How to solve a linear system in matrix form using Laplace transform? How to solve this linear system using Laplace transform?
$$\mathbf X'(t)=\left[\begin{array}{r,r,r}-3&0&2\\1&-1&0\\-2&-1&0\end{array}\right]\mathbf X(t); ~~~~~~~~\mathbf X(0)=\left[\begin{array}{r}4\\-1\\2\end{array}\right]$$
I am struggling with this problem. I tried by write it as   
$$\begin{cases}x_1' &= -3x_1+2x_3,\quad   x_1(0)=4\\
x_2'&= -1x_1+1x_2, \quad   x_2(0)=-1\\
x_3'&= -2x_1 -1x_2,    \quad x_3(0)=2 \end{cases}$$
Then Laplace transform both sides but I have hit a dead end. Is there any better way to solving this problem? I would be grateful for help. I did only simple problems so far. This is very complicated for me.  

We are given:
$$X'(t) = \begin{bmatrix} -3 & 0 & 2 \\ 1 & -1 & 0\\ -2 & -1 & 0\end{bmatrix} \begin{bmatrix} x(t) \\ y(t)\\ z(t)\end{bmatrix}, ~~ X(0) = \begin{bmatrix} 4 \\ -1\\ 2\end{bmatrix}$$
We can write this as:
$$\tag 1 \begin{align} x' &= -3x + 2z \\ y' &= x-y \\ z' &= -2x - y \end{align}$$
Taking the Laplace transform of $(1)$ yields:
$$\begin{align} s x(s) - x(0) &= -3 x(s) + 2 z(s) \\ s y(s) - y(0) &= x(s) - y(s) \\ s z(s) - z(0) &= -2 x(s) - y(s) \end{align}$$
This reduces to the system:
$$\begin{bmatrix} s+3 & 0 & -2 \\ -1 & s+1 & 0\\ 2 & 1 & s\end{bmatrix} \begin{bmatrix} x(s) \\ y(s)\\ z(s)\end{bmatrix} = \begin{bmatrix} -4 \\ 1\\ -2 \end{bmatrix}$$
All that is needed is to solve for $x(s),y(s),z(s)$ and then find the inverse Laplace Transform.
You could have also used many methods to solve this system, including eigenvalues/eigenvectors, matrix exponential, etc.
Update
If we find the inverse of the matrix on the left, we get:
$$\begin{bmatrix}
 \frac{s^2+s}{s^3+4 s^2+7 s+6} & -\frac{2}{s^3+4 s^2+7 s+6} & \frac{2 s+2}{s^3+4 s^2+7 s+6} \\
 \frac{s}{s^3+4 s^2+7 s+6} & \frac{s^2+3 s+4}{s^3+4 s^2+7 s+6} & \frac{2}{s^3+4 s^2+7 s+6} \\
 \frac{-2 s-3}{s^3+4 s^2+7 s+6} & \frac{-s-3}{s^3+4 s^2+7 s+6} & \frac{s^2+4 s+3}{s^3+4 s^2+7 s+6} \\
\end{bmatrix}$$
Multiplying that by the column vector on the right yields:
$$\begin{bmatrix} x(s) \\ y(s)\\ z(s)\end{bmatrix}  = \begin{bmatrix}
 -\frac{2 (2 s+2)}{s^3+4 s^2+7 s+6}-\frac{4 \left(s^2+s\right)}{s^3+4 s^2+7 s+6}-\frac{2}{s^3+4 s^2+7 s+6} \\
 -\frac{4 s}{s^3+4 s^2+7 s+6}+\frac{s^2+3 s+4}{s^3+4 s^2+7 s+6}-\frac{4}{s^3+4 s^2+7 s+6} \\
 \frac{-s-3}{s^3+4 s^2+7 s+6}-\frac{4 (-2 s-3)}{s^3+4 s^2+7 s+6}-\frac{2 \left(s^2+4 s+3\right)}{s^3+4 s^2+7 s+6} \\
\end{bmatrix}$$
We now want to find the inverse Laplace Transform of each component, for example:
$$x(t) = \mathscr{L^{-1}} \left(-\frac{2 (2 s+2)}{s^3+4 s^2+7 s+6}-\frac{4 \left(s^2+s\right)}{s^3+4 s^2+7 s+6}-\frac{2}{s^3+4 s^2+7 s+6} \right)$$
Thus,
$$x(t) = -e^{-2 t} \left(-\sqrt{2} e^t \sin \left(\sqrt{2} t\right)+2 e^t \cos \left(\sqrt{2} t\right)+2\right) $$
Hopefully, this gives you enough to do the other two, else you will not learn.
