State-Space Realization of a Matrix of Transfer Functions I am uncertain how to perform some steps in this state-space realization.
Given $G(s) =$
\begin{bmatrix}
    \frac{1}{s^3}-\frac{1}{s+1}       & \frac{1}{s^2}+\frac{1}{s+1} \\
    \frac{1}{s^2}       & \frac{1}{s}
\end{bmatrix}
I can split the matrix into
$$
G(s) = 
\begin{bmatrix}
    \frac{1}{s} \\
    1 
\end{bmatrix}
\frac{1}{s}
\begin{bmatrix}
    \frac{1}{s} & 1 
\end{bmatrix}
+
\begin{bmatrix}
    1 \\
    0 
\end{bmatrix}
\frac{1}{s+1}
\begin{bmatrix}
    -1 & 1 
\end{bmatrix}.
$$
Then I do not know how to achieve the next two steps
$$
G = 
\begin{bmatrix}
    0 & 1 \\
    1 & 0 \\
    0 & 1
\end{bmatrix}
\begin{bmatrix}
    0 & 1 \\
    1 & 0 
\end{bmatrix}
\begin{bmatrix}
    0 & 1 & 0 \\
    1 & 0 & 1 
\end{bmatrix}
+
\begin{bmatrix}
    -1 & -1 & 1 \\
     1 & 0 & 0 \\
     0 & 0 & 0
\end{bmatrix}
$$
And this step to the state-space realization
$$
\begin{bmatrix}
    A & B \\
    C & D 
\end{bmatrix}
=
\begin{bmatrix}
    0 & 1 & 0 & 0 & 0 & 0 \\
    0 & 0 & 1 & 0 & 0 & 1 \\
    0 & 0 & 0 & 0 & 1 & 0 \\
    0 & 0 & 0 & -1 & -1 & 1 \\
    1 & 0 & 0 & 1 & 0 & 0 \\
    0 & 1 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
The numerical matrices are in a compact notation form of the $A,B,C,D $ matrices.
 A: I think you are making a mistake where you are combining two parallel systems:
$$
\begin{cases}
\boldsymbol{\dot x}_1=\boldsymbol{A}_1 \boldsymbol{x}_1+\boldsymbol{B}_1 \boldsymbol{u}_1 \\
\boldsymbol{y}_1=\boldsymbol{C}_1 \boldsymbol{x}_1+\boldsymbol{D}_1 \boldsymbol{u}_1
\end{cases}
$$
$$
\begin{cases}
\boldsymbol{\dot x}_2=\boldsymbol{A}_2 \boldsymbol{x}_2+\boldsymbol{B}_2 \boldsymbol{u}_2 \\
\boldsymbol{y}_2=\boldsymbol{C}_2 \boldsymbol{x}_2+\boldsymbol{D}_2 \boldsymbol{u}_2
\end{cases}
$$
Then the corresponding compact matrix is 
$$
\begin{bmatrix}
\boldsymbol{A}_1&\boldsymbol{0}&|&\boldsymbol{B}_1 & \boldsymbol{0} \\
\boldsymbol{0}&\boldsymbol{A}_2&|&\boldsymbol{0}&\boldsymbol{B}_2 \\
\hline
\boldsymbol{C}_1&\boldsymbol{0}&|&\boldsymbol{D}_1 & \boldsymbol{0} \\
\boldsymbol{0}&\boldsymbol{C}_2&|&\boldsymbol{0}&\boldsymbol{D}_2 \\
\end{bmatrix}
$$
But if you cascade them, the result is 
$$
\begin{bmatrix}
\boldsymbol{A}_1&\boldsymbol{0}&|&\boldsymbol{B}_1  \\
\boldsymbol{B}_2\boldsymbol{C}_1&\boldsymbol{A}_2&|&\boldsymbol{B}_2 \boldsymbol{D}_1 \\
\hline
\boldsymbol{D}_2\boldsymbol{C}_1&\boldsymbol{C}_2&|& \boldsymbol{D}_2\boldsymbol{D}_1 \\
\end{bmatrix}
$$
Are you following the parallel and cascade rules?
If hard to follow, you can simply use the traditional methods:
$$
\begin{cases}
y_1=x_3-x_4+x_6+7_7 \\
y_2=x_2+x_5
\end{cases}
$$
where
$$
\begin{cases}
x_1=\frac1s u_1 \\
x_2=\frac1{s^2} u_1 =\frac1s x_1\\
x_3=\frac1{s^3} u_1 =\frac1s x_2\\
x_4=\frac1{s+1} u_1 \\
x_5=\frac1s u_2 \\
x_6=\frac1{s^2} u_2 =\frac1s x_5\\
x_7=\frac1{s+1} u_2
\end{cases}
$$
Now easy to build state-space representation:
$$
\frac{d }{d t}\boldsymbol{x}=
\frac{d }{d t}
\begin{bmatrix}
x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7
\end{bmatrix}
=
\begin{bmatrix}
0\\x_1\\x_2\\-x_4\\0\\x_5\\-x_7
\end{bmatrix}
+
\begin{bmatrix}
u_1\\0\\u_1\\u_2\\0\\u_2
\end{bmatrix}
$$
You can easily build $\boldsymbol{A}$ and $\boldsymbol{B}$ where
$$\boldsymbol{C}=\begin{bmatrix}
0&0&1&-1&0&1&1\\
0&1&0&0&1&0&0
\end{bmatrix}$$
$$\boldsymbol{D}=\begin{bmatrix}
0&0\\
0&0
\end{bmatrix}$$
