A simple model of the carbon dioxide molecule can be modeled by a system of three masses and two springs. The oxygen atoms have mass M and carbon has mass m. The springs (bonds) have spring constant $k$.
The corresponding mathematical model is $$ Mx'' = k(y-x)$$ $$my'' = -k(y-x) - k(y-z)$$ $$Mz'' = -k(z-y) $$
I want to model this such that there is a system of equations where $\vec{x} = Ax$. I set $x_1 = x - y$, $x_2 = y - z$, $x_3 = x_1'$, and $x_4 = x_2'$. But I don't know what to do next.
On
$$ \left[ \begin{array}{rrr} m_{o} x'' \\ m_{c} y'' \\ m_{o} z'' \end{array} \right] % = % k \left[ \begin{array}{rrr} -1 & 1 & 0 \\ 1 & -2 & 1 \\ 0 & 1 & -1 \end{array} \right] % \left[ \begin{array}{rrr} x \\ y \\ z \end{array} \right] $$ The key is to diagonalize $$ \mathbf{A} = \left[ \begin{array}{rrr} -1 & 1 & 0 \\ 1 & -2 & 1 \\ 0 & 1 & -1 \end{array} \right] $$ Eigenvalues: $$ \lambda \left( \mathbf{A} \right) = \left( -3, -1, 0 \right) $$ Matrix of eigenvectors: $$ \mathbf{P} = \left[ \begin{array}{rrr} 1 & -2 & 1 \\ -1 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array} \right] $$ Diagonalization: $$ \mathbf{P} \, \mathbf{A} \, \mathbf{P}^{-1} = \left[ \begin{array}{rrr} -3 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] $$
Take $x'=a, y'=b, z'=c$, so $a'=k(y-z)/M, b'=k(z-x)/m, c'=k(y-z)/M$. If you row reduce the matrix you might find that you can reduce the number of variables from 6.
I'm assuming you meant $x'=Ax$. Otherwise I'm not sure what you meant by $\bar{x}=Ax$