00:01
So, let us start with the concept which we are going to use here for this question to find the eigenvalues.
00:06
We have to use the characteristic equation determinant of a minus of lambda times pi eta is equals to 0, where lambda indicating the eigenvalues.
00:14
So, according to question, let's suppose the matrix is nothing but a having the values 1, 1, 1 by 25 and 1.
00:21
For that, in the first part, we have to find the eigenvalues.
00:24
So, first of all, let us apply the characteristic equation determinant of a minus of lambda i.
00:30
So, this should be equals to.
00:32
So, a is what 1 minus of lambda will be there, then here should be 1, then here should be 1 by 25 and here should be 1 minus of lambda.
00:43
So, this whole equation we have to equate it to 0.
00:48
Simplifying this, we get 1 minus lambda whole square minus of 1 by 25 is equals to 0.
00:53
This implies 1 square minus 2 lambda plus lambda square minus 1 by 25 is equals to 0.
00:59
So, this will give me lambda square minus 2 lambda 1 minus 1 by 25 will give me plus 24 by 25 is equals to 0.
01:08
So, we can make the factors like 6 by 5 multiplied by 4 by 5.
01:14
So, this will becomes lambda minus 6 by 5 and the second one will be lambda minus 4 by 5.
01:25
This will be equals to 0.
01:27
So, this will give me lambda 1 is equals to 6 by 5 and lambda 2 is equals to 4 by 5 as the eigenvalues.
01:34
So, this is going to be the first part answer.
01:37
Now, also we have to find the eigenvectors...