4
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\begin{document}
\begin{enumerate}
\item 
Applying 2 fold symmetry // X$_1$ :
\\
\\
\[
a=
  \begin{bmatrix}
    1 & 0 & 0 \\
    0 & -1 & 0 \\
    0 & 0 & -1 
  \end{bmatrix}
\]
\\
\[
\alpha =
  \begin{bmatrix}
    1 & 0 & 0 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 & 0 & 0 \\
    0 & 0 & 1 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & 0 & 0 \\
    0 & 0 & 0 & 0 & -1 & 0 \\
    0 & 0 & 0 & 0 & 0 & -1 \\
  \end{bmatrix}
\]
\\
Applying 3 fold symmetry // X$_3$: 
\\
\\
\[
a=
  \begin{bmatrix}
    \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
    \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
    0 & 0 & 1 
  \end{bmatrix}
\]
\\
\[
\alpha_1 =
  \begin{bmatrix}
    1 & 0 & 0 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 & 0 & 0 \\
    0 & 0 & 1 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & 0 & 0 \\
    0 & 0 & 0 & 0 & -1 & 0 \\
    0 & 0 & 0 & 0 & 0 & -1 \\
  \end{bmatrix}
\]
\\
For d$_{im}$,
\\
d' = a * d * \alpha
\[
    \begin{bmatrix}
    \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
    \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
    0 & 0 & 1 
    \end{bmatrix}

    \begin{bmatrix}
    \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
    \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
    0 & 0 & 1 
    \end{bmatrix}
\]
\end{enumerate}
\end{document}
|improve this question
  • 6
    The d' = a * d * \alpha is outside math mode, but it needs math mode: $d' = a * d * \alpha$. But the whole code is ugly. You should avoid \\ before and after \[…\]. You should not use X$_3$ but $X_3$ etc. – Schweinebacke Sep 5 '17 at 9:30
  • 1
    Or perhaps something like $\text{x}_3, if the intent is a non-italic variable. (My memory is fuzzy on the correct command to use, though. \mathrm instead?) – chepner Sep 5 '17 at 11:34
14

Adaptations:

  • math mode for d' = a * d * \alpha
  • no linebreak inside \[ ... \]
  • use align* instead of \[ ... \]
  • use \cdot instead of *
  • use linebreaks instead of \\
  • use $X_1$ instead of X$_1$
  • you can use \renewcommand{\arraystretch}{1.5} to increase the spacing of the matrices with the fractions (see How can I increase the line spacing in a matrix?) or write the fractions with $a/b$ as JPi suggested.

Code:

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}

\begin{document}
\begin{enumerate}
\item 
    Applying 2 fold symmetry // $X_1$:
    \begin{align*}
        a &=
          \begin{bmatrix}
            1 & 0 & 0 \\
            0 & -1 & 0 \\
            0 & 0 & -1
          \end{bmatrix}
        \\
        \alpha &=
        \begin{bmatrix}
            1 & 0 & 0 & 0 & 0 & 0 \\
            0 & 1 & 0 & 0 & 0 & 0 \\
            0 & 0 & 1 & 0 & 0 & 0 \\
            0 & 0 & 0 & 1 & 0 & 0 \\
            0 & 0 & 0 & 0 & -1 & 0 \\
            0 & 0 & 0 & 0 & 0 & -1 \\
        \end{bmatrix}
    \end{align*}

    Applying 3 fold symmetry // $X_3$: 
    \begin{align*}
        a &=
        {
        \renewcommand{\arraystretch}{1.5}
        \begin{bmatrix}
            \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
            \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
            0 & 0 & 1 
        \end{bmatrix}
        }
        \\
        \alpha_1 &=
        \begin{bmatrix}
            1 & 0 & 0 & 0 & 0 & 0 \\
            0 & 1 & 0 & 0 & 0 & 0 \\
            0 & 0 & 1 & 0 & 0 & 0 \\
            0 & 0 & 0 & 1 & 0 & 0 \\
            0 & 0 & 0 & 0 & -1 & 0 \\
            0 & 0 & 0 & 0 & 0 & -1 \\
        \end{bmatrix}
    \end{align*}

    For $d_{im}$,
    $d' = a \cdot d \cdot \alpha$
    {
    \renewcommand{\arraystretch}{1.5}
    \begin{align*}
         \begin{bmatrix}
             \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
             \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
             0 & 0 & 1 
         \end{bmatrix}
        \\
         \begin{bmatrix}
             \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
             \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
             0 & 0 & 1 
         \end{bmatrix}
    \end{align*}
    }
\end{enumerate}
\end{document}

Result:

enter image description here

|improve this answer
  • 3
    You've forgotten to list the adaption $X_1$ instead of X$_1$. Nevertheless +1 for listing all adaptions! – Schweinebacke Sep 5 '17 at 9:39
  • 3
    You may want to increase the spacing in the matrices or better yet: write $\sqrt{3}/2$ instead of $\frac{\sqrt{3}}{2}$. – JPi Sep 5 '17 at 10:03
  • Thank you Schweinebacke and JPi. I added your suggestions to the answer. – dexteritas Sep 5 '17 at 10:51
  • the width of the matrices in the first two blocks makes it easy to identify them as separate. but it wouldn't hurt to have a little vertical space between the two in the last block. add with an option after the double backslash separating them, say \\[3pt] and be sure not to leave a (typed) space between the second backslash and the [. – barbara beeton Sep 5 '17 at 15:43
2

I did not do much, just corrected some shells - for \alpha, it can only be used in a mathematical environment, either $\alpha$ or \[\alpha\] it is possible to print the \alpha character in a text, but then it is \textalpha. so I will avoid writing X$_1$ but rather $X_1$ to leave its coherence to the mathematical text. I also corrected some spaces generating in the mathematical formulas

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\begin{document}
\begin{enumerate}
\item 
Applying 2 fold symmetry // X$_1$ :
\\
\\
\[
a=
  \begin{bmatrix}
    1 & 0 & 0 \\
    0 & -1 & 0 \\
    0 & 0 & -1 
  \end{bmatrix}
\]
\\
\[
\alpha =
  \begin{bmatrix}
    1 & 0 & 0 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 & 0 & 0 \\
    0 & 0 & 1 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & 0 & 0 \\
    0 & 0 & 0 & 0 & -1 & 0 \\
    0 & 0 & 0 & 0 & 0 & -1 \\
  \end{bmatrix}
\]
\\
Applying 3 fold symmetry // X$_3$: 
\\
\\
\[
a=
  \begin{bmatrix}
    \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
    \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
    0 & 0 & 1 
  \end{bmatrix}
\]
\\
\[
\alpha_1 =
  \begin{bmatrix}
    1 & 0 & 0 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 & 0 & 0 \\
    0 & 0 & 1 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & 0 & 0 \\
    0 & 0 & 0 & 0 & -1 & 0 \\
    0 & 0 & 0 & 0 & 0 & -1 \\
  \end{bmatrix}
\]
\\
For d$_{im}$,
\\
$d' = a * d * \alpha$
\[
    \begin{bmatrix}
    \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
    \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
    0 & 0 & 1 
    \end{bmatrix}\\
    \begin{bmatrix}
    \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
    \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
    0 & 0 & 1 
    \end{bmatrix}
\]
\end{enumerate}
\end{document}
|improve this answer

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