Why a matrix equation is not vertically aligned?

I'm struggling to make all the matrices in my equation of same height (vertical alignment). I would appreciate if I could get some suggestions. I tried some solutions here and there but it's looks my question is different. \documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation} \label{eq:system_dynamics_l}
\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}
\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\dfrac{L}{2}
\end{bmatrix}\begin{bmatrix}
x \\$4mm$  \theta
\end{bmatrix}
=\begin{bmatrix}
0 \\$4mm$  0
\end{bmatrix}
\end{equation}
\end{document}
• try \tfrac for your fractions – Elements in Space Apr 26 '18 at 13:40
• why are you forcing over sized 1/2 via \dfrac{1}{2} ? (just use \frac) – David Carlisle Apr 26 '18 at 13:40

Because \dfrac is higher than other elements of your matrices. Here are two possibilities: use \tfrac or \vphantoms. \documentclass{article}
\usepackage{amsmath}
\begin{document}
Arguably simplest solution (thanks to David Carlisle)
\begin{equation} \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}
\end{equation}

Original proposal 1:
\begin{equation} \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}
\end{equation}

In case you want to keep the \verb|\dfrac|s, which is perfectly fine IMHO
\begin{equation} \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}
\end{equation}
\end{document}
• +1. However, instead of m_2g\tfrac{L}{2}, I'd write \tfrac{1}{2}m_2gL to improve legibility. – Mico Apr 26 '18 at 14:14
• @Mico I agree, of course, but I guess that is up to the OP to decide. I only focused on the alignment of the brackets. – user121799 Apr 26 '18 at 14:39
• I'd use \frac rather than \tfrac as shown by the question most of the time forcing \dfrac or \tfrac either does what \frac would do, or does the wrong thing. – David Carlisle Apr 26 '18 at 15:17
• @DavidCarlisle You are right, of course. Except if the OP would start adding \displaystyle somewhere. So with your permission I'd like to modify my answer to include your suggestion. – user121799 Apr 26 '18 at 15:20
• sure feel free:-) – David Carlisle Apr 26 '18 at 15:22

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.

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation} \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}
\end{equation}
\end{document} • +1 for the nicely-placed reference to one of the most memorable scenes in the movie "The Pricess Bride". :-) – Mico Apr 26 '18 at 14:15
• Just to be fussy: I'd put the redefinition of \arraystretch within equation. The environment grouping prevents unpleasant consequences. – campa Apr 26 '18 at 14:42
• @campa - right you are. – John Kormylo Apr 26 '18 at 15:00
• @JohnKormylo what is this"1.7"? – AlFagera Apr 28 '18 at 10:47
• @AlFagera - \arraystretch is a scale factor applied to the strut at the start of every row of the array row. 1.6 was too small. – John Kormylo Apr 28 '18 at 13:53

I propose a solution based on the medium-sized fractions from nccmath (in my opinion, for numerical coefficients, \dfrac is too large, and tfrac too small) and \vphantom{\mfrac{1}{2} at the relevant places. I also load cellspace to add some vertical padding to the matrix rows:

\documentclass{article}

\usepackage{amsmath, nccmath}
\usepackage{array}
\usepackage[math]{cellspace}
\setlength{\cellspacetoplimit}{2pt}
\setlength{\cellspacebottomlimit}{2pt}

\newcommand*{\mystrut}{\vphantom{\mfrac{1}{2}}}

\begin{document}

\begin{equation} \label{eq:system_dynamics_l}
\begin{bmatrix}
m_1+m_2 & \mfrac{1}{2}m_2 L \\
\mfrac{1}{2}m_2 L & \mfrac{1}{2}m_2 L^2
\end{bmatrix}
\begin{bmatrix}
\ddot{x} \mystrut \\ \ddot{\theta} \mystrut
\end{bmatrix}
+ \begin{bmatrix}
b & 0 \mystrut \\
0 & c \mystrut
\end{bmatrix}
\begin{bmatrix}
\dot{x} \mystrut \\ \dot{\theta} \mystrut
\end{bmatrix}
+\begin{bmatrix}
k & 0 \mystrut \\ 0 & m_2g\mfrac{L}{2}
\end{bmatrix}\begin{bmatrix}
x \mystrut \\ \theta \mystrut
\end{bmatrix}
=\begin{bmatrix}
0 \mystrut \\ 0 \mystrut
\end{bmatrix}
\end{equation}

\end{document} 