Consider the system defined by the following equation:
$$
\overset\cdots{y} \,+\, 6 \ddot{y} \,+\, 11 \dot{y} \,+\, 6 y \,=\, 6 u
$$
What is the diagonal canonical form of this system in state space?

Options:
A. $\begin{aligned}&\begin{bmatrix}\dot{x}_1\\\dot{x}_3\\\dot{x}_4\end{bmatrix}=\begin{bmatrix}-3&0&1\\0&-1&0\\0&0&-2\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}+\begin{bmatrix}1\\1\\0\end{bmatrix}u\\&y=\begin{bmatrix}4&0&3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\end{aligned}$
B. $\begin{aligned}&\begin{bmatrix}\dot{x}_1\\\dot{x}_3\\\dot{x}_4\end{bmatrix}=\begin{bmatrix}-1&0&0\\0&-2&0\\0&0&-3\end{bmatrix}\begin{bmatrix}x_1\\x_3\\x_3\end{bmatrix}+\begin{bmatrix}1\\1\\1\end{bmatrix}u\\&y=\begin{bmatrix}3&-6&3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\end{aligned}$
C. $\begin{aligned}&\begin{bmatrix}\dot{x}_1\\\dot{x}_2\\\dot{x}_4\end{bmatrix}=\begin{bmatrix}-1&0&1\\0&-2&1\\0&0&-3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}+\begin{bmatrix}0\\1\\2\end{bmatrix}u\\&y=\begin{bmatrix}2&1&3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\end{aligned}$
D. $\begin{aligned}&\begin{bmatrix}\dot{x}_1\\\dot{x}_2\\\dot{x}_3\end{bmatrix}=\begin{bmatrix}-3&0&1\\0&-1&0\\1&0&-2\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}+\begin{bmatrix}1\\0\\1\end{bmatrix}u\\&y=\begin{bmatrix}2&4&5\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\end{aligned}$
E. $\begin{aligned}&\begin{bmatrix}\dot{x}_1\\\dot{x}_2\\\dot{x}_3\end{bmatrix}=\begin{bmatrix}-2&1&0\\0&-3&0\\1&0&-1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}+\begin{bmatrix}0\\1\\1\end{bmatrix}u\\&y=\begin{bmatrix}1&2&3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\end{aligned}$
F. $\begin{aligned}&\begin{bmatrix}\dot{x}_1\\\dot{x}_3\\\dot{x}_4\end{bmatrix}=\begin{bmatrix}-2&0&0\\0&-3&0\\0&1&-1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}+\begin{bmatrix}1\\0\\0\end{bmatrix}u\\&y=\begin{bmatrix}0&3&3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\end{aligned}$
G. $\begin{aligned}&\begin{bmatrix}\dot{x}_1\\\dot{x}_4\\\dot{x}_3\end{bmatrix}=\begin{bmatrix}-6&0&0\\0&-4&0\\1&0&-2\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}+\begin{bmatrix}1\\2\\0\end{bmatrix}u\\&y=\begin{bmatrix}0&2&-3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\end{aligned}$
H. $\begin{aligned}&\begin{bmatrix}\dot{x}_1\\\dot{x}_2\\\dot{x}_4\end{bmatrix}=\begin{bmatrix}-1&1&0\\0&-3&0\\0&0&-2\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}+\begin{bmatrix}0\\0\\1\end{bmatrix}u\\&y=\begin{bmatrix}1&3&-6\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\end{aligned}$
I. $\begin{aligned}&\begin{bmatrix}\dot{x}_1\\\dot{x}_3\\\dot{x}_4\end{bmatrix}=\begin{bmatrix}-2&0&0\\0&-3&0\\0&0&-1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}+\begin{bmatrix}0\\0\\1\end{bmatrix}u\\&y=\begin{bmatrix}1&-1&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\end{aligned}$
J. $\begin{aligned}&\begin{bmatrix}\dot{x}_1\\\dot{x}_3\\\dot{x}_4\end{bmatrix}=\begin{bmatrix}-4&0&0\\0&-5&0\\0&0&-6\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}+\begin{bmatrix}1\\1\\1\end{bmatrix}u\\&y=\begin{bmatrix}3&0&-3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\end{aligned}$
Answer: B