大家好，我正在尝试将矩阵方程放在与方程组相同的行空间中。我不确定我如何尝试将第三个align*方程放在一起，但似乎行不通。
\begin{align*}
\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1\\
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
x_{11} \\
x_{12} \\
x_{13} \\
x_{21}\\
x_{22}\\
x_{23}\\
\end{bmatrix}
&=
\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1\\
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
u_{11} \\
u_{12} \\
u_{13} \\
u_{21}\\
u_{22}\\
u_{23}\\
\end{bmatrix}
\end{align*}
\begin{align*}
x_{11}x_{22}-x_{12}x_{21}&=0 \\
x_{11}x_{22}-x_{12}x_{21}&=0 \\
x_{11}x_{22}-x_{12}x_{21}&=0 \\
\end{align*}
答案1
也许是这样的？
\documentclass{article}
\usepackage[a4paper,margin=2.5cm]{geometry} % set page margins
\usepackage{amsmath} % for 'bmatrix' and 'aligned' environments
\begin{document}
\[
\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1\\
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x_{11} \\
x_{12} \\
x_{13} \\
x_{21} \\
x_{22} \\
x_{23}
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1\\
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
u_{11} \\
u_{12} \\
u_{13} \\
u_{21} \\
u_{22} \\
u_{23}
\end{bmatrix}
\qquad\qquad
\begin{aligned}
x_{11}x_{22}-x_{12}x_{21}&=0 \\
x_{11}x_{22}-x_{12}x_{21}&=0 \\
x_{11}x_{22}-x_{12}x_{21}&=0
\end{aligned}
\]
\end{document}
答案2
你是这个意思吗？
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{align*}
\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1\\
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
x_{11} \\
x_{12} \\
x_{13} \\
x_{21}\\
x_{22}\\
x_{23}\\
\end{bmatrix}
&=
\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1\\
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
u_{11} \\
u_{12} \\
u_{13} \\
u_{21}\\
u_{22}\\
u_{23}\\
\end{bmatrix} \\
x_{11}x_{22}-x_{12}x_{21}&=0 \\
x_{11}x_{22}-x_{12}x_{21}&=0 \\
x_{11}x_{22}-x_{12}x_{21}&=0 \\
\end{align*}
\end{document}
答案3
或者也许是这些解决方案之一？
\documentclass{article}
\usepackage{mathtools, nccmath}
\usepackage[showframe]{geometry}
\begin{document}
\vspace*{1cm}
\begin{alignat*}{2}
\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1\\
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
x_{11} \\
x_{12} \\
x_{13} \\
x_{21}\\
x_{22}\\
x_{23}\\
\end{bmatrix}
&=
\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1\\
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
u_{11} \\
u_{12} \\
u_{13} \\
u_{21}\\
u_{22}\\
u_{23}\\
\end{bmatrix}%
& \hspace{ 4em}%
\begin{matrix} x_{11}x_{22}-x_{12}x_{21}=0 \\
x_{11}x_{22}-x_{12}x_{21}=0 \\
x_{11}x_{22}-x_{12}x_{21}=0 \\
\end{matrix}
\end{alignat*}
\vskip1cm
\begin{align*}
\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1\\
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
x_{11} \\
x_{12} \\
x_{13} \\
x_{21}\\
x_{22}\\
x_{23}\\
\end{bmatrix}
&=
\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1\\
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
u_{11} \\
u_{12} \\
u_{13} \\
u_{21}\\
u_{22}\\
u_{23}\\
\end{bmatrix}%
& \hspace{ 4em}%
\begin{cases} x_{11}x_{22}-x_{12}x_{21}=0 \\
x_{11}x_{22}-x_{12}x_{21}=0 \\
x_{11}x_{22}-x_{12}x_{21}=0 \\
\end{cases}\hspace{-1em}
\end{align*}
\end{document}
