我尝试过很多方法来垂直对齐这些方程式，但都没有成功。
\documentclass[11pt,a4paper,oneside]{report}
\usepackage[T1]{fontenc}
\usepackage{fouriernc}
\usepackage{mathtools}
\usepackage{amsthm}
\begin{document}
first attempt
\begin{equation}
\begin{aligned}
&-\omega\cdot \alpha_{1}=\bigl|B_{1}A_{2}A_{3}\bigr|\cdot \alpha_{1}+\bigl|A_{1}B_{1}A_{3}\bigr|\cdot \alpha_{2}+\bigl|A_{1}A_{2}B_{1}\bigr|\cdot \alpha_{3}
\\
&-\omega\cdot \alpha_{2}=\bigl|B_{2}A_{2}A_{3}\bigr|\cdot \alpha_{1}+\bigl|A_{1}B_{2}A_{3}\bigr|\cdot \alpha_{2}+\bigl|A_{1}A_{2}B_{2}\bigr|\cdot \alpha_{3}
\\
&-\omega\cdot \alpha_{3}=\bigl|B_{3}A_{2}A_{3}\bigr|\cdot \alpha_{1}+\bigl|A_{1}B_{2}A_{3}\bigr|\cdot \alpha_{2}+\bigl|A_{1}A_{2}B_{2}\bigr|\cdot \alpha_{3}
\\
&-\theta\cdot \alpha_{1}=\bigl|A_{1}B_{2}B_{3}\bigr|\cdot \alpha_{1}+\bigl|B_{1}A_{1}B_{3}\bigr|\cdot \alpha_{2}+\bigl|B_{1}B_{2}A_{1}\bigr|\cdot \alpha_{3}
\\
&-\theta\cdot \alpha_{1}=\bigl|A_{1}B_{2}B_{3}\bigr|\cdot \alpha_{1}+\bigl|B_{1}A_{1}B_{3}\bigr|\cdot \alpha_{2}+\bigl|B_{1}B_{2}A_{1}\bigr|\cdot \alpha_{3}
\end{aligned}
\end{equation}
second attempt
\begin{alignat*}{3}
&-\omega\cdot \alpha_{1}=&&\bigl|B_{1}A_{2}A_{3}\bigr|\cdot \alpha_{1}+&&\bigl|A_{1}B_{1}A_{3}\bigr|\cdot \alpha_{2}+\bigl|A_{1}A_{2}B_{1}\bigr|\cdot \alpha_{3} \\
&-\omega\cdot \alpha_{2}=&&\bigl|B_{2}A_{2}A_{3}\bigr|\cdot \alpha_{1}+&&\bigl|A_{1}B_{2}A_{3}\bigr|\cdot \alpha_{2}+\bigl|A_{1}A_{2}B_{2}\bigr|\cdot \alpha_{3} \\
& -\theta\cdot \alpha_{3}=&&\bigl|B_{3}A_{2}A_{3}\bigr|\cdot \alpha_{1}+&&\bigl|A_{1}B_{2}A_{3}\bigr|\cdot \alpha_{2}+\bigl|A_{1}A_{2}B_{2}\bigr|\cdot \alpha_{3} \\
&-\theta\cdot \alpha_{1}=&&\bigl|A_{1}B_{2}B_{3}\bigr|\cdot \alpha_{1}+&&\bigl|B_{1}A_{1}B_{3}\bigr|\cdot \alpha_{2}+\bigl|B_{1}B_{2}A_{1}\bigr|\cdot \alpha_{3}
\end{alignat*}
third attempt
\begin{equation}
\begin{split}
&-\omega\cdot \alpha_{1}=\bigl|B_{1}A_{2}A_{3}\bigr|\cdot \alpha_{1}+\bigl|A_{1}B_{1}A_{3}\bigr|\cdot \alpha_{2}+\bigl|A_{1}A_{2}B_{1}\bigr|\cdot \alpha_{3}
\\
&-\omega\cdot \alpha_{2}=\bigl|B_{2}A_{2}A_{3}\bigr|\cdot \alpha_{1}+\bigl|A_{1}B_{2}A_{3}\bigr|\cdot \alpha_{2}+\bigl|A_{1}A_{2}B_{2}\bigr|\cdot \alpha_{3}
\\
&-\omega\cdot \alpha_{3}=\bigl|B_{3}A_{2}A_{3}\bigr|\cdot \alpha_{1}+\bigl|A_{1}B_{2}A_{3}\bigr|\cdot \alpha_{2}+\bigl|A_{1}A_{2}B_{2}\bigr|\cdot \alpha_{3}
\\
&-\theta\cdot \alpha_{1}=\bigl|A_{1}B_{2}B_{3}\bigr|\cdot \alpha_{1}+\bigl|B_{1}A_{1}B_{3}\bigr|\cdot \alpha_{2}+\bigl|B_{1}B_{2}A_{1}\bigr|\cdot \alpha_{3}
\\
&-\theta\cdot \alpha_{1}=\bigl|A_{1}B_{2}B_{3}\bigr|\cdot \alpha_{1}+\bigl|B_{1}A_{1}B_{3}\bigr|\cdot \alpha_{2}+\bigl|B_{1}B_{2}A_{1}\bigr|\cdot \alpha_{3}
\end{split}
\end{equation}
\end{document}
生产
第一次尝试仅对齐前三行，而不对齐最后两行（它们彼此对齐，但不与其余行对齐），而第二次尝试几乎修复了除第三列和第四列之外的所有内容，但我没有在中间获得编号。最后一个似乎与第一个类似。有人能帮我吗？谢谢。
答案1
您必须先决定，是要一个方程，因此只有一个数字，还是要多个方程和多个数字（有关此选择，请参阅文章末尾的注释）。第一次使用时，\begin{equation}使用\end{equation}环境ed（如aligned、gathered）。由于您有多个对齐点，因此应该使用环境alignedat。
这里是 MWE（包括序言）：
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\begin{alignedat}{8}
&-\omega &&\cdot \alpha_{1} &&= \bigl|B_{1}A_{2}A_{3}\bigr| &&\cdot \alpha_{1} &&+ \bigl|A_{1}B_{1}A_{3}\bigr| &&\cdot \alpha_{2} &&+ \bigl|A_{1}A_{2}B_{1}\bigr| &&\cdot \alpha_{3} \\
&-\omega &&\cdot \alpha_{2} &&= \bigl|B_{2}A_{2}A_{3}\bigr| &&\cdot \alpha_{1} &&+ \bigl|A_{1}B_{2}A_{3}\bigr| &&\cdot \alpha_{2} &&+ \bigl|A_{1}A_{2}B_{2}\bigr| &&\cdot \alpha_{3} \\
&-\theta &&\cdot \alpha_{3} &&= \bigl|B_{3}A_{2}A_{3}\bigr| &&\cdot \alpha_{1} &&+ \bigl|A_{1}B_{2}A_{3}\bigr| &&\cdot \alpha_{2} &&+ \bigl|A_{1}A_{2}B_{2}\bigr| &&\cdot \alpha_{3} \\
&-\theta &&\cdot \alpha_{1} &&= \bigl|A_{1}B_{2}B_{3}\bigr| &&\cdot \alpha_{1} &&+ \bigl|B_{1}A_{1}B_{3}\bigr| &&\cdot \alpha_{2} &&+ \bigl|B_{1}B_{2}A_{1}\bigr| &&\cdot \alpha_{3}
\end{alignedat}
\end{equation}
\end{document}
评论：
对于多个方程：\begin{equation}用替换\begin{alignat}{8}，\end{equation}用 替换\end{alignat}，并删除\begin{alignedat}{8}和\end{alignedat}。