# Why a matrix equation is not vertically aligned?

Because \dfrac is higher than other elements of your matrices. Here are two possibilities: use \tfrac or \vphantoms.

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Arguably simplest solution (thanks to David Carlisle)
$$\label{eq:system_dynamics_0} \begin{bmatrix} m_1+m_2 & \tfrac{1}{2}m_2 L \\[4mm] \tfrac{1}{2}m_2 L & \tfrac{1}{2}m_2 L^2 \end{bmatrix} \begin{bmatrix} \ddot{x} \\[4mm] \ddot{\theta} \end{bmatrix} + \begin{bmatrix} b & 0 \\[4mm] 0 & c \end{bmatrix}\begin{bmatrix} \dot{x} \\[4mm] \dot{\theta} \end{bmatrix} +\begin{bmatrix} k & 0 \\[4mm] 0 & m_2g\tfrac{L}{2} \end{bmatrix}\begin{bmatrix} x \\[4mm] \theta \end{bmatrix} =\begin{bmatrix} 0 \\[4mm] 0 \end{bmatrix}$$

Original proposal 1:
$$\label{eq:system_dynamics_l} \begin{bmatrix} m_1+m_2 & \tfrac{1}{2}m_2 L \\[4mm] \tfrac{1}{2}m_2 L & \tfrac{1}{2}m_2 L^2 \end{bmatrix} \begin{bmatrix} \ddot{x} \\[4mm] \ddot{\theta} \end{bmatrix} + \begin{bmatrix} b & 0 \\[4mm] 0 & c \end{bmatrix}\begin{bmatrix} \dot{x} \\[4mm] \dot{\theta} \end{bmatrix} +\begin{bmatrix} k & 0 \\[4mm] 0 & m_2g\tfrac{L}{2} \end{bmatrix}\begin{bmatrix} x \\[4mm] \theta \end{bmatrix} =\begin{bmatrix} 0 \\[4mm] 0 \end{bmatrix}$$

In case you want to keep the \verb|\dfrac|s, which is perfectly fine IMHO
$$\label{eq:system_dynamics_2} \begin{bmatrix} m_1+m_2 & \dfrac{1}{2}m_2 L \\[4mm] \dfrac{1}{2}m_2 L & \dfrac{1}{2}m_2 L^2 \end{bmatrix} \begin{bmatrix}\vphantom{\dfrac{1}{2}} \ddot{x} \\[4mm] \vphantom{\dfrac{1}{2}}\ddot{\theta} \end{bmatrix} + \begin{bmatrix} \vphantom{\dfrac{1}{2}}b & 0 \\[4mm] \vphantom{\dfrac{1}{2}}0 & c \end{bmatrix}\begin{bmatrix} \vphantom{\dfrac{1}{2}}\dot{x} \\[4mm] \vphantom{\dfrac{1}{2}}\dot{\theta} \end{bmatrix} +\begin{bmatrix} \vphantom{\dfrac{1}{2}}k & 0 \\[4mm] 0 & m_2g\dfrac{L}{2} \end{bmatrix}\begin{bmatrix} \vphantom{\dfrac{1}{2}}x \\[4mm] \vphantom{\dfrac{1}{2}} \theta \end{bmatrix} =\begin{bmatrix} \vphantom{\dfrac{1}{2}}0 \\[4mm] \vphantom{\dfrac{1}{2}}0 \end{bmatrix}$$
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The easiest way to handle Ratios Of Unusual Size (ROUS) is to increase \arraystretch (macro). Every row starts with a strut of height \arraystretch\ht\strutbox.

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$$\label{eq:system_dynamics_l} \def\arraystretch{1.7}% \begin{bmatrix} m_1+m_2 & \dfrac{1}{2}m_2 L \\ \dfrac{1}{2}m_2 L & \dfrac{1}{2}m_2 L^2 \end{bmatrix} \begin{bmatrix} \ddot{x} \\ \ddot{\theta} \end{bmatrix} + \begin{bmatrix} b & 0 \\ 0 & c \end{bmatrix}\begin{bmatrix} \dot{x} \\ \dot{\theta} \end{bmatrix} +\begin{bmatrix} k & 0 \\ 0 & m_2g\dfrac{L}{2} \end{bmatrix}\begin{bmatrix} x \\ \theta \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
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