Eigenvectors:For λ = i, we subtract i down the diagonal of the matrix.[-2-i...1.......
x₁' = (−2B − A + 2B) sin t + (−2A + 2A + B) cos t
x₁' = −A sin t + B cos t
x₂' = −5 (A cos t + B sin t) + 2 ((2A + B) cos t + (−A + 2B) sin t)
x₂' = (−5B − 2A + 4B) sin t + (−5A + 4A + 2B) cos t
x₂' = (−2A − B) sin t + (−A + 2B) cos t
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Eigenvalues:
We solve the characteristic equation λ^2 - 0λ + 1 = 0 ==> λ = -i, i.
Eigenvectors:
For λ = i, we subtract 'i' down the diagonal of the matrix.
[-2-i...1.|.0]
[-5...2-i.|.0]
Multiply the top row by 2-i:
[-5...2-i.|.0]
[-5...2-i.|.0].
This row reduces to
[1.(2-i)/5|0]
[0......0.|0]; this yields eigenvector (5, -2+i)^t.
For real solutions to the matrix DE, we isolate the real and imaginary parts of
[5....]e^(it)
[-2+i]
This simplifies to
[5....] (cos t + i sin t) =
[-2+i]
[5 cos t + i * 5 sin t]
[(-2 cos t - sin t) + i (cos t - 2 sin t)].
So, extracting real and imaginary parts yields
[5 cos t]
[-2 cos t - sin t], and
[5 sin t]
[cos t - 2 sin t], respectively.
Hence, a general solution (in real form) is
x1(t) = A(5 cos t) + B * (5 sin t)
x2(t) = A(-2 cos t - sin t) + B(cos t - 2 sin t).
