对齐环境中方程式中的下支撑

Question

在下面的代码中，我想在第二个带有 z 的方程式开头的“B”下括号下添加另一个下括号。第二个下括号应覆盖从 [x,y] 到“=”符号之前的部分。如下所示：

如有任何帮助，我们将不胜感激。

这是我的代码：

\documentclass{article}
\usepackage{amsmath}

\begin{document}
\begin{align}
\begin{bmatrix}
x\\
y
\end{bmatrix} & = \begin{bmatrix}
-(L_{1} + r_{3v})\sin\theta_{1} + r_{2v}\cos\theta_{1} & \sin\theta_{1} & \cos\theta_{1}\\
(L_{1} + r_{3v})\cos\theta_{1} + r_{2v}\sin\theta_{1} & -\cos\theta_{1} & \sin\theta_{1}
\end{bmatrix}
\begin{bmatrix}
\delta\theta_{1}\\
\delta r_{2} \\
\delta r_{3}
\end{bmatrix} +  \nonumber\\
& \qquad\qquad\qquad\qquad\qquad\qquad\qquad
\begin{bmatrix}
(L_{1} + r_{3v})\cos\theta_{1} + r_{2v}\sin\theta_{1} \\
(L_{1} + r_{3v})\sin\theta_{1} - r_{2v}\cos\theta_{1}
\end{bmatrix}\\[2ex]
\begin{bmatrix}
x\\
y
\end{bmatrix} & -
{\underbrace{\begin{bmatrix}
(L_{1} + r_{3v})\cos\theta_{1} + r_{2v}\sin\theta_{1} \\
(L_{1} + r_{3v})\sin\theta_{1} - r_{2v}\cos\theta_{1}
\end{bmatrix}}_B} = \nonumber\\
& \qquad\qquad{\underbrace{\begin{bmatrix}
-(L_{1} + r_{3v})\sin\theta_{1} + r_{2v}\cos\theta_{1} & \sin\theta_{1} & \cos\theta_{1}\\
(L_{1} + r_{3v})\cos\theta_{1} + r_{2v}\sin\theta_{1} & -\cos\theta_{1} & \sin\theta_{1}
\end{bmatrix}}_{H}}
{\underbrace{\begin{bmatrix}
\delta\theta_{1}\\
\delta r_{2} \\
\delta r_{3}
\end{bmatrix}}_{\bf{u}}}
\end{align}

\end{document}

Answer 1

改变 的位置&并添加 的另一层\underbrace：

\documentclass{article}
\usepackage{amsmath}

\begin{document}
\begin{align}
&\begin{bmatrix}
x\\
y
\end{bmatrix} = \begin{bmatrix}
-(L_{1} + r_{3v})\sin\theta_{1} + r_{2v}\cos\theta_{1} & \sin\theta_{1} & \cos\theta_{1}\\
(L_{1} + r_{3v})\cos\theta_{1} + r_{2v}\sin\theta_{1} & -\cos\theta_{1} & \sin\theta_{1}
\end{bmatrix}
\begin{bmatrix}
\delta\theta_{1}\\
\delta r_{2} \\
\delta r_{3}
\end{bmatrix} +  \nonumber\\
& \hspace{14em}
\begin{bmatrix}
(L_{1} + r_{3v})\cos\theta_{1} + r_{2v}\sin\theta_{1} \\
(L_{1} + r_{3v})\sin\theta_{1} - r_{2v}\cos\theta_{1}
\end{bmatrix}\\[2ex]
&{\underbrace{\begin{bmatrix}
x\\
y
\end{bmatrix} \mathrel{-}% just for getting alignment
\underbrace{\begin{bmatrix}
(L_{1} + r_{3v})\cos\theta_{1} + r_{2v}\sin\theta_{1} \\
(L_{1} + r_{3v})\sin\theta_{1} - r_{2v}\cos\theta_{1}
\end{bmatrix}}_B}_{\mathbf{z}}} = \nonumber\\
& \hspace{4em}
{\underbrace{\begin{bmatrix}
-(L_{1} + r_{3v})\sin\theta_{1} + r_{2v}\cos\theta_{1} & \sin\theta_{1} & \cos\theta_{1}\\
(L_{1} + r_{3v})\cos\theta_{1} + r_{2v}\sin\theta_{1} & -\cos\theta_{1} & \sin\theta_{1}
\end{bmatrix}}_{H}}
{\underbrace{\begin{bmatrix}
\delta\theta_{1}\\
\delta r_{2} \\
\delta r_{3}
\end{bmatrix}}_{\mathbf{u}}}
\end{align}

\end{document}

请避免\bf：正确的语法是\mathbf{u}，而不是。使用{\bf{u}}而不是一堆，这样更简单。\qquad\hspace