00:01
We're going to solve the initial value problem for the vector value function x.
00:08
We have the differential equation, the vector differential equation x derivative equal a x, where a is a constant matrix negative 1, negative 4, 2, 5, and x at t equals 0, or x at 0 is 1, negative 4.
00:25
So formally this differential equation should be written maybe this way in order to emphasize the dependence of x with respect to variable t.
00:37
So we want this equation to be true for any value of t.
00:42
And the initial condition, initial value is the vector 1, 84, which is the value of vector value function x at t equals 0.
00:54
So what we do first is we calculate the eigenvalues and associated eigenvectors of matrix a.
01:11
So eigenvalues of a and associated eigenvectors.
01:21
Okay so we write down the characteristic polynomial of a that is the determinant of a minus lambda times identity 2 by 2 matrix let's call it i2 so that will be the determinant of the the 2 by 2 matrix, negative 1 minus lambda, then negative 4, then 2, and then 5 minus lambda, and that is negative 1 minus lambda times 5 minus lambda minus 2 times negative then that is negative 5 plus lambda minus 5 lambda plus lambda squared plus 8.
02:33
So, pa of lambda is equal to lambda squared minus 4 lambda.
02:46
8 minus 5 is 3.
02:48
And now we see we can factor out this polynomial as, okay, before that, let me say this is the characteristic polynomial of a.
03:17
So the characteristic equation will be pa of lambda equals zero.
03:22
So the solutions of the characteristic equation, which are just the zeros of the characteristic polynomial, are the eigenvalues of a.
03:30
So we got to solve the equation pa of lambda equals 0.
03:43
That is lambda squared minus 4 lambda plus 3 equals 0.
04:01
And what we're going to do here is factorize this polynomial, second degree polynomial.
04:08
So we have that lambda squared minus 4 lambda plus 3 equals 0 is equivalent to the factorization is lambda minus 1 times lambda minus 3.
04:23
That equals 0.
04:24
That is because if we develop this product here, we get just this polynomial here.
04:32
And so we know from this factorization that lambda can be 1 or 3.
04:38
So lambda is equal to 1 over lambda is 3.
04:51
So then the first answer you get to enter is the eigenvalues of a.
05:02
The eigenvalues of a are 1 and 3.
05:11
You get to enter the two values separated by a comma.
05:17
But that's the first answer, let's say.
05:21
And because the two eigenvalues are different, then we what we need now is to calculate corresponding eigenvector eigenvector for lambda 1 equal 1 okay good so we got to solve the equation a x the muse in a letter az equal lambda lambda 1 z for z different from 0 and that we are sure we can solve we must be able to find an unzero vector z for which a z equal lambda 1 z because by definition vector must exist that's another definition of eigenvalue of the matrix and so we get to that equation that is a is one and yet the wanted for 25 time that implies linear system negative z1 minus 4 4z2 equals z1 and 2z1 plus 5z2 equals z2.
07:19
Negative 4z2 equals 2z1 and the second one is 2z1 equal negative 4z2.
07:47
That's just the same equation.
07:59
Two of sides we get z1 equal negative negative 2, 1 also is 0, and we get the 0, 0 vector, which is never an eigenvector for an eigenvalue...