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.  
 A: 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.
A: You have a system $\mathbf X'=\mathbf{AX}$. Call $\mathbf x=\mathscr L\{\mathbf X\}$. Then if you apply Laplace transform to the initial differential equations you will obtain $s\mathbf x-\mathbf x(0)=\mathbf{Ax}$. This implies that 
$$(s\mathbf I-\mathbf A)\mathbf x=\mathbf x(0)\Longrightarrow \mathbf x = (s\mathbf I-\mathbf A)^{-1}\mathbf x(0)\Longrightarrow \mathbf X = \mathscr L^{-1}\{(s\mathbf I-\mathbf A)^{-1}\mathbf x(0)\}.$$ Which is more or less the situation described by Amzoti (he expanded the system of equations as you wanted to do, here we are using matrix exponential with Laplace transform).
So:


*

*step 1: Write  $(s\mathbf I-\mathbf A).$  

*step2: Find $(s\mathbf I-\mathbf A)^{-1}$, this is a problem of linear algebra actually: finding inverses. You can check this out on a linear algebra book.  

*step 3: Multiply this inverse by the initial condition (again you should know how to multiply a matrix by a vector).

*step 4: Check if you can apply inverse of Laplace transform (you could use partial fractions for each entry of your matrix, generally this is the most common problem when applying this method).

*step 5: Apply inverse of Laplace transform.

