I think I might just be having trouble with formatting my answer, because I'm fairly sure my work is right up until this point.
The question asks to find the general solution to $$X'= \begin{bmatrix} 3 & 2 \\ -1 & 1 \\ \end{bmatrix} X$$
I found that the eigenvalues and eigenvectors are $$\lambda_1=2+i,v_1=\begin{bmatrix} -1-i \\ 1\\ \end{bmatrix} \\ \lambda_1=2-i,v_1=\begin{bmatrix} -1+i \\ 1\\ \end{bmatrix}. $$ According to the formula, I should be getting this as the general solution: $$x(t)=c_1e^{2t}\left(\begin{bmatrix} -1 \\ 1\\ \end{bmatrix}\cos(t)-\begin{bmatrix} -1 \\ 0\\ \end{bmatrix}\sin(t)\right)+c_2e^{2t}\left(\begin{bmatrix} -1 \\ 1\\ \end{bmatrix}\sin(t)+\begin{bmatrix} 1 \\ 0\\ \end{bmatrix}\sin(t)\right) $$
What's confusing me is that Wolfram Alpha is giving me this, which looks like a different answer completely:
$$x(t)=\begin{bmatrix} 2c_2e^{2t}\sin(t) + c_1e^{2t}(\sin(t)+\cos(t))\\ c_2e^{2t}(\cos(t)-\sin(t))-c_1e^{2t}\sin(t) \\ \end{bmatrix} $$
I understand that it's in a different format, but is it still equal to my answer above?