00:01
Ok, the first step here is to find the eigenvalues.
00:04
To do that, we solve this determinant.
00:08
3 minus lambda minus 2, 1, 0 minus lambda equals 0.
00:17
In other words, let's take this number and this number, take away lambda from both of those and put it in that format.
00:25
And this will be 3 minus lambda times minus lambda minus minus 2 equals 0.
00:35
In other words, ad minus bc.
00:38
And this will be minus 3 lambda plus lambda squared plus 2 equals 0.
00:52
So lambda squared minus 3 lambda plus 2 is 0.
00:58
That will factorize nicely.
01:00
Lambda minus 1, lambda minus 2, which implies my two eigenvalues are 1 and 2.
01:15
Now to find the eigenvectors, what we need to do is take the matrix we began with, 3 minus 2, 1, 0, times that by eigenvector xy has to equal this number here, 1 times xy.
01:36
This comes from how we define eigenvectors.
01:43
Now from that, i can see that 3x minus 2y has to equal x and x is equal to y.
01:54
Both of those give the same result, that x equals y, which means my eigenvector has the form of 1, 1.
02:08
It could be 2, 2, 3, 3, your choice.
02:11
I'll always pick the lowest one possible.
02:13
So 1, 1 is an eigenvector.
02:19
And then same thing, 3 minus 2, 1, 0, x, y.
02:26
This time take the 2 from here.
02:31
So it's twice xy.
02:33
From that we have 3x minus 2y equals 2x, and then we have x equals 2y.
02:45
And those both give the same result, that x equals 2y.
02:50
So again, any vector of the form a, 2a is going to work.
02:56
So simple choice, choose 1, 2.
03:00
Sorry, not 1, 2, 2, 1.
03:08
The x value is twice the y value, so 2, 1.
03:11
So these then are my two eigenvectors.
03:19
Now next we have this format, p minus 1, ap equals d...