00:01
Okay, we want to solve this matrix differential equation.
00:04
So x prime is going to be like a t derivative, and then a a is this matrix.
00:11
So the easiest way to do this is to start by finding the eigenvalues of a.
00:20
So we would have to expand this determinant, which will give us an equation for lambda.
00:30
Lambda will be the eigenvalues.
00:35
Okay, so this is an easy determinant.
00:38
To expand because as they point out in the problem, there are a lot of zeros.
00:47
The factor out to 4 minus lambda right away.
00:51
And we get a 3 by 3 determinant left over.
00:56
And it's easy to see that the only non -zero terms factors are going to look like this.
01:08
Okay.
01:09
So that means that my possible eigenvalues are 4 minus 4, 5.
01:16
And minus.
01:24
So now we need the eigenvectors.
01:33
So the way to get them is to write out an x that has, you know, some unknown vector substituted into the equation for each lambda.
01:52
So our eigenvectors come out to be this.
02:16
I'm going to call those v1 or va, vb, vc, and vd.
02:25
Okay, we can call them anything we want, but i'm doing it this way to not confuse it with the x's.
02:32
It's easy to show that each of these vectors is the associated eigenvector with the particular eigenvalue.
02:43
We can see that they're linearly independent, which is what we need...