
Yefim S. answered 08/03/20
Math Tutor with Experience
Your equation is right, only igenvalue not 2 λ but λ = - 2 of multiplicity 2, but igenvector (- 1, 1, 0) right and for igenvalue λ = 1 igenvector is (1, - 1,1).
Now we have 1 independent solution: (1, - 1, 1)et and second independent solution: (-1, 1, 0)e-2t .
To get 3rd independent solution: (A + 2E)(x, y, z) = (-1, 1, 0); 4x + 4y + 3z = -1, - 4x - 4y - 3z = 1, x + y + z = 0
x = 0, y = - 1/4, z = 1/4. We get one more igenvector: (0, -1, 1).
Now solution:x(t) = C1(1,- 1, 1)et + C2(- 1, 1, 0)e-2t + C3[(- 1, 1, 0)t + (0, - 1, 1)]e-2t.
Here rows mean columns