Mulicolumn and Multirow alignment and spacing












1















In my first encounter with multicolumn and multirow, I have attempted to make a table containing math, including matrices.



Firstly, I have an error in my column names; I'm unsure where I am going wrong and have been unable to find similar problems. I wish for my titles to be "Design" and then "Results" as my multicolumn heading, but they both sit in the same box. (Edit: I'm also getting an error "Extra alignment tab has been changed to cr." which I think may be the source of this problem?)



Secondly, the matrices in the cells seem crowded, is it possible to modify the cell size whilst retaining centering?



Finally, in my first column I have two matrices which sit each in their own multirow, they are currently left aligned (or aligned with each other) is it possible again to have them centered within their (multiple) cell?



I appreciate any help but will continue to attempt to resolve the issues. Please see the code below:



documentclass{article}
usepackage{multirow}
usepackage{amsmath}
begin{document}

begin{center}
begin{tabular}{ c|c c }
Design &multicolumn{4}{c|}{Results}\
hline
multirow{2}{11em}{$xi_3 = begin{Bmatrix}
-1 &0 &1 \
1/3 &1/3 &1/3 \
end{Bmatrix}$}
&$M(xi_3)=begin{bmatrix}
3 &0 \
0 &2
end{bmatrix}$
&$d(x,xi_3)= 1+frac{3x^2}{2}$ \
&$begin{vmatrix}M(xi_{3})end{vmatrix}=6$
&$d(x,xi_{3})=2.5$\
multirow{2}{11em}{$xi_3^{'} = begin{Bmatrix}
-1 &1 \
1/2 &1/2 \
end{Bmatrix}$}
&$M(xi_3^{'})=begin{bmatrix}
3 &0 \
0 &2
end{bmatrix}$
&$d(x,xi_3^{'})= 1+frac{3x^2}{2}$ \
&$begin{vmatrix}M(xi_{3}^{'})end{vmatrix}=8$
&$d(x,xi_{3}^{'})=3$\
end{tabular}
end{center}

end{document}









share|improve this question



























    1















    In my first encounter with multicolumn and multirow, I have attempted to make a table containing math, including matrices.



    Firstly, I have an error in my column names; I'm unsure where I am going wrong and have been unable to find similar problems. I wish for my titles to be "Design" and then "Results" as my multicolumn heading, but they both sit in the same box. (Edit: I'm also getting an error "Extra alignment tab has been changed to cr." which I think may be the source of this problem?)



    Secondly, the matrices in the cells seem crowded, is it possible to modify the cell size whilst retaining centering?



    Finally, in my first column I have two matrices which sit each in their own multirow, they are currently left aligned (or aligned with each other) is it possible again to have them centered within their (multiple) cell?



    I appreciate any help but will continue to attempt to resolve the issues. Please see the code below:



    documentclass{article}
    usepackage{multirow}
    usepackage{amsmath}
    begin{document}

    begin{center}
    begin{tabular}{ c|c c }
    Design &multicolumn{4}{c|}{Results}\
    hline
    multirow{2}{11em}{$xi_3 = begin{Bmatrix}
    -1 &0 &1 \
    1/3 &1/3 &1/3 \
    end{Bmatrix}$}
    &$M(xi_3)=begin{bmatrix}
    3 &0 \
    0 &2
    end{bmatrix}$
    &$d(x,xi_3)= 1+frac{3x^2}{2}$ \
    &$begin{vmatrix}M(xi_{3})end{vmatrix}=6$
    &$d(x,xi_{3})=2.5$\
    multirow{2}{11em}{$xi_3^{'} = begin{Bmatrix}
    -1 &1 \
    1/2 &1/2 \
    end{Bmatrix}$}
    &$M(xi_3^{'})=begin{bmatrix}
    3 &0 \
    0 &2
    end{bmatrix}$
    &$d(x,xi_3^{'})= 1+frac{3x^2}{2}$ \
    &$begin{vmatrix}M(xi_{3}^{'})end{vmatrix}=8$
    &$d(x,xi_{3}^{'})=3$\
    end{tabular}
    end{center}

    end{document}









    share|improve this question

























      1












      1








      1


      1






      In my first encounter with multicolumn and multirow, I have attempted to make a table containing math, including matrices.



      Firstly, I have an error in my column names; I'm unsure where I am going wrong and have been unable to find similar problems. I wish for my titles to be "Design" and then "Results" as my multicolumn heading, but they both sit in the same box. (Edit: I'm also getting an error "Extra alignment tab has been changed to cr." which I think may be the source of this problem?)



      Secondly, the matrices in the cells seem crowded, is it possible to modify the cell size whilst retaining centering?



      Finally, in my first column I have two matrices which sit each in their own multirow, they are currently left aligned (or aligned with each other) is it possible again to have them centered within their (multiple) cell?



      I appreciate any help but will continue to attempt to resolve the issues. Please see the code below:



      documentclass{article}
      usepackage{multirow}
      usepackage{amsmath}
      begin{document}

      begin{center}
      begin{tabular}{ c|c c }
      Design &multicolumn{4}{c|}{Results}\
      hline
      multirow{2}{11em}{$xi_3 = begin{Bmatrix}
      -1 &0 &1 \
      1/3 &1/3 &1/3 \
      end{Bmatrix}$}
      &$M(xi_3)=begin{bmatrix}
      3 &0 \
      0 &2
      end{bmatrix}$
      &$d(x,xi_3)= 1+frac{3x^2}{2}$ \
      &$begin{vmatrix}M(xi_{3})end{vmatrix}=6$
      &$d(x,xi_{3})=2.5$\
      multirow{2}{11em}{$xi_3^{'} = begin{Bmatrix}
      -1 &1 \
      1/2 &1/2 \
      end{Bmatrix}$}
      &$M(xi_3^{'})=begin{bmatrix}
      3 &0 \
      0 &2
      end{bmatrix}$
      &$d(x,xi_3^{'})= 1+frac{3x^2}{2}$ \
      &$begin{vmatrix}M(xi_{3}^{'})end{vmatrix}=8$
      &$d(x,xi_{3}^{'})=3$\
      end{tabular}
      end{center}

      end{document}









      share|improve this question














      In my first encounter with multicolumn and multirow, I have attempted to make a table containing math, including matrices.



      Firstly, I have an error in my column names; I'm unsure where I am going wrong and have been unable to find similar problems. I wish for my titles to be "Design" and then "Results" as my multicolumn heading, but they both sit in the same box. (Edit: I'm also getting an error "Extra alignment tab has been changed to cr." which I think may be the source of this problem?)



      Secondly, the matrices in the cells seem crowded, is it possible to modify the cell size whilst retaining centering?



      Finally, in my first column I have two matrices which sit each in their own multirow, they are currently left aligned (or aligned with each other) is it possible again to have them centered within their (multiple) cell?



      I appreciate any help but will continue to attempt to resolve the issues. Please see the code below:



      documentclass{article}
      usepackage{multirow}
      usepackage{amsmath}
      begin{document}

      begin{center}
      begin{tabular}{ c|c c }
      Design &multicolumn{4}{c|}{Results}\
      hline
      multirow{2}{11em}{$xi_3 = begin{Bmatrix}
      -1 &0 &1 \
      1/3 &1/3 &1/3 \
      end{Bmatrix}$}
      &$M(xi_3)=begin{bmatrix}
      3 &0 \
      0 &2
      end{bmatrix}$
      &$d(x,xi_3)= 1+frac{3x^2}{2}$ \
      &$begin{vmatrix}M(xi_{3})end{vmatrix}=6$
      &$d(x,xi_{3})=2.5$\
      multirow{2}{11em}{$xi_3^{'} = begin{Bmatrix}
      -1 &1 \
      1/2 &1/2 \
      end{Bmatrix}$}
      &$M(xi_3^{'})=begin{bmatrix}
      3 &0 \
      0 &2
      end{bmatrix}$
      &$d(x,xi_3^{'})= 1+frac{3x^2}{2}$ \
      &$begin{vmatrix}M(xi_{3}^{'})end{vmatrix}=8$
      &$d(x,xi_{3}^{'})=3$\
      end{tabular}
      end{center}

      end{document}






      multirow multicolumn






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      asked Dec 21 '18 at 21:53









      BenBen

      303




      303






















          2 Answers
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          1














          You do not have to take into account the number pf columns of the matrices inside the tabular. I propose here some simplifications of the code and some improvemennts to the layout, using the array environment, and the makecell and nccmath packages:



          documentclass{article}
          usepackage{amsmath, nccmath}
          usepackage{multirow, makecell}

          begin{document}

          [
          begin{array}{c|cc}
          text{Design} &multicolumn{2}{c|}{text{Results}}\%
          hline
          multirowcell{3}{ xi_3 = begin{Bmatrix}
          -1 &0 &1 \
          1/3 &1/3 &1/3 \
          end{Bmatrix} }
          &Gape[6pt][2pt]{M(xi_3)=begin{bmatrix}
          3 &0 \
          0 &2
          end{bmatrix}}
          & d(x,xi_3)= 1+mfrac{3x^2}{2} \[1ex]%
          & begin{vmatrix}M(xi_{3})end{vmatrix}=6
          & d(x,xi_{3})=2.5 \
          multirowcell{3}{xi'_3 = begin{Bmatrix}
          -1 &1 \
          1/2 &1/2 \
          end{Bmatrix}}
          &Gape[10pt][2pt]{M(xi'_3)=begin{bmatrix}
          3 &0 \
          0 &2
          end{bmatrix}}
          &d(x,xi_3')= 1+mfrac{3x^2}{2} \%
          &begin{vmatrix}M(xi'_{3})end{vmatrix}=8
          &d(x,xi'_{3})=3 \
          end{array} ]

          end{document}


          enter image description here






          share|improve this answer































            2














            I would avoid multirow altogether. With array the entries are automatically typeset in math mode, to which I add displaystyle.



            Note that x_{3}^{'} is wrong and it should be x_{3}'.



            documentclass{article}
            usepackage{amsmath}
            usepackage{booktabs}
            usepackage{array}

            begin{document}

            begin{center}
            $begin{array}{ *{3}{>{displaystyle}l} }
            toprule
            multicolumn{1}{c}{text{Design}} & multicolumn{2}{c}{text{Results}} \
            midrule
            smash[b]{xi_{3} = begin{Bmatrix}
            -1 &0 &1 \
            1/3 &1/3 &1/3 \
            end{Bmatrix}}
            &M(xi_{3})=begin{bmatrix}
            3 &0 \
            0 &2
            end{bmatrix}
            &d(x,xi_{3})=1+frac{3x^2}{2}
            \ addlinespace
            &lvert M(xi_{3})rvert=6
            &d(x,xi_{3})=2.5
            \ midrule
            smash[b]{xi_{3}' = begin{Bmatrix}
            -1 &1 \
            1/2 &1/2 \
            end{Bmatrix}}
            &M(xi_{3}')=begin{bmatrix}
            3 &0 \
            0 &2
            end{bmatrix}
            &d(x,xi_{3}')=1+frac{3x^2}{2}
            \ addlinespace
            &lvert M(xi_{3}')rvert=8
            &d(x,xi_{3}')=3
            \
            bottomrule
            end{array}$
            end{center}

            end{document}


            enter image description here






            share|improve this answer























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              2 Answers
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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1














              You do not have to take into account the number pf columns of the matrices inside the tabular. I propose here some simplifications of the code and some improvemennts to the layout, using the array environment, and the makecell and nccmath packages:



              documentclass{article}
              usepackage{amsmath, nccmath}
              usepackage{multirow, makecell}

              begin{document}

              [
              begin{array}{c|cc}
              text{Design} &multicolumn{2}{c|}{text{Results}}\%
              hline
              multirowcell{3}{ xi_3 = begin{Bmatrix}
              -1 &0 &1 \
              1/3 &1/3 &1/3 \
              end{Bmatrix} }
              &Gape[6pt][2pt]{M(xi_3)=begin{bmatrix}
              3 &0 \
              0 &2
              end{bmatrix}}
              & d(x,xi_3)= 1+mfrac{3x^2}{2} \[1ex]%
              & begin{vmatrix}M(xi_{3})end{vmatrix}=6
              & d(x,xi_{3})=2.5 \
              multirowcell{3}{xi'_3 = begin{Bmatrix}
              -1 &1 \
              1/2 &1/2 \
              end{Bmatrix}}
              &Gape[10pt][2pt]{M(xi'_3)=begin{bmatrix}
              3 &0 \
              0 &2
              end{bmatrix}}
              &d(x,xi_3')= 1+mfrac{3x^2}{2} \%
              &begin{vmatrix}M(xi'_{3})end{vmatrix}=8
              &d(x,xi'_{3})=3 \
              end{array} ]

              end{document}


              enter image description here






              share|improve this answer




























                1














                You do not have to take into account the number pf columns of the matrices inside the tabular. I propose here some simplifications of the code and some improvemennts to the layout, using the array environment, and the makecell and nccmath packages:



                documentclass{article}
                usepackage{amsmath, nccmath}
                usepackage{multirow, makecell}

                begin{document}

                [
                begin{array}{c|cc}
                text{Design} &multicolumn{2}{c|}{text{Results}}\%
                hline
                multirowcell{3}{ xi_3 = begin{Bmatrix}
                -1 &0 &1 \
                1/3 &1/3 &1/3 \
                end{Bmatrix} }
                &Gape[6pt][2pt]{M(xi_3)=begin{bmatrix}
                3 &0 \
                0 &2
                end{bmatrix}}
                & d(x,xi_3)= 1+mfrac{3x^2}{2} \[1ex]%
                & begin{vmatrix}M(xi_{3})end{vmatrix}=6
                & d(x,xi_{3})=2.5 \
                multirowcell{3}{xi'_3 = begin{Bmatrix}
                -1 &1 \
                1/2 &1/2 \
                end{Bmatrix}}
                &Gape[10pt][2pt]{M(xi'_3)=begin{bmatrix}
                3 &0 \
                0 &2
                end{bmatrix}}
                &d(x,xi_3')= 1+mfrac{3x^2}{2} \%
                &begin{vmatrix}M(xi'_{3})end{vmatrix}=8
                &d(x,xi'_{3})=3 \
                end{array} ]

                end{document}


                enter image description here






                share|improve this answer


























                  1












                  1








                  1







                  You do not have to take into account the number pf columns of the matrices inside the tabular. I propose here some simplifications of the code and some improvemennts to the layout, using the array environment, and the makecell and nccmath packages:



                  documentclass{article}
                  usepackage{amsmath, nccmath}
                  usepackage{multirow, makecell}

                  begin{document}

                  [
                  begin{array}{c|cc}
                  text{Design} &multicolumn{2}{c|}{text{Results}}\%
                  hline
                  multirowcell{3}{ xi_3 = begin{Bmatrix}
                  -1 &0 &1 \
                  1/3 &1/3 &1/3 \
                  end{Bmatrix} }
                  &Gape[6pt][2pt]{M(xi_3)=begin{bmatrix}
                  3 &0 \
                  0 &2
                  end{bmatrix}}
                  & d(x,xi_3)= 1+mfrac{3x^2}{2} \[1ex]%
                  & begin{vmatrix}M(xi_{3})end{vmatrix}=6
                  & d(x,xi_{3})=2.5 \
                  multirowcell{3}{xi'_3 = begin{Bmatrix}
                  -1 &1 \
                  1/2 &1/2 \
                  end{Bmatrix}}
                  &Gape[10pt][2pt]{M(xi'_3)=begin{bmatrix}
                  3 &0 \
                  0 &2
                  end{bmatrix}}
                  &d(x,xi_3')= 1+mfrac{3x^2}{2} \%
                  &begin{vmatrix}M(xi'_{3})end{vmatrix}=8
                  &d(x,xi'_{3})=3 \
                  end{array} ]

                  end{document}


                  enter image description here






                  share|improve this answer













                  You do not have to take into account the number pf columns of the matrices inside the tabular. I propose here some simplifications of the code and some improvemennts to the layout, using the array environment, and the makecell and nccmath packages:



                  documentclass{article}
                  usepackage{amsmath, nccmath}
                  usepackage{multirow, makecell}

                  begin{document}

                  [
                  begin{array}{c|cc}
                  text{Design} &multicolumn{2}{c|}{text{Results}}\%
                  hline
                  multirowcell{3}{ xi_3 = begin{Bmatrix}
                  -1 &0 &1 \
                  1/3 &1/3 &1/3 \
                  end{Bmatrix} }
                  &Gape[6pt][2pt]{M(xi_3)=begin{bmatrix}
                  3 &0 \
                  0 &2
                  end{bmatrix}}
                  & d(x,xi_3)= 1+mfrac{3x^2}{2} \[1ex]%
                  & begin{vmatrix}M(xi_{3})end{vmatrix}=6
                  & d(x,xi_{3})=2.5 \
                  multirowcell{3}{xi'_3 = begin{Bmatrix}
                  -1 &1 \
                  1/2 &1/2 \
                  end{Bmatrix}}
                  &Gape[10pt][2pt]{M(xi'_3)=begin{bmatrix}
                  3 &0 \
                  0 &2
                  end{bmatrix}}
                  &d(x,xi_3')= 1+mfrac{3x^2}{2} \%
                  &begin{vmatrix}M(xi'_{3})end{vmatrix}=8
                  &d(x,xi'_{3})=3 \
                  end{array} ]

                  end{document}


                  enter image description here







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Dec 21 '18 at 22:44









                  BernardBernard

                  172k775203




                  172k775203























                      2














                      I would avoid multirow altogether. With array the entries are automatically typeset in math mode, to which I add displaystyle.



                      Note that x_{3}^{'} is wrong and it should be x_{3}'.



                      documentclass{article}
                      usepackage{amsmath}
                      usepackage{booktabs}
                      usepackage{array}

                      begin{document}

                      begin{center}
                      $begin{array}{ *{3}{>{displaystyle}l} }
                      toprule
                      multicolumn{1}{c}{text{Design}} & multicolumn{2}{c}{text{Results}} \
                      midrule
                      smash[b]{xi_{3} = begin{Bmatrix}
                      -1 &0 &1 \
                      1/3 &1/3 &1/3 \
                      end{Bmatrix}}
                      &M(xi_{3})=begin{bmatrix}
                      3 &0 \
                      0 &2
                      end{bmatrix}
                      &d(x,xi_{3})=1+frac{3x^2}{2}
                      \ addlinespace
                      &lvert M(xi_{3})rvert=6
                      &d(x,xi_{3})=2.5
                      \ midrule
                      smash[b]{xi_{3}' = begin{Bmatrix}
                      -1 &1 \
                      1/2 &1/2 \
                      end{Bmatrix}}
                      &M(xi_{3}')=begin{bmatrix}
                      3 &0 \
                      0 &2
                      end{bmatrix}
                      &d(x,xi_{3}')=1+frac{3x^2}{2}
                      \ addlinespace
                      &lvert M(xi_{3}')rvert=8
                      &d(x,xi_{3}')=3
                      \
                      bottomrule
                      end{array}$
                      end{center}

                      end{document}


                      enter image description here






                      share|improve this answer




























                        2














                        I would avoid multirow altogether. With array the entries are automatically typeset in math mode, to which I add displaystyle.



                        Note that x_{3}^{'} is wrong and it should be x_{3}'.



                        documentclass{article}
                        usepackage{amsmath}
                        usepackage{booktabs}
                        usepackage{array}

                        begin{document}

                        begin{center}
                        $begin{array}{ *{3}{>{displaystyle}l} }
                        toprule
                        multicolumn{1}{c}{text{Design}} & multicolumn{2}{c}{text{Results}} \
                        midrule
                        smash[b]{xi_{3} = begin{Bmatrix}
                        -1 &0 &1 \
                        1/3 &1/3 &1/3 \
                        end{Bmatrix}}
                        &M(xi_{3})=begin{bmatrix}
                        3 &0 \
                        0 &2
                        end{bmatrix}
                        &d(x,xi_{3})=1+frac{3x^2}{2}
                        \ addlinespace
                        &lvert M(xi_{3})rvert=6
                        &d(x,xi_{3})=2.5
                        \ midrule
                        smash[b]{xi_{3}' = begin{Bmatrix}
                        -1 &1 \
                        1/2 &1/2 \
                        end{Bmatrix}}
                        &M(xi_{3}')=begin{bmatrix}
                        3 &0 \
                        0 &2
                        end{bmatrix}
                        &d(x,xi_{3}')=1+frac{3x^2}{2}
                        \ addlinespace
                        &lvert M(xi_{3}')rvert=8
                        &d(x,xi_{3}')=3
                        \
                        bottomrule
                        end{array}$
                        end{center}

                        end{document}


                        enter image description here






                        share|improve this answer


























                          2












                          2








                          2







                          I would avoid multirow altogether. With array the entries are automatically typeset in math mode, to which I add displaystyle.



                          Note that x_{3}^{'} is wrong and it should be x_{3}'.



                          documentclass{article}
                          usepackage{amsmath}
                          usepackage{booktabs}
                          usepackage{array}

                          begin{document}

                          begin{center}
                          $begin{array}{ *{3}{>{displaystyle}l} }
                          toprule
                          multicolumn{1}{c}{text{Design}} & multicolumn{2}{c}{text{Results}} \
                          midrule
                          smash[b]{xi_{3} = begin{Bmatrix}
                          -1 &0 &1 \
                          1/3 &1/3 &1/3 \
                          end{Bmatrix}}
                          &M(xi_{3})=begin{bmatrix}
                          3 &0 \
                          0 &2
                          end{bmatrix}
                          &d(x,xi_{3})=1+frac{3x^2}{2}
                          \ addlinespace
                          &lvert M(xi_{3})rvert=6
                          &d(x,xi_{3})=2.5
                          \ midrule
                          smash[b]{xi_{3}' = begin{Bmatrix}
                          -1 &1 \
                          1/2 &1/2 \
                          end{Bmatrix}}
                          &M(xi_{3}')=begin{bmatrix}
                          3 &0 \
                          0 &2
                          end{bmatrix}
                          &d(x,xi_{3}')=1+frac{3x^2}{2}
                          \ addlinespace
                          &lvert M(xi_{3}')rvert=8
                          &d(x,xi_{3}')=3
                          \
                          bottomrule
                          end{array}$
                          end{center}

                          end{document}


                          enter image description here






                          share|improve this answer













                          I would avoid multirow altogether. With array the entries are automatically typeset in math mode, to which I add displaystyle.



                          Note that x_{3}^{'} is wrong and it should be x_{3}'.



                          documentclass{article}
                          usepackage{amsmath}
                          usepackage{booktabs}
                          usepackage{array}

                          begin{document}

                          begin{center}
                          $begin{array}{ *{3}{>{displaystyle}l} }
                          toprule
                          multicolumn{1}{c}{text{Design}} & multicolumn{2}{c}{text{Results}} \
                          midrule
                          smash[b]{xi_{3} = begin{Bmatrix}
                          -1 &0 &1 \
                          1/3 &1/3 &1/3 \
                          end{Bmatrix}}
                          &M(xi_{3})=begin{bmatrix}
                          3 &0 \
                          0 &2
                          end{bmatrix}
                          &d(x,xi_{3})=1+frac{3x^2}{2}
                          \ addlinespace
                          &lvert M(xi_{3})rvert=6
                          &d(x,xi_{3})=2.5
                          \ midrule
                          smash[b]{xi_{3}' = begin{Bmatrix}
                          -1 &1 \
                          1/2 &1/2 \
                          end{Bmatrix}}
                          &M(xi_{3}')=begin{bmatrix}
                          3 &0 \
                          0 &2
                          end{bmatrix}
                          &d(x,xi_{3}')=1+frac{3x^2}{2}
                          \ addlinespace
                          &lvert M(xi_{3}')rvert=8
                          &d(x,xi_{3}')=3
                          \
                          bottomrule
                          end{array}$
                          end{center}

                          end{document}


                          enter image description here







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                          answered Dec 21 '18 at 22:30









                          egregegreg

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