen vector right Eigen vector corresponding to to the given Eigen value. Lambda. Okay, okay. That we know quite well from the linear algebra. Okay, so let's summarize what we have done so far. Okay, let's re summarize it. Okay, now we can claim that. Okay, what do I have approved so far? Is Okay vector X. Of the tie some the non zero vector K times E. To learn the T. Okay, is a solution of the given system X. Prime is equal to their backs. Right? If and only. Okay, summarizing what we have done so far. We can conclude, okay, this type of this is, is the nontrivial solution. Right. More precisely, nontrivial solution of the given system lambda is an Eigen value of. Okay. Eigen value of K. Or Eigen value of the coefficient matrix A. And then K is on K. Is chris ponding Eigen vector? Ok, K. Is chris ponding I again back to Okay, for the Eigen value. Lambda. Okay, okay, okay. That's the conclusion we can make so far. Okay, so the problem of solving problem absolving this given linear force toward the homogeneous system, re constant coefficients is now reduced to find the Eigen values of the coefficient A. And the corresponding Eigen vectors way. Okay, looks like it is reduced to the problem. Okay, okay. Then depending on the the nature of the the Eigen value, lambda and the corresponding Eigen vectors to the top of the solution for a the type of the solution for this problem will be different. Okay,
