Code:
$$x(t) =
\left[ \begin{array}{c}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)\\
i_l(t)
\end{array}\right]
$$
Recall that the input voltage $V_g$ is an independent voltage source and the load impedence is $Z_l = R_l + sL_l$. Apparently, the load is short-circuited by $S_2$ in Mode 1.\\
The circuit equations in Mode 1 can be written in the state space form $K\cdot x = A_1\cdot x + B_1\cdot u$, that is,
$$\left[ \begin{array}{ccccc}
L_1 & 0 & 0 & 0 & 0\\
0 & L_2 & 0 & 0 & 0\\
0 & 0 & C_1 & 0 & 0\\
0 & 0 & 0 & C_2 & 0\\
0 & 0 & 0 & 0 & L_l
end{array}\right] \frac{d}{dt} \left[ \begin{array}{c}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)\\
i_l(t)
\end{array}\right] = \left[ \begin{array}{ccccc}
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
-1 & 0 & 0 & 0 & 0\\
0 & -1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & -R_l
\end{array}\right] \left[ \begin{array}{c}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)\\
i_l(t)
\end{array}\right]$$
Error:
! Extra \right.
l.25 end{array}\right]
\frac{d}{dt} \left[ \begin{array}{c}

\documentclass{...}and ending with\end{document}. – May 31 '15 at 17:07end{array}. – Sigur May 31 '15 at 17:18