How to solve IVP using Laplace transform(of matrix)? $$x' = \begin{bmatrix} 1 &amp; 0 &amp; 0 \\ 2 &amp; 1 &amp; -2 \\ 3&amp;2 &amp; 1\end{bmatrix} x, ~~ x{(0)} = \begin{bmatrix} 2 \\ -1 \\ 1\end{bmatrix}$$

I am having very hard time solving this question using Laplace transform. I don't know how to do it for $3\times 3$ matrix. Would anyone be able to show me step by step how to do it for one of the cases? I have no clue how to do the inverse.
We are given:

$$X'(t) = \begin{bmatrix} 1 &amp; 0 &amp; 0 \\ 2 &amp; 1 &amp; -2\\ 3 &amp; 2 &amp; 1\end{bmatrix} \begin{bmatrix} x(t) \\ y(t)\\ z(t)\end{bmatrix}, ~~ X(0) = \begin{bmatrix} 2 \\ -1\\ 1\end{bmatrix}$$

We can write this as:

$$\tag 1 \begin{align} x' &amp;= x  \\ y' &amp;= 2x+y-2z \\ z' &amp;= 3x +2y + z \end{align}$$

Taking the Laplace transform of $(1)$ yields:

$$\begin{align} s x(s) - x(0) &amp;= x(s) \\ s y(s) - y(0) &amp;= 2x(s) + y(s) - 2z(s) \\ s z(s) - z(0) &amp;= 3 x(s) + 2y(s) + z(s)\end{align}$$

This reduces to the system:

$$\begin{bmatrix} s-1 &amp; 0 &amp; 0 \\ -2 &amp; s-1 &amp; 2\\ -3 &amp; -2 &amp; s-1\end{bmatrix} \begin{bmatrix} x(s) \\ y(s)\\ z(s)\end{bmatrix} = \begin{bmatrix} 2 \\ -1\\ 1 \end{bmatrix}$$

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


The inverse of the matrix is:


$$\begin{bmatrix} \frac{1}{s-1} &amp; 0 &amp; 0 \\
 \frac{2 (s-4)}{s^3-3 s^2+7 s-5} &amp; \frac{s-1}{s^2-2 s+5} &amp; -\frac{2}{s^2-2 s+5} \\
 \frac{3 s+1}{s^3-3 s^2+7 s-5} &amp; \frac{2}{s^2-2 s+5} &amp; \frac{s-1}{s^2-2 s+5} \end{bmatrix}$$

The inverse matrix times the column vector yields:

$$\begin{bmatrix} x(s) \\y(s)\\z(s) \end{bmatrix} = 
\begin{bmatrix} \frac{2}{s-1} \\ \frac{2 s}{s^2-2s+5}-\frac{3}{s-1} \\
 \frac{s (s+2)+5}{(s-1) (s^2-2s+5)}\end{bmatrix}$$

Next, just find the inverse Laplace Transform of $x(s), y(s), z(s)$ and you are done.

I will assume you can take it from here.
Since now it is not clear why I would post this (partial) answer, I would like to mention that I strongly feel that the accepted answer is not useful because it was quite probably derived using a computer algebra system. This is not a sin, however, most of the students are supposed to solve problems like this with just pencil and paper within quite limited time. In my answer I show how to do it without the extra unnecessary steps presented in the accepted answer.

When someone solves the problem of the form $\dot x=Ax$ by the Laplace transform method, it immediately implies that the application of the Laplace transform yields
$$
s X-x(0)=AX, 
$$
where $X=\mathcal L(x)$, or system
$$
(sI-A)X=x(0),
$$
which theoretically is solved by simply
$$
X=(sI-A)^{-1}x(0).
$$
This is a nice formula, but should never be used for actual calculations by hand. Much better to use usual elimination of variables procedure. In the given example, the system actually reads
$$
(s-1)x\quad \quad\quad\quad\qquad\qquad\quad=2\\
-2x+(s-1)y+\qquad 2z=-1\\
-3x\quad\,-2y+(s-1)z=1.
$$
From the first equation we immediately find 
$$
x=\frac{2}{s-1}.
$$
Now, what is left is
$$
-\frac{4}{s-1}+(s-1)y+\qquad 2z=-1\\
-\frac{6}{s-1}\quad\,-2y+(s-1)z=1.
$$
Multiply the first equation by 2 and the second by $(s-1)$ and add, you get
$$
(s^2-2s+5)z=\frac{s^2+2s+5}{s-1}.
$$
To find $y$ multiply the first equation in the same system by $(s-1)$ and the second by $-2$, add, and you find
$$
(s^2-2s+5)y=\frac{-s^2+4s-15}{s-1}.
$$
Now the whole system is solved (it took me around 5 minutes on paper).