Is this SISO (single input single output) or MIMO (multiple instead of single) system? If I transform wave equation for vibrating string Mx′′+Cx′+Kx=b(t) in linear system using 
$x_1(t)=x(t)$ and $x_2(t)=x_1^{'}(t)$ vibrating string equation becomes $Md_tx_2(t)+Cx_2(t)+Kx_1(t)=b(t)$. That is:
\begin{align}
\frac{d}{dt}\left[
   \begin{array}{c}
     x_1 \\
     x_2 \\
   \end{array}
 \right]&= \left[
   \begin{array}{cc}
     0 & 1 \\
     -M^{-1}K& -M^{-1}C \\
   \end{array}
 \right]\left[
   \begin{array}{c}
     x_1 \\
     x_2 \\
   \end{array}
 \right]+\left[\begin{array}{c}
   0 \\
   M^{-1}
 \end{array}\right]b(t)
\end{align}
We think of $x(t)$ as the output, $b(t)$ as the input and define output equation as:
$y=\left[\begin{array}{cc}
   1&0
 \end{array}\right]\left[\begin{array}{c}
   x_1 \\
   x_2
 \end{array}\right]$
So now, in the standard control theory notation of first equation:
$x'=Ax+Bu$ $y=Cx$
We make the following identifications:
$A= \left[
   \begin{array}{cc}
     0 & 1 \\
     -M^{-1}K& -M^{-1}C \\
   \end{array}
 \right], $ a $B=\left[\begin{array}{c}
   0 \\
   M^{-1}
 \end{array}\right]$
$C=\left[\begin{array}{cc}
   1&0
   \end{array}\right]$
$y(t)=x_1(t);u(t)=b(t).$
Shouldn't this be a MIMO system ( I conclude it from dimension of B)? I'm asking because I got information that this is an SISO system, but I can't figure it out?
 A: I don't understand why are you basing your definition as either SISO or MIMO on the dimensionality of $B$.
The same physical system (viberating string in your case), can be either SISO or MIMO depending on your configuration.
The question of classifying a system as SISO or MIMO depends on your control parameters, and the parameters which you "read out" or sample.
From the way I understand your question your control system scheme includes one control parameter $b(t)$, which indicates applied force, and one output parameter $x$ which indicates displacement.
Hence this is a SISO system.
If for instance you would have been interested also in the velocity of the string, you could have defined a second output variable $\dot{x}$. In this case you would've had:
$$
\begin{bmatrix}
y_1 & y_2
\end{bmatrix}=
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
x_1 \\ x_2
\end{bmatrix}
$$
In this case, the same system wouldn't have been SISO any more. The way to understand whether your SISO or MIMO isn't according to the dimensionality, but according to the non-zero entries in $B$ and $C$.
A: Although it can be generalized to accommodate an n-dimensional non-linear state space system, I'll describe for you a 2 dimensional linear state space system as you have posed in your specific example. This way to make matters simpler and more illustrative.
For this system, the $B$ vector is a 2 X 1 vector and $b(t)$ is a scalar. The $B$ vector is properly called the input coupling vector and its purpose is to couple the input, $b(t)$ into the state equations.
Similarly, the $C$ vector is a 1 X 2 vector and $y(t)$ is a scalar. The $C$ vector is properly called the output coupling vector and its purpose is to couple the state equations into the output, $y(t)$.
Since $b(t)$ and $y(t)$ are both scalars, the system is properly called Single Input, Single Output (SISO).
If $b(t)$ were rather a vector there would be 2 inputs thus making the system a multiple input system. Although $B$ does not directly determine whether the system is single or multiple input, it must conform to the dimensionality of $b(t)$ and in your example it would have to be 2 X 2 with the dimension of $b(t)$ taken as 2 X 1.
Likewise If $y(t)$ were rather a vector there would be 2 outputs thus making the system a multiple output system. Although $C$ does not directly determine whether the system is single or multiple output, it must conform to the dimensionality of $y(t)$ and in your example it would have to be 2 X 2 with the dimension of $y(t)$ taken as 2 X 1.
To summarize it's the dimensions of the input and output vectors that strictly define a single or multiple description of the system. The coupling matrices (or vectors) however must conform to their dimensions and properly describe how they couple to the state equations within.
