Model reduction of states - State space model I'm going to do a state space model reduction and I wonder how it will be if I have a $3\times 3$ $A$ matrix?
Let's say that I have this state space model:
$$\begin{bmatrix}
\dot{x_1}\\ 
\dot{x_2}\\ 
\dot{x_3}
\end{bmatrix}= \begin{bmatrix}
-2 & 1  & 0\\ 
-3 & 0 &1 \\ 
-4 & 0 & 0
\end{bmatrix} \begin{bmatrix}
x_1\\ 
x_2\\ 
x_3
\end{bmatrix}+\begin{bmatrix}
0\\ 
1\\ 
2
\end{bmatrix}u \\ y = \begin{bmatrix}
1 & 0 &0 
\end{bmatrix}\begin{bmatrix}
x_1\\ 
x_2\\ 
x_3
\end{bmatrix}+ \begin{bmatrix}
0
\end{bmatrix}u$$
To reduce one state e.g $x_3$ I need to split up the matrices into pieces.
$$\begin{bmatrix}
\dot{x}_1\\ 
\dot{x}_0
\end{bmatrix} = \begin{bmatrix}
A_{11} & A_{12} \\ 
A_{21} & A_{22}
\end{bmatrix}\begin{bmatrix}
x_1\\ 
x_0
\end{bmatrix} + \begin{bmatrix}
B_1\\ 
B_2
\end{bmatrix} u \\ y = \begin{bmatrix}
C_1& C_2
\end{bmatrix}\begin{bmatrix}
x_1\\ 
x_0
\end{bmatrix} + \begin{bmatrix}
D
\end{bmatrix}u$$
I assume that $\dot{x}_0 = 0$. Then I can write:
$$x_0 =  - A_{22}^{-1}A_{21}x_1 - A_{22}^{-1}B_0u$$
Then I write:
$$\dot{x}_1 = (A_{11} - A_{12}A_{22}^{-1}A_{21})x_1 +(B_1 - A_{12}A_{22}^{-1}B_2)u \\ y = (C_1 - C_2A_{22}^{-1}A_{21})x_1 +(D - C_2A_{22}^{-1}B_2)u$$
My question is, if I want to reduce state $x_3 = x_0$. How should I distribute $A$ to $A_{11}, A_{12}, A_{21}, A_{22}$ and $C$ to $C_1, C_2$ and $B$ to $B_1, B_2$ ?
 A: In order to simplify the state space model when $x_2 = x_3$ you can just substitute in $x_2 = x_3 = z$ which gives
$$
\begin{bmatrix}
\dot{x}_1 \\ \dot{z} \\ \dot{z}
\end{bmatrix} = 
\begin{bmatrix}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{bmatrix}
\begin{bmatrix}
x_1 \\ z \\ z
\end{bmatrix} + 
\begin{bmatrix}
B_1 \\ B_2 \\ B_3
\end{bmatrix} u. \tag{1}
$$
Equating both expressions for $\dot{z}$ gives
$$
(A_{21} - A_{31})\,x_1 + (A_{22} + A_{23} - A_{32} - A_{33})\,z + (B_2 - B_3)\,u = 0. \tag{2}
$$
Solving this for $z$ gives
$$
z = (A_{22} + A_{23} - A_{32} - A_{33})^{-1} \left((A_{31} - A_{21})\,x_1 + (B_3 - B_2)\,u\right) \tag{3}
$$
and therefore the dynamics of $x_1$ can also be written as
$$
\dot{x}_1 = \left(A_{11} + (A_{12} + A_{13})\,(A_{22} + A_{23} - A_{32} - A_{33})^{-1} (A_{31} - A_{21})\right)\,x_1 + \left(B_1 + (A_{12} + A_{13})\,(A_{22} + A_{23} - A_{32} - A_{33})^{-1} (B_3 - B_2)\right)\,u. \tag{4}
$$
However equating both expressions for $\dot{z}$ and solving for $z$ does not actually ensure that the two expressions for it are actually equal to the time derivative of $z$. In order to reduce the size of the following expressions I will define the following matrices
$$
A_z = (A_{22} + A_{23} - A_{32} - A_{33})^{-1} (A_{31} - A_{21}) \tag{5}
$$
$$
B_z = (A_{22} + A_{23} - A_{32} - A_{33})^{-1} (B_3 - B_2) \tag{6}
$$
such that $z = A_z\,x_1 + B_z\,u$. Taking the time derivative of this gives
$$
\dot{z} = A_z\,\dot{x}_1 + B_z\,\dot{u}. \tag{7}
$$
Substituting $(4)$ into $(7)$ gives
$$
\dot{z} = A_z\,\left(A_{11} + (A_{12} + A_{13})\,A_z\right)\,x_1 + A_z\,\left(B_1 + (A_{12} + A_{13})\,B_z\right)\,u + B_z\,\dot{u} \tag{8}
$$
which should be equal to one of the expressions for $\dot{z}$ from $(1)$. But since both are equivalent I will just use the first, which gives
$$
\dot{z} = \left(A_{21} + (A_{22} + A_{23})\,A_z\right)\,x_1 + \left(B_2 + (A_{22} + A_{23})\,B_z\right)\,u. \tag{9}
$$
I am not sure what your end goal would be but by equating the right hand sides of $(8)$ and $(9)$ you could either solve for $x_1$, $u$ or $\dot{u}$. But since the context of the question is model reduction I would assume you would want to solve for $x_1$. Doing so gives
$$
x_1 = \left(A_{21} + M\,A_z - A_z\,A_{11}\right)^{-1} \left(\left(A_z\,B_1 - M\,B_z - B_2\right)\,u + B_z\,\dot{u}\right) \tag{10}
$$
with 
$$
M = A_{22} + A_{23} - A_z\,(A_{12} + A_{13}). \tag{11}
$$
However equation $(10)$ might have multiple solutions if the dimension of $x_1$ is larger than that that of $z$. But if their dimensions are of equal size and the matrix inverse in $(10)$ exists, then plugging all this into the definition of $y$ gives that $y$ is a direct linear function of $u$ and $\dot{u}$, which would make it a non-causal system (so a system with no proper transfer function).
