Creating additional space in tabular for \dfrac, while vertically centering multirow and other cells

Question

\documentclass{article}
\usepackage[a4paper, portrait, margin=0.5in]{geometry}
\usepackage[table]{xcolor}
\usepackage[most]{tcolorbox}
\usepackage{amsmath}
\usepackage{multirow}

\begin{document}
\large

\renewcommand{\arraystretch}{1.5}
\newcolumntype{P}[1]{>{\raggedright\arraybackslash}p{#1}}
\newcolumntype{M}[1]{>{\raggedright\arraybackslash}m{#1}}
\begin{center}
    \begin{tabular}{ | M{1.5cm} | M{9.5cm} | M{1cm} | M{4.5cm} |}
    \hline
    \centering \multirow{6}{*}{1} &
    $f(3) = 100$
    \par \medskip
    $a=\dfrac{50}{21}-\dfrac{13}{42}b$ & \hfill \textbf{A} & B
    \tabularnewline
    \cline{2-4}
    & $3x^2-x-4=(3x-4)(x+1)$ & \hfill \textbf{C} & D
    \tabularnewline
    \cline{2-4}
    & \vspace{0.15cm} $f(-1) = 0$ or $f(\dfrac{4}{3}) = 0$
    \par \medskip
    $b=-\dfrac{2}{3}a$ & \hfill \textbf{E} & F
    \tabularnewline
    \cline{2-4}
    & $a=3, \; b=-2$ & \hfill \textbf{G} & H
    \tabularnewline
    \hline
\end{tabular}
\end{center}

\end{document}

This code gives the following output:

The bottom \dfrac at rows AB and EF are getting cut off by the \cline. Is there a way to create additional space for the fractions while maintaining vertical centering of the other cells in the row? (The cells with the \dfrac should also be vertically centered)

The vertical centering of the multirow column (labelled "1") is passable but I am a perfectionist. Can anyone also help to ensure vertical centering of the multirow column?

Thanks in advance!

Answer 1

Like this?

%\documentclass{article}
%\usepackage[a4paper, portrait, margin=0.5in]{geometry}
\documentclass[margin=3mm]{standalone}
\usepackage{xcolor}
\usepackage{tabularray}
\UseTblrLibrary{amsmath}

\begin{document}
    \begin{tblr}{hlines, vlines,
                 colspec={Q[c,m, wd=15mm] Q[l,m, wd=95mm, mode=dmath] 
                          Q[r,m, wd=10mm, font=\bfseries] Q[l, m, wd=45mm]},
                 rowsep=7pt,
                 }
\SetCell[r=4]{c}    1
    &   \begin{aligned}
            f(3) & = 100  \\
            a    & =\dfrac{50}{21} - \dfrac{13}{42}b
        \end{aligned}   &   A   &   B       \\
    &   3x^2-x-4=(3x-4)(x+1)
                        &   C   &   D       \\
    &   \begin{aligned}
            f(-1) & = 0 \text{ or } f\biggl(\dfrac{4}{3}\biggr) = 0  \\
                b & = -\dfrac{2}{3}a
        \end{aligned}   &   E   &   F       \\
    &   a=3,  b=-2      &   G   &   H       \\
    \end{tblr}
\end{document}

Addendum:
The simplest way to left align all equations in cells (as you request in comment below) and preserve simple table code and desired vertical spaces around cells contents, is use a array environment. For example:

\begin{array}{@{}l}
            f(3)  = 100 \\
            a     =\dfrac{50}{21} - \dfrac{13}{42}b
\end{array}

For shorter writing it may be handy to define new environmentČ

\newenvironment{la}%
    {
    \begin{array}{@{} l}
    }
    {
    \end{array}
    }

The MWE, which contain both ways of arrays writing is:

%\documentclass{article}
%\usepackage[a4paper, portrait, margin=0.5in]{geometry}
\documentclass[margin=3mm]{standalone}
\usepackage{xcolor}
\usepackage{tabularray}
\UseTblrLibrary{amsmath}
\newenvironment{la}%
    {
    \begin{array}{@{} l}
    }
    {
    \end{array}
    }

\begin{document}
    \begin{tblr}{hlines, vlines,
                 colspec={Q[c,m, wd=15mm] Q[l,m, wd=95mm, mode=dmath]
                          Q[r,m, wd=10mm, font=\bfseries] Q[l, m, wd=45mm]},
                 row{1} = {font=\bfseries, mode=text},
                 row{2-Z} ={rowsep=7pt},
                 }
no. & equations         & right & left      \\
\SetCell[r=4]{c}    1
    &   \begin{array}{@{} l}
            f(3) = 100  \\
            a    = \dfrac{50}{21} - \dfrac{13}{42}b
        \end{array}     &   A   &   B       \\
    &   3x^2-x-4=(3x-4)(x+1)
                        &   C   &   D       \\
    &   \begin{la}
            f(-1) = 0 \quad\text{or}\quad f\biggl(\dfrac{4}{3}\biggr) = 0 \\
                b = -\dfrac{2}{3}a\hfill
        \end{la}        &   E   &   F       \\
    &   a=3,  b=-2      &   G   &   H       \\
    \end{tblr}
\end{document}

Answer 2

The package nicematrix helps to solve several issues, with a simpler code.

The environment NiceTabularX allows to keep the table within the text area. \begin{NiceTabularX}{<total width>}{ X ....

\Block{<nro of rows - nro of columns>}{<content>} will center the content in the allocated space (e.g a 3x1 cell in this example) and allows to use \\ inside.

hvlines will draw all lines.

\documentclass{article}
\usepackage[a4paper, portrait, margin=0.5in, showframe]{geometry}% added showframe <<<<<<<<<<<<<
\usepackage[table]{xcolor}
\usepackage[most]{tcolorbox}
\usepackage{amsmath}
\usepackage{multirow}

\usepackage{nicematrix} % added <<<<<<<<<<<

\begin{document}
\large

\renewcommand{\arraystretch}{1.5}

\newcolumntype{C}[1]{>{\centering\arraybackslash}m{#1}} % centered column
\newcolumntype{B}[1]{>{\bfseries\centering\arraybackslash}m{#1}} % bold column

\newcolumntype{M}[1]{>{\raggedright\arraybackslash}m{#1}}

\noindent\begin{NiceTabularX}{\linewidth}{X C{9.5cm} B{1cm} C{4.5cm} }[hvlines] 
\Block{8-1}{1}  &\Block{3-1}{$\begin{aligned}
                                f(3)    & = 100 \\ 
                                    a   &=\dfrac{50}{21}-\dfrac{13}{42}b
                            \end{aligned}$ }                                                        &\Block{3-1}{A} &\Block{3-1}{B} \\                      
                &                                                                                   &               &       \\              
                &                                                                                   &               &       \\                  
                & $3x^2-x-4=(3x-4)(x+1)$                                                            &C              & D     \\
                &\Block{3-1}{$\begin{aligned}
                        f(-1)   & = 0 \text{ or } f\biggl(\dfrac{4}{3}\biggr) = 0\\
                            b   & = -\dfrac{2}{3}a
                \end{aligned}$}                                                                     &\Block{3-1}{E} & \Block{3-1}{F} \\                 
                &                                                                                   &               &       \\
                &                                                                                   &               &       \\                                          
                & $a=3, \; b=-2$                                                                    &G              & H     \\
\end{NiceTabularX}
        
\end{document}

Answer 3

Here is another way to do this using nicematrix. You must compile twice.

• \NiceMatrixOptions{cell-space-limits = 2pt} will increase spacing to accommodate your big fractions.
• column types wc, wl and wr allow you to set the column width.
• Option [hvlines] for horizontal and vertical lines (but respect the blocks).
• To center the 1 in the first column, use \Block{4-1}{1} in cell 1-1. {4-1} indicates a block of 4 rows and 1 column.
• \Block[l]{}{line1\\line2} permits multiline cells. This is known as a "mono-cell" block. To increase vertical spacing, I used \\[1ex].

\documentclass{article}
\usepackage[a4paper, margin=0.5in]{geometry}

\usepackage{nicematrix}

\begin{document}
\large
\renewcommand{\arraystretch}{1.5}

\begin{center}
\NiceMatrixOptions{cell-space-limits = 2pt}
\begin{NiceTabular}{wc{1.5cm}wl{9.5cm}wr{1cm}wl{4.5cm}}[hvlines]
\Block{4-1}{1} & \Block[l]{}{$f(3)=100$\\[1ex]$a=\dfrac{50}{21}-\dfrac{13}{42}b$} & \textbf{A} & B \\
 & $3x^2-x-4=(3x-4)(x+1)$ & \textbf{C} & D \\
 & \Block[l]{}{$f(-1) = 0$ or $f(\dfrac{4}{3}) = 0$\\[1ex]$b=-\dfrac{2}{3}a$} & \textbf{E} & F \\
 & $a=3, \; b=-2$ & \textbf{G} & H
\end{NiceTabular}
\end{center}

\end{document}