3

I start with this. You see that we have a few equations that all are refereed to just one number that is $(1)$ put exactly in the middle. Here, I want to put a text instead of that number and so I did the following MWF by manipulating the codes:

\documentclass[12pt]{book}
\usepackage{amsmath,amssymb,amsthm,amsfonts} 
\begin{document}
\begin{flalign*}
(a)~ &  \det (A_1+A', A_2)=\det (A_1,A_2)+\det (A'_1,A_2) &\\
(b)~ &  \det(A_1, A_2+A'_2)=\det (A_1,A_2)+\det (A_1, A'_2) \\[-1.8ex]
     & \hspace*{9.5cm}\text{(666)}\\[-1.8ex]
(c)~ &  \det (cA_1,A_2)=c\det (A_1,A_2) \\
(d)~& \det(A_1, cA_2)=c\det (A_1, A_2)
\end{flalign*}
\end{document}

to find:

enter image description here

Is there another way for that? Please guide me if this question is a duplicate and feel free to add needed tags.

Thanks for your every seconds. enter image description here

2

A solution with minimal changes:

\documentclass[12pt]{book}
\usepackage{amsmath,amssymb,amsthm,amsfonts} 
\begin{document}

\begin{equation}
\tag{Damien 666}
\begin{array}{ll}
\text{(a)}~ &  \det (A_1+A', A_2)=\det (A_1,A_2)+\det (A'_1,A_2) \\
\text{(b)}~ &  \det(A_1, A_2+A'_2)=\det (A_1,A_2)+\det (A_1, A'_2) \\
\text{(c)}~ &  \det (cA_1,A_2)=c\det (A_1,A_2) \\
\text{(d)}~& \det(A_1, cA_2)=c\det (A_1, A_2)
\end{array}
\end{equation}
\end{document}

enter image description here

1
  • Thanks for your time.
    – Mikasa
    Sep 8 at 7:25
2

If you compile this:

\documentclass[12pt]{book}
\usepackage{amsmath,amssymb,amsthm,amsfonts} 
\begin{document}
\begin{equation}
\begin{aligned}
(a)~ &  \det (A_1+A', A_2)=\det (A_1,A_2)+\det (A'_1,A_2) \\
(b)~ &  \det(A_1, A_2+A'_2)=\det (A_1,A_2)+\det (A_1, A'_2) \\
(c)~ &  \det (cA_1,A_2)=c\det (A_1,A_2) \\
(d)~& \det(A_1, cA_2)=c\det (A_1, A_2)
\end{aligned}
\tag{666}
\end{equation}
\end{document}

then you will get this:

enter image description here

1
  • Thanks Jose but, I preferred the statements get left.
    – Mikasa
    Sep 8 at 7:25
2

Here's a solution that embeds an enumerate environment in a minipage environment. This approach will let you cross-reference the enumerated items via the standard \label-\ref mechanism.

enter image description here

\documentclass[12pt]{book}
\usepackage{enumitem} 
\usepackage{amsmath} % for '\tag' macro

\begin{document}
\begin{equation} 
\begin{minipage}{0.9\textwidth}
\begin{enumerate}[label=(\alph*),left=0pt,itemsep=2pt]
\item  $\det(A_1+A', A_2)=\det (A_1,A_2)+\det (A'_1,A_2)$
\item  $\det(A_1, A_2+A'_2)=\det (A_1,A_2)+\det (A_1, A'_2)$
\item  $\det(cA_1,A_2)=c\det (A_1,A_2)$
\item  $\det(A_1, cA_2)=c\det (A_1, A_2)$
\end{enumerate}
\end{minipage}
\tag{666}
\end{equation}
\end{document}
1
  • Thank you for the new way! It opens my eyes to see the codes better! :)
    – Mikasa
    Sep 10 at 16:52
2

You can do like that, but I'd prefer subequations.

\documentclass[12pt]{book}
\usepackage{amsmath,amssymb,amsthm,amsfonts}

\usepackage{lipsum} % for context

\begin{document}

\setcounter{equation}{665}

\lipsum[10][1-5]
\begin{flalign}
\makebox[2em][l]{(a)} & \det (A_1+A', A_2)=\det (A_1,A_2)+\det (A'_1,A_2) & \notag \\
\makebox[2em][l]{(b)} & \det(A_1, A_2+A'_2)=\det (A_1,A_2)+\det (A_1, A'_2) \notag \\[-1.8ex]
    \\[-1.8ex]
\makebox[2em][l]{(c)} & \det (cA_1,A_2)=c\det (A_1,A_2) \notag \\
\makebox[2em][l]{(d)} & \det(A_1, cA_2)=c\det (A_1, A_2) \notag
\end{flalign}
\lipsum[10][1-5]

\begin{subequations}
\lipsum[10][1-5]
\begin{align}
& \det (A_1+A', A_2)=\det (A_1,A_2)+\det (A'_1,A_2)\\
& \det(A_1, A_2+A'_2)=\det (A_1,A_2)+\det (A_1, A'_2) \\
& \det (cA_1,A_2)=c\det (A_1,A_2) \\
& \det(A_1, cA_2)=c\det (A_1, A_2)
\end{align}
\lipsum[10][1-5]
\end{subequations}

\end{document}

enter image description here

1
  • Thank you very much! :)
    – Mikasa
    Sep 10 at 16:51

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