Jordan Canonical Form matrix
It would be \$ \left(
\begin{array}{cccc}
 0 &amp; 0 &amp; 1 &amp; 1 \\
\end{array}
\right)^T\$. 

The reasoning is as follows. From the output equation, we know that

$$ X_1(s)=\frac{1}{(s+1)^3}U(s) \ \ \ X_2(s)=\frac{1}{(s+1)^2}U(s) \ \ \ X_3(s)=\frac{1}{s+1}U(s) \ \ \ X_4(s)=\frac{1}{s}U(s)$$

This can be simplified as: 
$$ X_1(s)=\frac{1}{s+1}X_2(s) \ \ \ X_2(s)=\frac{1}{s+1}X_3(s) \ \ \ X_3(s)=\frac{1}{s+1}U(s) \ \ \ X_4(s)=\frac{1}{s}U(s)$$

Now we are beginning to see that the input only appears in the last two equations.

Converting these to the time domain, using the inverse Laplace transform, we get the following equations:

$$ \dot{x}_1+x_1=x_2$$
$$ \dot{x}_2+x_2=x_3$$
$$ \dot{x}_3+x_3=u$$
$$ \dot{x}_4=u$$

which is the desired Jordan form.