1

I found this code here (stackexchange) and modified it for the matrix I needed to make. But I want to have thoses brackets[ ]. I tried to change blockarray for \left[ and \right]. Can anybody help me please? I needed \usepackage{blkarray} for this.

 \[ \newcounter{rowct}\newcounter{colct}%\setlength\extrarowheight{1cm}
    \begin{blockarray}{*{8}{c}}
    \begin{block}{r*{7}{>{\refstepcounter{colct}\mathclap{\mathbf\thecolct}}c}}
     & & & & & & & \\%[-1ex]
     \end{block}
     \begin{block}{ >{\refstepcounter{rowct}\mathbf\therowct}r( >{\hspace{0.2em}}*{7}{c} <{\hspace{0.2em}})}
      & 0 & 1 & 1 & 1 & 1 & 0 & 0 \rule{0pt}{3.5ex} \\
      & 1 & 0 & 1 & 1 & 0 & 1 & 0 \\
      & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
      & 1 & 1 & 0 & 0 & 1 & 1 & 1 \\
      & 1 & 0 & 0 & 1 & 0 & 0 & 1\\
      & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\
      & 0 & 0 & 0 & 1 & 1 & 0 & 0\\[1ex]
     \end{block}
    \end{blockarray}
    \]

enter image description here

Aezir
  • 13

2 Answers2

0

Just replace the pair ( ) in the second block with a pair of [ ]:

\documentclass{article}
\usepackage{mathtools}
\usepackage{blkarray}

\begin{document}

[ \newcounter{rowct}\newcounter{colct}%\setlength\extrarowheight{1cm} \begin{blockarray}{{8}{c}} \begin{block}{r{7}{>{\refstepcounter{colct}\mathclap{\mathbf\thecolct}}c}} & & & & & & & \%[-1ex] \end{block} \begin{block}{ >{\refstepcounter{rowct}\mathbf\therowct}r[ >{\hspace{0.2em}}*{7}{c} <{\hspace{0.2em}}]} & 0 & 1 & 1 & 1 & 1 & 0 & 0 \rule{0pt}{3.5ex} \ & 1 & 0 & 1 & 1 & 0 & 1 & 0 \ & 1 & 1 & 0 & 0 & 0 & 0 & 0 \ & 1 & 1 & 0 & 0 & 1 & 1 & 1 \ & 1 & 0 & 0 & 1 & 0 & 0 & 1\ & 0 & 1 & 0 & 1 & 0 & 0 & 0 \ & 0 & 0 & 0 & 1 & 1 & 0 & 0\[1ex] \end{block} \end{blockarray} ]

\end{document}

enter image description here

Bernard
  • 271,350
0

You can do that easily with nicematrix.

\documentclass{article}
\usepackage{nicematrix}

\begin{document}

$\begin{bNiceMatrix}% [ first-row, code-for-first-row = \mathbf{\arabic{jCol}}, first-col, code-for-first-col = \mathbf{\arabic{iRow}}, ] & & & & & & & \ & 0 & 1 & 1 & 1 & 1 & 0 & 0 \ & 1 & 0 & 1 & 1 & 0 & 1 & 0 \ & 1 & 1 & 0 & 0 & 0 & 0 & 0 \ & 1 & 1 & 0 & 0 & 1 & 1 & 1 \ & 1 & 0 & 0 & 1 & 0 & 0 & 1 \ & 0 & 1 & 0 & 1 & 0 & 0 & 0 \ & 0 & 0 & 0 & 1 & 1 & 0 & 0 \ \end{bNiceMatrix}$

\end{document}

Output of the above code

F. Pantigny
  • 40,250