Solution by Characteristic Value

Heres a problem that I have solved:

Find the general solution of the equation x' = Ax, where A is the given matrix:

[-2 4]

[1 1]

Using characteristic values, this matrix turns into:

[-2-λ 4]

[1 1-λ]

Simplifies to (λ + 3)(λ - 2) = 0

λ1 = 2

λ2 = -3

For λ1, (A - λI)k = 0

[-4 4][k1]=[0]

[1 -1][k2]=[0]

Solving: c1e^(2t)[1]

[1]

Set this up the same way with λ2:

[1 4][k1]=[0]

[1 4][k2]=[0]

Solving: c2e^(-3t)[4 ]

[-1]

Thus the solution is x(t) = c1c1e^(2t)[1] + c2e^(-3t)[4 ]

[1] [-1]

How do I go about solving for the general solution of the matrix:

[-2 0]

[0 -2]

I run into a brick wall pretty quick on this one.