00:01
In this problem, we have the matrix a as 3, 0, minus 2, 0, 4, 0, minus 2, 0, 3.
00:11
So, the first step is to find the eigenvalues for which we solve the determinant a minus lambda i and equate that to 0.
00:21
So, this is nothing but 3 minus lambda 0, minus 2, 0, 4 minus lambda 0, minus 2, 0, 3 minus lambda.
00:34
This must be 0.
00:35
Opening this, this is nothing but 3 minus lambda, 4 minus lambda, 3 minus lambda, minus 2, 2 times 4 minus lambda.
00:49
This must be equals to 0.
00:51
So, taking 4 minus lambda common, we have 3 minus lambda square, which is 9 plus lambda square minus 6 lambda.
01:00
And then we have minus 4.
01:03
This must be equals to 0.
01:05
We have 4 minus lambda into lambda square minus 6 lambda plus 5 equals to 0.
01:13
Or this is 4 minus lambda, lambda minus 5, lambda minus 1 equals to 0.
01:19
Or the 3 eigenvalues are 1, 4, and 5.
01:24
Next step is to find the eigenvectors.
01:29
So, for the first eigenvalue, which is 1, the eigenvector will be found by solving a minus lambda times the eigenvector.
01:40
This must be equals to 0 vector.
01:43
So, a minus lambda i.
01:45
So, this is 2, 0, minus 2.
01:48
This is 0, this becomes 3.
01:50
This is 0, this is minus 2, 0 times x, y, z.
01:56
This must be 0, 0, 0.
01:59
From here we get 2x minus 2z equals to 0.
02:04
And we get 3y equals to 0.
02:07
From simply here we get y is 0 and here we get x as z.
02:11
So, the first eigenvector v1 is x is z, taking z as 1, y is 0 itself, and again z is 1.
02:21
So, this is the first eigenvector...