00:01
In this question we are given this matrix 5 4 2 then we have 4 5 2 and then 2 2 2.
00:13
Let this matrix be a.
00:15
We need to find out the eigenvalues and eigenvectors.
00:19
So let us start with eigenvalues.
00:21
For that we will consider the characteristic equation which is determinant of a minus lambda i equal to 0.
00:31
This will be 5 minus lambda 4 2 then we have 4 5 minus lambda then 2 then 2 2 and this is 2 minus lambda this is equal to 0.
00:46
So determinant will be 5 minus lambda into 5 minus lambda into 2 minus lambda minus 4 then minus 4 into 4 into 2 minus lambda minus 4 then we have plus 2 into 8 minus 2 into 5 minus lambda equal to 0.
01:09
On simplifying this we will have this is 5 minus lambda this will be 10 this is minus 5 lambda minus 2 lambda.
01:20
So this is minus 7 lambda plus lambda square minus 4 then taking 4 common so this will be 16 into 2 minus lambda minus 1 plus 2 into 8 minus 10 minus 2 lambda equal to 0.
01:40
Now on simplifying all of this we will get that the characteristic equation is equal to lambda cube.
01:50
This is minus lambda cube plus 12 lambda square minus 21 lambda plus 10 equal to 0.
01:59
Now by hidden trial method if we put lambda equal to 1 in this equation so this will become 0.
02:07
That means this equation is satisfied.
02:09
So we can say that lambda minus 1 is one of the factor.
02:14
So we will be left with lambda square minus 11 lambda plus 10 equal to 0.
02:21
So splitting the middle term this can be written as minus 10 lambda minus lambda plus 10 equal to 0.
02:32
That means this will be lambda minus 1 into lambda minus 10 and again we will have lambda minus 1 equal to 0.
02:40
So lambda will be equal to 1 1 and 10.
02:44
These three will be the eigenvalues.
02:46
Now we need to calculate the corresponding eigenvectors as well.
02:50
So for lambda 1 equal to 1 we will consider the equation a minus lambda 1 i.
02:57
That means a minus i into v1 equal to 0.
03:00
So this will be equal to 5 minus 1 that is 4 4.
03:08
This is 2.
03:10
This is again 4 4 2 and this will be 2 2 1.
03:15
Let v1 be x y z and this is the 0 matrix.
03:20
Considering the augmented matrix.
03:23
So this will be 4 4 2 4 4 2 and this is 2 2 1 and here we have 0 0 0...