I have been trying to solve this system of equations using Laplace transforms for a while. It is very easy to solve it using eigenvalues and eigenvectors, but when I tried to do it using Laplace I ended up with crazy polynomials for X(s), Y(s), and Z(s); I assume that I must have done something wrong, so what is the proper method to solve such a system using Laplace Transform. Any ideas?
$x'= x- y+4z; x(0)=1$
$y'=3x+2y- z; y(0)=1$
$z'=2x+ y- z; z(0)=0$
Denote $$A=\begin{pmatrix}1 &-1 & 4\\ 3 & 2 & -1\\ 2 & 1 & -1\end{pmatrix}$$ Then your equation is $Y'=AY$. You know that the solution will be $Y=e^{At}C$ with $C^T=\begin{pmatrix}1 & 1 & 0\end{pmatrix}$. So you want to find $e^{At}=\mathcal{L}^{-1}\left((sI-A)^{-1}\right)$. Now, let's find $(sI-A)^{-1}$ using the adjoint matrix: $$sI-A=\begin{pmatrix}s-1 &1 & -4\\ -3 & s-2 & 1\\ -2 & -1 & s+1\end{pmatrix} \hspace{10pt}\Rightarrow\hspace{2pt} \operatorname{Adj}(sI-A)=\begin{pmatrix}s^2-s-1& 3-s & 4s-7\\ 3s+1 & s^2-9 & 13-s\\ 2s-1 & s-3 & s^2-3s+5\end{pmatrix}$$ And so $\det(sI-A)=s^3-2s^2-5s+6$ (by multiplying, for example, the first row in $sI-A$ by first column in the adjoint) which factors as $\det(sI-A)=(s+2)(s-1)(s-3)$. As $(sI-A)^{-1}=\frac1{\det(sI-A)}\operatorname{Adj}(sI-A)$, what remains is to multiply, and calculate the reverse transform.
Hope this helps.