- #1
Justin90
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1.
Find the corresponding Eigenline
A= (3 -3 2 -4)
A=(a b c d)
k2-(a+d)k+(ad-bc)=0
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 3y)
Hence 3y=0 and 2x-7y=0
Eigenvector for k=-2
(3 -3 2 -4)(x y) = -2(x y)
(3x -3y 2x -4y)=(-2x -2y)
Hence 5x-3y=0 and 2x-2y=0
This is where I get confused because my notes say that 3y=0 and 2x-7y=0 need to reduce to a single equation, and similarly 5x-3y=0 and 2x-2y=0 need to reduce to a single equation to give the eigenline for the given eigenvalue?
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 3y)
Hence 3y=0 and 2x-7y=0
Eigenvector for k=-2
(3 -3 2 -4)(x y) = -2(x y)
(3x -3y 2x -4y)=(-2x -2y)
Hence 5x-3y=0 and 2x-2y=0
This is where I get confused because my notes say that 3y=0 and 2x-7y=0 need to reduce to a single equation, and similarly 5x-3y=0 and 2x-2y=0 need to reduce to a single equation to give the eigenline for the given eigenvalue?