Thanks for that! Couldn't see the wood through the trees. Got it now.
K^2+k-6=0
(k-2)(k+3)=0
So k=2 and k=-3
So eventual both equations reduce to x=y/2
Can't believe I missed that.
[b]1.
Find the corresponding Eigenline
A= (3 -3 2 -4)
Homework Equations
A=(a b c d)
k2-(a+d)k+(ad-bc)=0
The Attempt at a Solution
k2-(3-(-4))k+(3(-4)-(-3)2)=0
k2+k-6=0
(k+2)(k-3)
So k=-2 and k=3
Eigenvector for k=3
(3 -3 2 -4)(x y) = 3(x y)
(3x -3y 2x -4y)= (3x...