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跨多个方程进行对齐

tag-icon • tag-icon horizontal-alignment matrices
跨多个方程进行对齐

现在，我的文档如下所示： 在此处输入图片描述

或者像这样：

在此处输入图片描述

我很烦恼，因为方程式到处都是，而且排列不均匀。例如，我希望每行都与左侧对齐问题 1或者问题 2。 有什么建议么？

\documentclass[paper=a4, fontsize=11pt]{scrartcl} % A4 paper an
\usepackage{amsmath,amsthm,amssymb}
\usepackage{enumitem}
\begin{document}

\begin{enumerate}[label=\bfseries Problem \arabic*:]

\item %PROBLEM 1

Find $\textbf{A}+\textbf{B}$, $\textbf{AB}$, $\textbf{BA}$.\\\\


a) $\textbf{A} = \begin{bmatrix}  1 & 2  \\   3  & -1  \end{bmatrix}\,\,\,\,\,$ and $\,\,\,\,\,\textbf{B} = \begin{bmatrix} 2 & 1  \\1  & 1 \end{bmatrix}$. \,\,\,\,\,\textbf{A} is $2\times2$ and \textbf{B} is $2\times2$.

 \[\textbf{A}+\textbf{B}=\begin{bmatrix}  1 & 2  \\ 3  & -1\end{bmatrix} + \begin{bmatrix} 2 & 1  \\1  & 1 \end{bmatrix}=\begin{bmatrix} 1+2 & 2+1  \\3+1  & -1+1 \end{bmatrix}=\begin{bmatrix} 3 & 3  \\4  & 0 \end{bmatrix}\]

 \[\textbf{AB}=\begin{bmatrix}  1 & 2  \\ 3  & -1\end{bmatrix} \cdot \begin{bmatrix} 2 & 1  \\1  & 1 \end{bmatrix}=\begin{bmatrix} 1\cdot2+2\cdot1 & 1\cdot 1+2\cdot1  \\3\cdot2+(-1)\cdot1  & 3\cdot1+(-1)\cdot1 \end{bmatrix}=\begin{bmatrix} 4 & 3  \\5  & 2 \end{bmatrix}\]

 \[\textbf{BA}= \begin{bmatrix} 2 & 1  \\1  & 1 \end{bmatrix}\cdot \begin{bmatrix}  1 & 2  \\ 3  & -1\end{bmatrix}=\begin{bmatrix} 2\cdot1+1\cdot3 & 2\cdot 2+1\cdot(-1)  \\1\cdot1+1\cdot3  & 1\cdot2+1\cdot(-1) \end{bmatrix}=\begin{bmatrix} 5 & 3  \\4  & 1 \end{bmatrix}\]\\




b) $\textbf{A} = \begin{bmatrix} 2 & 3 & 1\\ 0  & -1 & 2 \end{bmatrix}\,\,\,\,\,$and 
$\,\,\,\,\,\textbf{B} = \begin{bmatrix} 2 & -2  \\ 4  & 3 \\1 & 5 \end{bmatrix}$. \,\,\,\,\,\,\textbf{A} is $2\times3$ and \textbf{B} is $3\times2$.     


    $\textbf{A}+\textbf{B}$ is undefined because the two matrices are incompatible for addition. 

 \[ \textbf{AB}=\begin{bmatrix} 2 & 3 & 1\\ 0  & -1 & 2 \end{bmatrix}\cdot\begin{bmatrix} 2 & -2  \\ 4  & 3 \\1 & 5 \end{bmatrix}=\begin{bmatrix} 2\cdot2+3\cdot4+1\cdot1 & 2\cdot (-2)+3\cdot3+1\cdot 5  \\0\cdot2+(-1)\cdot4 +2\cdot 1 & 0\cdot(-2)+(-1)\cdot3 +2\cdot5 \end{bmatrix}=\begin{bmatrix} 17 & 10  \\-2  & 7 \end{bmatrix} \]


 \[ \textbf{BA}=\begin{bmatrix} 2 & -2  \\ 4  & 3 \\1 & 5 \end{bmatrix}\cdot \begin{bmatrix} 2 & 3 & 1\\ 0  & -1 & 2 \end{bmatrix}=\begin{bmatrix} 2\cdot2+(-2)\cdot0 & 2\cdot 3+(-2)\cdot(1) &2\cdot1+(-2)\cdot 2  \\4\cdot2+3\cdot0&4\cdot3 +3\cdot (-1) & 4\cdot1+3\cdot2\\ 1\cdot2+5\cdot0&1\cdot3+5\cdot(-1)&1\cdot1+5\cdot2 \end{bmatrix}=\begin{bmatrix} 4 & 8&-2  \\8  & 9&10\\2&-2&11 \end{bmatrix} \]


\item %PROBLEM 2
  Let $\textbf{a}' = \begin{bmatrix}
       3 & 6 &-3 & 5 &  9& 2  \\
     \end{bmatrix}$. Find \textbf{a}$'$\textbf{a} and \textbf{aa}$'$.


 \textbf{a}$'$ is the transpose of $\textbf{a} = \begin{bmatrix}
       3 \\ 6 \\-3 \\ 5 \\  9\\ 2  \\
     \end{bmatrix}$

 \[\textbf{a}'\textbf{a} = \begin{bmatrix}
       3 & 6 &-3 & 5 &  9& 2  \\
     \end{bmatrix} \cdot \begin{bmatrix}
       3 \\ 6 \\-3 \\ 5 \\  9\\ 2  \\
     \end{bmatrix} = \begin{bmatrix}
       3^2 + 6^2 + (-3)^2 + 5^2 +  9^2 + 2^2  \\
     \end{bmatrix} =[164] \]

   \[\textbf{aa}' = \begin{bmatrix}
       3 \\ 6 \\-3 \\ 5 \\  9\\ 2  \\
     \end{bmatrix}\cdot \begin{bmatrix}
       3 & 6 &-3 & 5 &  9& 2  \\
     \end{bmatrix}  = \begin{bmatrix}
       3^2& 3\cdot 6 & 3\cdot (-3) & 3\cdot 5 &  3\cdot 9 & 3\cdot 2  \\
  6\cdot 3& 6^2 & 6\cdot (-3) & 6\cdot 5 &  6\cdot 9 & 6\cdot 2  \\
    (-3)\cdot 3& (-3)\cdot 6 & (-3)^2 & (-3)\cdot 5 &  (-3)\cdot 9 & (-3)\cdot 2  \\
      5\cdot 3& 5\cdot 6 & 5\cdot (-3) & 5^2 &  5\cdot 9 & 5\cdot 2  \\
        9\cdot 3& 9\cdot 6 & 9\cdot (-3) & 9\cdot 5 &  9^2 & 9\cdot 2  \\
         2\cdot 3& 2\cdot 6 & 2\cdot (-3) & 2\cdot 5 &  2\cdot 9 & 2^2  \\
     \end{bmatrix} \\=\begin{bmatrix}
       9& 18 & 9 & 15 &  27 & 6  \\
  18& 36 & -18 & 30 &  54 & 12  \\
    -9& -18 & 9 & -15 &  27 & -6  \\
      15& 30 & 15 & 10 &  45 & 10 \\
        27& 54 & 27 & 45 &  81 & 18  \\
         6& 12 & 6 & 10 &  18 & 4  \\
     \end{bmatrix} \]   







   Let $\textbf{b}' = \begin{bmatrix}
       1 & 1 & 1 & 1 & 1 & 1  \\
     \end{bmatrix}\,\,\,\,\,$
     and
    $\,\,\,\,\,\textbf{c}' = \begin{bmatrix}
       1 & -1 & 1 & -1 & 1 & -1  \\
     \end{bmatrix}$. Find \textbf{b}$'$\textbf{c} and \textbf{bc}$'$.


  \end{enumerate}
\end{document}

答案1

我基本上从您的 MWE 中实施了两项更改。首先，我在调用align=left中使用了enumerate。然后，我将的实例替换为\[ ... \]。$ ...$请注意，在某些情况下，您可能必须在分隔符\displaystyle内调用$，但这里并非如此。

此外，对于换行的数学表达式，我添加了一些幻影以获得适当的缩进。

\documentclass[paper=a4, fontsize=11pt]{scrartcl} % A4 paper an
\usepackage{amsmath,amsthm,amssymb}
\usepackage{enumitem}
\begin{document}

\begin{enumerate}[label=\bfseries Problem \arabic*:, align=left]
\parskip 1em
\item %PROBLEM 1

Find $\textbf{A}+\textbf{B}$, $\textbf{AB}$, $\textbf{BA}$.\\

\item[a)] $\textbf{A} = \begin{bmatrix}  1 & 2  \\   3  & -1  \end{bmatrix}\,\,\,\,\,$ and $\,\,\,\,\,\textbf{B} = \begin{bmatrix} 2 & 1  \\1  & 1 \end{bmatrix}$. \,\,\,\,\,\textbf{A} is $2\times2$ and \textbf{B} is $2\times2$.

 $\textbf{A}+\textbf{B}=\begin{bmatrix}  1 & 2  \\ 3  & -1\end{bmatrix} + \begin{bmatrix} 2 & 1  \\1  & 1 \end{bmatrix}=\begin{bmatrix} 1+2 & 2+1  \\3+1  & -1+1 \end{bmatrix}=\begin{bmatrix} 3 & 3  \\4  & 0 \end{bmatrix}$

 $\textbf{AB}=\begin{bmatrix}  1 & 2  \\ 3  & -1\end{bmatrix} \cdot \begin{bmatrix} 2 & 1  \\1  & 1 \end{bmatrix}=\begin{bmatrix} 1\cdot2+2\cdot1 & 1\cdot 1+2\cdot1  \\3\cdot2+(-1)\cdot1  & 3\cdot1+(-1)\cdot1 \end{bmatrix}=\begin{bmatrix} 4 & 3  \\5  & 2 \end{bmatrix}$

 $\textbf{BA}= \begin{bmatrix} 2 & 1  \\1  & 1 \end{bmatrix}\cdot \begin{bmatrix}  1 & 2  \\ 3  & -1\end{bmatrix}=\begin{bmatrix} 2\cdot1+1\cdot3 & 2\cdot 2+1\cdot(-1)  \\1\cdot1+1\cdot3  & 1\cdot2+1\cdot(-1) \end{bmatrix}=\begin{bmatrix} 5 & 3  \\4  & 1 \end{bmatrix}$


\item[b)] $\textbf{A} = \begin{bmatrix} 2 & 3 & 1\\ 0  & -1 & 2 \end{bmatrix}\,\,\,\,\,$and 
$\,\,\,\,\,\textbf{B} = \begin{bmatrix} 2 & -2  \\ 4  & 3 \\1 & 5 \end{bmatrix}$. \,\,\,\,\,\,\textbf{A} is $2\times3$ and \textbf{B} is $3\times2$.     


    $\textbf{A}+\textbf{B}$ is undefined because the two matrices are incompatible for addition. 

$\textbf{AB}=\begin{bmatrix} 2 & 3 & 1\\ 0  & -1 & 2 \end{bmatrix}\cdot\begin{bmatrix} 2 & -2  \\ 4  & 3 \\1 & 5 \end{bmatrix}=\begin{bmatrix} 2\cdot2+3\cdot4+1\cdot1 & 2\cdot (-2)+3\cdot3+1\cdot 5  \\0\cdot2+(-1)\cdot4 +2\cdot 1 & 0\cdot(-2)+(-1)\cdot3 +2\cdot5 \end{bmatrix}=\phantom{\textbf{AB}=}\begin{bmatrix} 17 & 10  \\-2  & 7 \end{bmatrix}$


 $\textbf{BA}=\begin{bmatrix} 2 & -2  \\ 4  & 3 \\1 & 5 \end{bmatrix}\cdot \begin{bmatrix} 2 & 3 & 1\\ 0  & -1 & 2 \end{bmatrix}=\begin{bmatrix} 2\cdot2+(-2)\cdot0 & 2\cdot 3+(-2)\cdot(1) &2\cdot1+(-2)\cdot 2  \\4\cdot2+3\cdot0&4\cdot3 +3\cdot (-1) & 4\cdot1+3\cdot2\\ 1\cdot2+5\cdot0&1\cdot3+5\cdot(-1)&1\cdot1+5\cdot2 \end{bmatrix}=\phantom{\textbf{BA}=}\begin{bmatrix} 4 & 8&-2  \\8  & 9&10\\2&-2&11 \end{bmatrix} $


\item %PROBLEM 2
  Let $\textbf{a}' = \begin{bmatrix}
       3 & 6 &-3 & 5 &  9& 2  \\
     \end{bmatrix}$. Find \textbf{a}$'$\textbf{a} and \textbf{aa}$'$.


 \textbf{a}$'$ is the transpose of $\textbf{a} = \begin{bmatrix}
       3 \\ 6 \\-3 \\ 5 \\  9\\ 2  \\
     \end{bmatrix}$

 \[\textbf{a}'\textbf{a} = \begin{bmatrix}
       3 & 6 &-3 & 5 &  9& 2  \\
     \end{bmatrix} \cdot \begin{bmatrix}
       3 \\ 6 \\-3 \\ 5 \\  9\\ 2  \\
     \end{bmatrix} = \begin{bmatrix}
       3^2 + 6^2 + (-3)^2 + 5^2 +  9^2 + 2^2  \\
     \end{bmatrix} =[164] \]

   $\textbf{aa}' = \begin{bmatrix}
       3 \\ 6 \\-3 \\ 5 \\  9\\ 2  \\
     \end{bmatrix}\cdot \begin{bmatrix}
       3 & 6 &-3 & 5 &  9& 2  \\
     \end{bmatrix}  = \begin{bmatrix}
       3^2& 3\cdot 6 & 3\cdot (-3) & 3\cdot 5 &  3\cdot 9 & 3\cdot 2  \\
  6\cdot 3& 6^2 & 6\cdot (-3) & 6\cdot 5 &  6\cdot 9 & 6\cdot 2  \\
    (-3)\cdot 3& (-3)\cdot 6 & (-3)^2 & (-3)\cdot 5 &  (-3)\cdot 9 & (-3)\cdot 2  \\
      5\cdot 3& 5\cdot 6 & 5\cdot (-3) & 5^2 &  5\cdot 9 & 5\cdot 2  \\
        9\cdot 3& 9\cdot 6 & 9\cdot (-3) & 9\cdot 5 &  9^2 & 9\cdot 2  \\
         2\cdot 3& 2\cdot 6 & 2\cdot (-3) & 2\cdot 5 &  2\cdot 9 & 2^2  \\
     \end{bmatrix} \\\phantom{\textbf{aa}'}=\begin{bmatrix}
       9& 18 & 9 & 15 &  27 & 6  \\
  18& 36 & -18 & 30 &  54 & 12  \\
    -9& -18 & 9 & -15 &  27 & -6  \\
      15& 30 & 15 & 10 &  45 & 10 \\
        27& 54 & 27 & 45 &  81 & 18  \\
         6& 12 & 6 & 10 &  18 & 4  \\
     \end{bmatrix} $







   Let $\textbf{b}' = \begin{bmatrix}
       1 & 1 & 1 & 1 & 1 & 1  \\
     \end{bmatrix}\,\,\,\,\,$
     and
    $\,\,\,\,\,\textbf{c}' = \begin{bmatrix}
       1 & -1 & 1 & -1 & 1 & -1  \\
     \end{bmatrix}$. Find \textbf{b}$'$\textbf{c} and \textbf{bc}$'$.


  \end{enumerate}
\end{document}

在此处输入图片描述

答案2

这里有一个解决方案：由于您加载了amsthmI 定义problem为example-style新定理。这满足了问题的自动编号和交叉引用。此外，问题以斜体排版。如果您喜欢罗马字体，请将示例样式更改为其他definition样式。

此外，由于定理之间的垂直间距似乎太紧，我在软件包的帮助下etoolbox，更准确地说是它的\AtBeginEnvironment命令，添加了一些垂直间距。

至于对齐，我使用环境align*。在我看来，如果某些矩阵的列右对齐，它们看起来会更好（那些具有一些负系数的矩阵）。对于这些，我使用了环境bmatrix*frommathtools和选项[r]。不要加载amsmath：mathtools会为您完成。

\documentclass[paper=a4, fontsize=11pt]{scrartcl} % A4 paper

\usepackage{mathtools, amsthm, amssymb}
\usepackage{enumitem}
\usepackage{etoolbox}

\theoremstyle{example}
\newtheorem{problem}{Problem}
\AtBeginEnvironment{problem}{\vspace{1ex}}{}{}

\begin{document}

\begin{problem}\label{pb1}% PROBLEM 1
Find $\mathbf{A}+\mathbf{B}$, $\mathbf{AB}$, $\mathbf{BA}$.
\end{problem}

 \begin{enumerate}[label = \alph*), wide = 0em]
\item $\mathbf{A} = \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}$\enspace and \enspace$\mathbf{B} = \begin{bmatrix} 2 & 1 \\1 & 1 \end{bmatrix}$. $ \mathbf{A} $ is $2\times2$ and $ \mathbf{B} $ is $2\times2$.

\begin{align*}
    \mathbf{A}+\mathbf{B} &= \begin{bmatrix*}[r] 1 & 2 \\ 3 & -1\end{bmatrix*} + \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix*}[r] 1+2 & 2+1 \\ 3+1 & -1+1 \end{bmatrix*}= \begin{bmatrix} 3 & 3 \\ 4 & 0 \end{bmatrix}\\[1ex]
%
  \mathbf{AB} &= \begin{bmatrix*}[r] 1 & 2 \\ 3 & -1\end{bmatrix*} \cdot \begin{bmatrix} 2 & 1 \\1 & 1 \end{bmatrix}= \begin{bmatrix} 1\cdot2+2\cdot1 & 1\cdot 1+2\cdot1 \\3\cdot2+(-1)\cdot1 & 3\cdot1+(-1)\cdot1 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ 5 & 2 \end{bmatrix} \\[1ex]
%
  \mathbf{BA} &= \begin{bmatrix} 2 & 1 \\1 & 1 \end{bmatrix}\cdot \begin{bmatrix*}[r] 1 & 2 \\ 3 & -1\end{bmatrix*} = \begin{bmatrix} 2\cdot1+1\cdot3 & 2\cdot 2+1\cdot(-1) \\1\cdot1+1\cdot3 & 1\cdot2+1\cdot(-1) \end{bmatrix} = \begin{bmatrix} 5 & 3 \\ 4 & 1 \end{bmatrix} \\
\end{align*}

\item $\mathbf{A} = \begin{bmatrix} 2 & 3 & 1\\ 0 & -1 & 2 \end{bmatrix}$\enspace and\enspace$ \mathbf{B} = \begin{bmatrix} 2 & -2 \\ 4 & 3 \\1 & 5 \end{bmatrix}$. $ \mathbf{A} $ is $2\times3$ and $ \mathbf{B} $ is $3\times2$.

    $\mathbf{A}+\mathbf{B}$ is undefined because the two matrices are incompatible for addition.

\begin{align*}
   \mathbf{AB} = \begin{bmatrix*}[r] 2 & 3 & 1 \\ 0 & -1 & 2 \end{bmatrix*} \cdot
   \smash{\begin{bmatrix*}[r] 2 & -2 \\ 4 & 3 \\ 1 & 5 \end{bmatrix*}}
   & = \begin{bmatrix}
   2\cdot 2+3\cdot 4+1\cdot 1 & 2\cdot (-2)+3\cdot 3+1\cdot 5 \\
   0\cdot2+(-1)\cdot 4 +2\cdot 1 & 0\cdot (-2)+(-1)\cdot 3 +2\cdot 5
   \end{bmatrix}\\
    & = \begin{bmatrix*}[r] 17 & 10 \\-2 & 7 \end{bmatrix*}\\[1.5ex]
%
   \mathbf{BA} = \begin{bmatrix*}[r] 2 & -2 \\ 4 & 3 \\1 & 5 \end{bmatrix*}\cdot
   \begin{bmatrix*}[r] 2 & 3 & 1\\ 0 & -1 & 2 \end{bmatrix*}
   & = \begin{bmatrix}
   2\cdot 2+(-2)\cdot 0 & 2\cdot 3+(-2)\cdot (-1) & 2\cdot 1+(-2)\cdot 2 \\4\cdot 2+3\cdot 0&4\cdot3 +3\cdot (-1) & 4\cdot1+3\cdot2 \\
   1\cdot 2+5\cdot 0& 1\cdot3 + 5\cdot (-1)&1\cdot 1+5\cdot 2
   \end{bmatrix}\\
    & = \begin{bmatrix*}[r] 4 & 8 & -2 \\
    8 & 9 & 10\\ 2 & -2 & 11
    \end{bmatrix*}
\end{align*}

\end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{problem}\label{pb2} %PROBLEM 2
  Let $\mathbf{a}' = \begin{bmatrix}3 & 6 &-3 & 5 & 9& 2 \\ \end{bmatrix}$. Find $\mathbf{a}'\mathbf{a}$ and $\mathbf{aa}'$.
\end{problem}

$\mathbf{a}'$ is the transpose of $\mathbf{a} = \begin{bmatrix}3 \\ 6 \\-3 \\ 5 \\ 9\\ 2 \end{bmatrix}$.

\begin{align*}
  \mathbf{a}' \mathbf{a} & = \begin{bmatrix} 3 & 6 &-3 & 5 & 9& 2 \end{bmatrix} \cdot \begin{bmatrix}
       3 \\ 6 \\-3 \\ 5 \\ 9\\ 2 \end{bmatrix} = \begin{bmatrix} 3^2 + 6^2 + (-3)^2 + 5^2 + 9^2 + 2^2
     \end{bmatrix} = [164] \\
%
  \mathbf{aa}' & = \begin{bmatrix} 3 \\ 6 \\-3 \\ 5 \\ 9\\ 2 \end{bmatrix}\cdot \begin{bmatrix} 3 & 6 &-3 & 5 & 9& 2 \end{bmatrix}
   = \begin{bmatrix}
       3^2& 3\cdot 6 & 3\cdot (-3) & 3\cdot 5 & 3\cdot 9 & 3\cdot 2 \\
  6\cdot 3& 6^2 & 6\cdot (-3) & 6\cdot 5 & 6\cdot 9 & 6\cdot 2 \\
    (-3)\cdot 3& (-3)\cdot 6 & (-3)^2 & (-3)\cdot 5 &( -3)\cdot 9 & (-3)\cdot 2 \\
      5\cdot 3& 5\cdot 6 & 5\cdot (-3) & 5^2 & 5\cdot 9 & 5\cdot 2 \\
        9\cdot 3& 9\cdot 6 & 9\cdot (-3) & 9\cdot 5 & 9^2 & 9\cdot 2 \\
         2\cdot 3& 2\cdot 6 & 2\cdot (-3) & 2\cdot 5 & 2\cdot 9 & 2^2
     \end{bmatrix} \\
      & = \begin{bmatrix*}[r]
       9& 18 & 9 & 15 & 27 & 6 \\
  18& 36 & -18 & 30 & 54 & 12 \\
    -9& -18 & 9 & -15 & 27 & -6 \\
      15& 30 & 15 & 10 & 45 & 10 \\
        27& 54 & 27 & 45 & 81 & 18 \\
         6& 12 & 6 & 10 & 18 & 4
     \end{bmatrix*}
\end{align*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{problem}\label{pb3}%% PROBLEM 3
       Let $\mathbf{b}' = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix}$\enspace and\enspace$ \mathbf{c}' = \begin{bmatrix} 1 & -1 & 1 & -1 & 1 & -1 \end{bmatrix}$. Find $\mathbf{b}'\mathbf{c}$ and $\mathbf{bc}'$.
\end{problem}

\end{document}

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