In this exercise the state space representation of the imaged system is asked for. $$G_1(s) = \frac{s-1}{s+2} = 1 - \frac{3}{s+2} G_2(s)=\frac{1}{s-1}$$ I can see that $G_1(s)$ is "able to leap" (hope it is the correct translation of sprungfähig), because nominator and denominator have the same order.
So for the system matrix I get $$A = \begin{pmatrix} -2 & 3 \\1 & 0 \end{pmatrix}$$
That should be correct.
But I am not sure with B and C.
Can I get both by looking at the image? Because that is what I did and it looks plausible.
$$B = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$$
$$C = \begin{pmatrix} 0 \\1 \end{pmatrix}$$
And does the output y look like this, because of the leapable ability? $$ y= \begin{pmatrix} 0 & 1 \end{pmatrix} x + d = \begin{pmatrix} 0 & 1 \end{pmatrix} x + 1$$
Differential equations: $$\frac{dx_1}{dt} = -2x_1+3x_2-3r$$ $$\frac{dx_2}{dt} = x_1+r$$
Matrix B
Matrix B is the control matrix, and determines how the system input affects the state change. If the state change is not dependent on the system input, then B will be the zero matrix.
Matrix C
Matrix C is the output matrix, and determines the relationship between the system state and the system output.
$$ G_1G_2 = \frac{1}{s+2} $$ Then, the transfer function from $r$ to $y$ is given by $$ G_{yr} = \frac{G_1 G_2}{1+G_1G_2} = \frac{1}{s+3} $$ A state space representation for this is $$ \left[ \begin{array}{c|c} A &B\\ \hline C& D \end{array}\right]= \left[ \begin{array}{c|c} -3 &1\\ \hline 1& 0 \end{array}\right] $$