Well her goes: You subtract the eigenvalues on the main diagonal and you get the equation for λ by zeroing out the determinant: (-2-λ)^{2} + 4 = 0 → λ = -2 ± -2i

Plugging the eigenvalue, -2 + 2i, back into the matrix, you get row 1 as -2i 1 and row 2 as -4 2i. This implies an eigenvector, ξ, of (1 2i)^{T}

Since we have a complex eigenvalue we will use the form (ξ)e^{-2t}(cos(2t) + isin(2t))

Separating the real and imaginary parts, we obtain:

e^{-2t} | cos(2t) | + e^{-2t} | sin(2t) | = |x|

| -2sin(2t)| | 2cos(2t)| |y|

I hope that helps. I checked X', but not Y' equation.

JACQUES D.

07/30/20

Fizaa A.

Should there be constants and I in the answers or not?07/30/20