# How can I create a table like this

\documentclass[12pt,a4paper]{article}
\usepackage{amsmath,amssymb}

\begin{document}

$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline & & & \begin{array}{l} \text{Basis Inverse} \\ B_1^{-1} \end{array} & & & & & \\ \hline B & X_B & \begin{array}{l} \beta_0 \\ Z^{\prime} \end{array} & \begin{array}{l} \beta_1 \\ S_1 \end{array} & \begin{array}{l} \beta_2 \\ S_2 \end{array} & \begin{array}{c} y_1 \\ c_k-Z_k \end{array} & \begin{array}{c} \text { Min Ratio } \\ \frac{X_B}{y_1} \end{array} & x_1 & x_2 \\ \hline Z^{\prime} & 0 & 1 & 0 & 0 & -1 & - & -1 & -1 \\ \hline S_1 & 6 & 0 & 1 & 0 & 2 & 3 & 2 & 5 \\ \hline S_2 & 2 & 0 & 0 & 1 & 1 & 2 & 1 & 1 \\ \hline \end{array}$

\end{document}


You can use \multicolumn in an array (same as tabular). The tricky bit was dividing the subcolumns evenly.

\documentclass[12pt,a4paper]{article}
\usepackage{amsmath,amssymb}
\usepackage{mathtools}% for mathmakebox
\usepackage{bm}% bold math

\newsavebox{\tempboxa}
\newsavebox{\tempboxb}

\begin{document}

\savebox\tempboxa{\textbf{\begin{tabular}{@{}c@{}}
Table
\end{tabular}}}% measure width

\savebox\tempboxb{$\begin{array}{@{}c@{}} \textbf{Basis Inverse} \\ \bm{B_1^{-1}} \end{array}$}% measure width

$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline & & \multicolumn{3}{c|}{\usebox\tempboxb} & & & \multicolumn{2}{c|}{\usebox\tempboxa}\\ \hline \bm{B} & \bm{X_B} & \mathmakebox[\dimexpr 0.333\wd\tempboxb-1.333\arraycolsep]{\begin{array}{@{}l@{}} \bm{\beta_0} \\ \bm{Z^{\prime}} \end{array}} & \mathmakebox[\dimexpr 0.333\wd\tempboxb-1.333\arraycolsep]{\begin{array}{@{}l@{}} \bm{\beta_1} \\ \bm{S_1} \end{array}} & \mathmakebox[\dimexpr 0.333\wd\tempboxb-1.333\arraycolsep]{\begin{array}{@{}l@{}} \bm{\beta_2} \\ \bm{S_2} \end{array}} & \begin{array}{@{}c@{}} \bm{y_1} \\ \bm{c_k-Z_k} \end{array} & \begin{array}{@{}c@{}} \textbf {Min Ratio} \\ \bm{\frac{X_B}{y_1}} \end{array} & \mathmakebox[\dimexpr 0.5\wd\tempboxa-\arraycolsep]{\bm{x_1}} & \mathmakebox[\dimexpr 0.5\wd\tempboxa-\arraycolsep]{\bm{x_2}} \\ \hline Z^{\prime} & 0 & 1 & 0 & 0 & -1 & - & -1 & -1 \\ \hline S_1 & 6 & 0 & 1 & 0 & 2 & 3 & 2 & 5 \\ \hline S_2 & 2 & 0 & 0 & 1 & 1 & 2 & 1 & 1 \\ \hline \end{array}$

\end{document}


• It's not running correctly. Mar 2 at 17:29

Here is a possibility using nicematrix. The 1×1 blocks allow you to use \\ in a cell.

You must compile twice each time you make a change.

\documentclass[12pt,a4paper]{article}
\usepackage{amsmath,amssymb,nicematrix,bm}

\begin{document}

$\NiceMatrixOptions{cell-space-limits = 5pt} \begin{NiceArray}{cc*3{w{c}{1cm}}cc*2{w{c}{1cm}}}[hvlines] % 2 centered columns, 3 fixed-width centered columns, 2 centered columns, 2 fixed-width centered columns & & \Block{1-3}{\textup{Basis Inverse}\\B_1^{-1}} & & & & & \Block{1-2}{\textup{Additional}\\\textup{Table}}\\ \bm{B} & \bm{X_B} & \Block{1-1}{\bm{\beta_0}\\\bm{Z'}} & \Block{1-1}{\bm{\beta_1}\\\bm{S_1}} & \Block{1-1}{\bm{\beta_2}\\\bm{S_2}} & \Block{1-1}{\bm{y_1}\\\bm{c_k-Z_k}} & \Block{1-1}{\textbf{Min Ratio}\\\dfrac{\bm{X_B}}{\bm{y_1}}} & \bm{x_1} & \bm{x_2} \\ Z' & 0 & 1 & 0 & 0 & -1 & - & -1 & -1 \\ S_1 & 6 & 0 & 1 & 0 & 2 & 3 & 2 & 5 \\ S_2 & 2 & 0 & 0 & 1 & 1 & 2 & 1 & 1 \end{NiceArray}$

\end{document}

\documentclass[12pt,a4paper]{article}

\begin{document}
$\renewcommand{\arraystretch}{2.2} \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \multirow[b]{2}{*}{B} & \multirow[b]{2}{*}{X_B} & \multicolumn{5}{|c|}{ Basis Inverse B_1^{-1}} & \multirow[b]{2}{*}{\begin{tabular}{c} y_1 \\[-0.5em] C_k - Z_k \end{tabular}} & \multirow[b]{2}{*}{\begin{tabular}{c} Min Ratio \\[-0.5em] \dfrac{X_B}{y_1} \end{tabular}} & \multicolumn{3}{|c|}{ Additional table } \\ \hline & & \begin{tabular}{c} \beta_0 \\ Z \end{tabular} & \begin{tabular}{l} \beta_1 \\ S_1 \end{tabular} & \begin{tabular}{l} \beta_2 \\ S_2 \end{tabular} & \begin{tabular}{l} \beta_3 \\ S_3 \end{tabular} & \begin{tabular}{l} \beta_4 \\ S_4 \end{tabular} & & & x_1 & x_2 & x_3 \\ \hlineZ & 0 & 1 & 0 & 0 & 0 & 0 & & - & -19500 & -23000 & -6250 \\ \hlineS_1 & 70 & 0 & 1 & 0 & 0 & 0 & & - & 10 & 10 & 50 \\ \hlineS_2 & 12 & 0 & 0 & 1 & 0 & 0 & & - & 3 & 7 & 2 \\ \hlineS_3 & 25 & 0 & 0 & 0 & 1 & 0 & & - & 5 & 10 & 10 \\ \hlineS_4 & 16 & 0 & 0 & 0 & 0 & 1 & & - & 3 & 8 & 5 \\ \hline \end{tabular}$

\end{document}

• well, is that an answer? Then please descibe what you have changed and show your result. Mar 2 at 20:33