I got -2, 1, 3 as my eigenvalues but when I put the equation into wolframalpha it shows 5, 3, and -2 as the exponents over the e so I might have...

Hello!

The system has the form `x'=Ax,` where `A` is the matrix with the given constant coefficients.

If `lambda` is an eigenvalue of `A` and `r` is the corresponding eigenvector, then by definition `Ar=lambda r.` So if we take `x(t)=e^(lambda t)*r,` then `x'(t)=lambda*e^(lambda t)*r,` so such `x` is the solution of the system `x'=Ax.`

Actually, if there are 3 different real eigenvalues `lambda_1,` `lambda_2` and `lambda_3` and their corresponding eigenvectors are ` r_1,` `r_2` and `r_3,` then the general solution for `x'=Ax` is

` x(t)=C_1*e^(lambda_1 t)*r_1+C_2*e^(lambda_2 t)*r_2+C_3*e^(lambda_3 t)*r_3`

for any constants `C_1,` `C_2,` `C_3.`

So we have to find not only eigenvalues but also eigenvectors. I agree with you about eigenvalues, they are `lambda_1=1,` `lambda_2=-2` and `lambda_3=3.` To find an eigenvector corresponding to `lambda` , we have to solve `(A-lambda*I)r = 0.` In our case r's may be found in the form `(1, a, b).`

For example, for `lambda_1=1`

`(A-lambda_1*I)r=(A-I)*(1, a, b)=(-a+4b, 3+a-b, 2+a-2b)=0.`

Solving this simple linear system for `a` and `b` we obtain `a=-4` and `b=-1,` so the eigenvector is `r_1=(1, -4, -1).`

(I don't know how to draw matrices here, so vectors are written horizontal).

The same way we find `r_2=(1,-1,-1)` and `r_3=(1,2,1).` This gives us the general solution of the original system of equations.