Consider the system defined by the following equation:
$$
\frac{Y(s)}{U(s)} = \frac{s + 3}{s^2 + 3s + 2}
$$
What is the controllable canonical form of its state-space representation?

Options:
A. $$
\begin{array} {c} {{{{\left[ \begin{array} {l} {{{{\dot{x}}_{1} ( t )}}} \\ {{{{\dot{x}}_{2} ( t )}}} \\ \end{array} \right]=\left[ \begin{array} {l} {{0}} & {{1}} \\ {{-4}} & {{-1}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]+\left[ \begin{array} {l} {{1}} \\ {{0}} \\ \end{array} \right] u ( t )}}}} \\ {{{{y ( t )=\left[ \begin{array} {l} {{1}} & {{3}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]}}}} \\ \end{array}
$$
B. $$
\begin{array} {c} {{{{\left[ \begin{array} {l} {{{{\dot{x}}_{1} ( t )}}} \\ {{{{\dot{x}}_{2} ( t )}}} \\ \end{array} \right]=\left[ \begin{array} {l} {{0}} & {{1}} \\ {{-3}} & {{-3}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]+\left[ \begin{array} {l} {{1}} \\ {{0}} \\ \end{array} \right] u ( t )}}}} \\ {{{{y ( t )=\left[ \begin{array} {l} {{2}} & {{2}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]}}}} \\ \end{array}
$$
C. $$
\begin{array} {c} {{{{\left[ \begin{array} {l} {{{{\dot{x}}_{1} ( t )}}} \\ {{{{\dot{x}}_{2} ( t )}}} \\ \end{array} \right]=\left[ \begin{array} {l} {{0}} & {{1}} \\ {{-1}} & {{-4}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]+\left[ \begin{array} {l} {{0}} \\ {{1}} \\ \end{array} \right] u ( t )}}}} \\ {{{{y ( t )=\left[ \begin{array} {l} {{2}} & {{1}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]}}}} \\ \end{array}
$$
D. $$
\begin{array} {c} {{{{\left[ \begin{array} {l} {{{{\dot{x}}_{1} ( t )}}} \\ {{{{\dot{x}}_{2} ( t )}}} \\ \end{array} \right]=\left[ \begin{array} {l} {{0}} & {{1}} \\ {{-2}} & {{-2}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]+\left[ \begin{array} {l} {{1}} \\ {{0}} \\ \end{array} \right] u ( t )}}}} \\ {{{{y ( t )=\left[ \begin{array} {l} {{4}} & {{1}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]}}}} \\ \end{array}
$$
E. $$
\begin{array} {c} {{{{\left[ \begin{array} {l} {{{{\dot{x}}_{1} ( t )}}} \\ {{{{\dot{x}}_{2} ( t )}}} \\ \end{array} \right]=\left[ \begin{array} {l} {{0}} & {{1}} \\ {{-2}} & {{-3}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]+\left[ \begin{array} {l} {{0}} \\ {{1}} \\ \end{array} \right] u ( t )}}}} \\ {{{{y ( t )=\left[ \begin{array} {l} {{3}} & {{1}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]}}}} \\ \end{array}
$$
F. $$
\begin{array} {c} {{{{\left[ \begin{array} {l} {{{{\dot{x}}_{1} ( t )}}} \\ {{{{\dot{x}}_{2} ( t )}}} \\ \end{array} \right]=\left[ \begin{array} {l} {{0}} & {{2}} \\ {{-1}} & {{-2}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]+\left[ \begin{array} {l} {{1}} \\ {{0}} \\ \end{array} \right] u ( t )}}}} \\ {{{{y ( t )=\left[ \begin{array} {l} {{3}} & {{0}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]}}}} \\ \end{array}
$$
G. $$
\begin{array} {c} {{{{\left[ \begin{array} {l} {{{{\dot{x}}_{1} ( t )}}} \\ {{{{\dot{x}}_{2} ( t )}}} \\ \end{array} \right]=\left[ \begin{array} {l} {{0}} & {{1}} \\ {{-1}} & {{-3}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]+\left[ \begin{array} {l} {{0}} \\ {{1}} \\ \end{array} \right] u ( t )}}}} \\ {{{{y ( t )=\left[ \begin{array} {l} {{3}} & {{2}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]}}}} \\ \end{array}
$$
H. $$
\begin{array} {c} {{{{\left[ \begin{array} {l} {{{{\dot{x}}_{1} ( t )}}} \\ {{{{\dot{x}}_{2} ( t )}}} \\ \end{array} \right]=\left[ \begin{array} {l} {{0}} & {{1}} \\ {{-3}} & {{-1}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]+\left[ \begin{array} {l} {{0}} \\ {{1}} \\ \end{array} \right] u ( t )}}}} \\ {{{{y ( t )=\left[ \begin{array} {l} {{1}} & {{2}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]}}}} \\ \end{array}
$$
I. $$
\begin{array} {c} {{{{\left[ \begin{array} {l} {{{{\dot{x}}_{1} ( t )}}} \\ {{{{\dot{x}}_{2} ( t )}}} \\ \end{array} \right]=\left[ \begin{array} {l} {{1}} & {{0}} \\ {{-2}} & {{-1}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]+\left[ \begin{array} {l} {{0}} \\ {{1}} \\ \end{array} \right] u ( t )}}}} \\ {{{{y ( t )=\left[ \begin{array} {l} {{1}} & {{4}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]}}}} \\ \end{array}
$$
J. $$
\begin{array} {c} {{{{\left[ \begin{array} {l} {{{{\dot{x}}_{1} ( t )}}} \\ {{{{\dot{x}}_{2} ( t )}}} \\ \end{array} \right]=\left[ \begin{array} {l} {{1}} & {{0}} \\ {{-3}} & {{-2}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]+\left[ \begin{array} {l} {{1}} \\ {{0}} \\ \end{array} \right] u ( t )}}}} \\ {{{{y ( t )=\left[ \begin{array} {l} {{2}} & {{3}} \\ \end{array} \right] \left[ \begin{array} {l} {{{x_{1} ( t )}}} \\ {{{x_{2} ( t )}}} \\ \end{array} \right]}}}} \\ \end{array}
$$
Answer: E