I am sure this is a very basic procedure, but I am stuck. Can someone please take me through how you determine the solution of the following? Thanks, Frostking
(1 - x) (3 - x) = 1
also
(2 - x) (- 2 - x) = -5
I need these to determine Eigenvalues and I am lost.
Thanks very much for showing me that I did not have to use i on that one. Could you please tell me how to if I do, say on this one:
I have ( 1 - lambda) (- 3 - lambda) + 5 = 0
So I get -3 + 2 lambda + lambda^2 + 5 = 0
lambda^2 + 2 lambda + 2 = 0
I know from the book that the answer is -1 + i and - 1 - i but I do not know how to get this. Do I just plug values into the quad formula? Thanks for your time. Frostking