00:01
Hi, today we are going to solve the problem which we are given that x equals h by 2, 1 0 minus 1 equals h by 2 sigma 2.
00:16
First, in part a we have to find the characteristic equation, that is, determinant of a minus lambda i equal 0.
00:35
We can return characteristic equation of sx as determinant of hutch by 2 1 minus lambda 0 minus 1 minus lambda equals 0.
00:53
It can be written as h h h by 2 into 1 minus lambda into minus 1 minus lambda minus 0.
01:02
Equals 0.
01:04
On taking minus commonly outside and using the formula of a square minus b square we get h by 2 into minus 1 into 1 minus lambda square equals 0.
01:18
On further simplification we get lambda equals plus or minus 1.
01:24
Therefore eigen values are 1 minus 1.
01:31
Next we have to write the eigenvector that is h by 2 into 1 minus lambda 0 minus 1 minus lambda x y equals 0 0.
01:48
First on taking lambda equals 1 we get h by 2 0 0 minus 2 x y 0 0 0 0.
02:01
So first we have to consider x equals 1 then we get h by 2 into minus 2 into y equals 0.
02:13
It implies y equal 0.
02:17
Therefore e plus equals 1 comma 0...