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* [ ] MikTeX `版本号`
* [ ] CTeX
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## 我的问题
\begin{equation}\label{eq13}
\begin{array}{c}
D_{\alpha \beta \gamma}=\left[\begin{array}{lll}
D_{11} & D_{12} & D_{13} \\
D_{21} & D_{22} & D_{23} \\
D_{31} & D_{32} & D_{33}
\end{array}\right]= \\
\left[\begin{array}{lll}
\cos \beta \cos \gamma & -\cos \alpha \sin \gamma+\sin \alpha \sin \beta \cos \gamma & \sin \alpha \sin \gamma+\cos \alpha \sin \beta \cos \gamma \\
\cos \beta \sin \gamma & \cos \alpha \cos \gamma+\sin \alpha \sin \beta \sin \gamma & -\sin \alpha \cos \gamma+\cos \alpha \sin \beta \sin \gamma \\
-\sin \beta & \sin \alpha \cos \beta & \cos \alpha \cos \beta
\end{array}\right]
\end{array}
\end{equation}
1 回答
0
将`&`换成&
```
\documentclass{article}
\usepackage{amsmath,mathtools}
\begin{document}
\begin{align*}
D_{\alpha \beta \gamma}=\left[\begin{array}{lll}
D_{11} & D_{12} & D_{13} \\
D_{21} & D_{22} & D_{23} \\
D_{31} & D_{32} & D_{33}
\end{array}\right] &=
\left[\begin{array}{l}
\cos \beta \cos \gamma \\
\cos \beta \sin \gamma \\
-\sin \beta
\end{array}\right.\\
&\mathrel{\phantom{=}}
\left.\begin{array}{ll}
-\cos \alpha \sin \gamma+\sin \alpha \sin \beta \cos \gamma & \sin \alpha \sin \gamma+\cos \alpha \sin \beta \cos \gamma \\
\cos \alpha \cos \gamma+\sin \alpha \sin \beta \sin \gamma & -\sin \alpha \cos \gamma+\cos \alpha \sin \beta \sin \gamma \\
\sin \alpha \cos \beta & \cos \alpha \cos \beta
\end{array}\right]
\end{align*}
\begin{align*}
D_{\alpha \beta \gamma}&=\left[\begin{array}{lll}
D_{11} & D_{12} & D_{13} \\
D_{21} & D_{22} & D_{23} \\
D_{31} & D_{32} & D_{33}
\end{array}\right] \\
&=\begin{bsmallmatrix}
\cos \beta \cos \gamma & -\cos \alpha \sin \gamma+\sin \alpha \sin \beta \cos \gamma & \sin \alpha \sin \gamma+\cos \alpha \sin \beta \cos \gamma \\
\cos \beta \sin \gamma & \cos \alpha \cos \gamma+\sin \alpha \sin \beta \sin \gamma & -\sin \alpha \cos \gamma+\cos \alpha \sin \beta \sin \gamma \\
-\sin \beta & \sin \alpha \cos \beta & \cos \alpha \cos \beta
\end{bsmallmatrix}
\end{align*}
\end{document}
```
-
将 – sikouhjw 2020-05-15 15:13 回复
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