Convert vector differential equation's order 
Here is a 2nd order vector differential equation: $$\overrightarrow{Y}''= \begin{pmatrix}a & b \\c & d \end{pmatrix} \overrightarrow{Y}$$ Don't work it out, but write it as a vector differential eqn in $1$st order in higher dimensions.

I am not sure where to begin.
How does one convert from 2nd to 1st order? Hints are appreciated.
 A: HINT
Here's a simple example. If you have the second order differential equation $$ y'' =y$$ you can write it as the pair of first order differential equations $$ u' = y \\ y' = u$$ which is the vector diff-eq $$ \vec w' =  A\vec w$$ where $\vec w=(u,y)^T$ and $$ A = \begin{pmatrix}0&1\\1&0\end{pmatrix}$$ 
A: Let 
$$
\vec{x} = 
\begin{pmatrix}
x_1 \\ 
x_2 \\ 
\end{pmatrix} = 
\begin{pmatrix}
y_1' \\ 
y_2' \\ 
\end{pmatrix} = \vec{y}'. 
$$
Then 
$$
\vec{x}' = 
\begin{pmatrix}
x_1' \\ 
x_2' \\ 
\end{pmatrix}
=
\begin{pmatrix}
a & b \\ 
c & d \\ 
\end{pmatrix}
\begin{pmatrix}
y_1\\ 
y_2 \\ 
\end{pmatrix} = 
\begin{pmatrix}
a & b \\ 
c & d \\ 
\end{pmatrix} \vec{y}. 
$$
In coordinates, 
$$
\begin{pmatrix}
\vec{x}' \\ 
\vec{y}' \\ 
\end{pmatrix}
= 
\begin{pmatrix}
x_1' \\ 
x_2' \\ 
y_1' \\ 
y_2' \\ 
\end{pmatrix} 
= \begin{pmatrix}
0 & 0 & a & b \\ 
0 & 0 & c & d \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\    
\end{pmatrix}
\begin{pmatrix}
x_1 \\ 
x_2 \\ 
y_1 \\ 
y_2 \\ 
\end{pmatrix} 
= \begin{pmatrix}
\vec{x} \\ 
\vec{y} \\ 
\end{pmatrix}, 
$$
or 
$$
\begin{pmatrix}
\vec{x}' \\ 
\vec{y}' \\ 
\end{pmatrix}
= 
\begin{pmatrix}
0 & A \\ 
I & 0 \\ 
\end{pmatrix}
\begin{pmatrix}
\vec{x} \\ 
\vec{y} \\ 
\end{pmatrix}, 
$$
where 
$$
A= 
\begin{pmatrix}
a & b \\ 
c & d \\ 
\end{pmatrix} 
\mbox{ and } 
I = 
\begin{pmatrix}
1 & 0 \\ 
0 & 1 \\ 
\end{pmatrix}. 
$$
