00:01
We have a matrix equals 1 0 minus 1 0 1 0 minus 1 0 1.
00:10
Now to find agent value of a, determinant a minus lambda i equals 0 1 minus lambda 0 minus 1 0 1 minus lambda 0 minus 1 0 1 minus lambda equals 0 1 minus lambda times 1 minus lambda square minus 0 minus 1 0 minus minus 1 times 1 minus lambda equals 0.
00:49
So, this implies 1 minus lambda cube minus 1 minus lambda equals 0.
00:58
So, this implies 1 minus lambda times 1 minus lambda square minus 1 equals 0.
01:07
So, this implies 1 minus lambda times 1 plus lambda square minus 2 lambda minus 1 equals 0.
01:16
So, this implies 1 minus lambda times lambda times lambda minus 2 equals 0.
01:24
So, this implies lambda equals 1 comma 0 comma 2.
01:28
Now let x y z be the agent vector corresponding to the agent value lambda equals 0, then a minus 0 i x y z equals 0.
02:03
So, we get x minus z equals 0 y equals 0 minus x plus z equals 0.
02:13
So, this implies x equals z.
02:16
Now if we take z equals 1, then x equals 1 and y equals 0.
02:28
Therefore, x y z equals 1 0 1 is an agent vector for lambda equals 0.
02:46
So, let x y z be the agent vector of a corresponding to the agent value lambda equals 1, then a minus 1 i x y z equals 0 that is 1 minus 1 0 minus 1 0 1 minus 1 0 minus 1 0 1 minus 1 x y z equals 0.
03:39
So, we get 0 0 minus 1 0 0 0 minus 1 0 0 x y z equals 0.
03:52
So, this implies minus z equals 0 and minus x equals 0 that is z equals x equals 0.
04:02
Therefore, x y z equals matrix 0 1 0 is an agent vector for lambda equals 1.
04:19
Now, agent vector corresponding to lambda equals 2 a minus 2 i x y z equals 0 matrix 1 minus 2 0 minus 1 0 1 minus 2 0 minus 1 0 1 minus 2 x y z equals 0 minus 1 0 minus 1 0 1 0 minus 1 0 minus 1 x y z equals 0.
05:08
So, this implies minus x minus z equals 0 y equals 0 minus x minus z equals 0 z equals minus x.
05:21
Now, take x equals 1 then z equals minus 1 y equals 0...