Empty elements as dots in matrix











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6
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I need to create a determinant (vmatrix) as shown in this picture.enter image description here So, I wrote a simple LaTEX file for testing purposes as follows:



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

begin{document}

begin{equation*}
begin{vmatrix}
0 &cdot &cdot &0 &a_{1} alpha_{1}+ cdots +k_{1} alpha_{n} &cdot &cdot &a_{1} kappa_{1}+ cdots +k_{1} kappa_{n} \
cdot &cdot &cdot &cdot &cdot &cdot &cdot &cdot \
0 &cdot &cdot &0 &a_{n} alpha_{1}+ cdots +k_{n} alpha_{n} &cdot &cdot &a_{n} kappa_{1}+ cdots +k_{n} kappa_{n} \
-1 &cdot &cdot &0 &alpha_{1} &cdot &cdot &kappa_{1} \
cdot &cdot &cdot &cdot &cdot &cdot &cdot &cdot \
0 &cdot &cdot &-1 &alpha_{n} &cdot &cdot &kappa_{n}
end{vmatrix}
end{equation*}

end{document}


When the above file gets compiled with TeXLive on a Linux desktop computer, it produces the following ugly result:enter image description here I thought using hdotsfor{} for the horizontal line, but the spacing between the dots is too close. If there is a simple way to tell hdotsfor{} to skip every dot, I believe that would do. However, perhaps someone here knows a much better way.










share|improve this question






















  • hdotsfor has an optional argument which is a multiplier for the amount of space between consecutive dots, so you could try hdotsfor[2]{<number of rows>}. I'm not sure if it'd look good though, since they wouldn't align with the dots in your first row.
    – Circumscribe
    2 days ago















up vote
6
down vote

favorite












I need to create a determinant (vmatrix) as shown in this picture.enter image description here So, I wrote a simple LaTEX file for testing purposes as follows:



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

begin{document}

begin{equation*}
begin{vmatrix}
0 &cdot &cdot &0 &a_{1} alpha_{1}+ cdots +k_{1} alpha_{n} &cdot &cdot &a_{1} kappa_{1}+ cdots +k_{1} kappa_{n} \
cdot &cdot &cdot &cdot &cdot &cdot &cdot &cdot \
0 &cdot &cdot &0 &a_{n} alpha_{1}+ cdots +k_{n} alpha_{n} &cdot &cdot &a_{n} kappa_{1}+ cdots +k_{n} kappa_{n} \
-1 &cdot &cdot &0 &alpha_{1} &cdot &cdot &kappa_{1} \
cdot &cdot &cdot &cdot &cdot &cdot &cdot &cdot \
0 &cdot &cdot &-1 &alpha_{n} &cdot &cdot &kappa_{n}
end{vmatrix}
end{equation*}

end{document}


When the above file gets compiled with TeXLive on a Linux desktop computer, it produces the following ugly result:enter image description here I thought using hdotsfor{} for the horizontal line, but the spacing between the dots is too close. If there is a simple way to tell hdotsfor{} to skip every dot, I believe that would do. However, perhaps someone here knows a much better way.










share|improve this question






















  • hdotsfor has an optional argument which is a multiplier for the amount of space between consecutive dots, so you could try hdotsfor[2]{<number of rows>}. I'm not sure if it'd look good though, since they wouldn't align with the dots in your first row.
    – Circumscribe
    2 days ago













up vote
6
down vote

favorite









up vote
6
down vote

favorite











I need to create a determinant (vmatrix) as shown in this picture.enter image description here So, I wrote a simple LaTEX file for testing purposes as follows:



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

begin{document}

begin{equation*}
begin{vmatrix}
0 &cdot &cdot &0 &a_{1} alpha_{1}+ cdots +k_{1} alpha_{n} &cdot &cdot &a_{1} kappa_{1}+ cdots +k_{1} kappa_{n} \
cdot &cdot &cdot &cdot &cdot &cdot &cdot &cdot \
0 &cdot &cdot &0 &a_{n} alpha_{1}+ cdots +k_{n} alpha_{n} &cdot &cdot &a_{n} kappa_{1}+ cdots +k_{n} kappa_{n} \
-1 &cdot &cdot &0 &alpha_{1} &cdot &cdot &kappa_{1} \
cdot &cdot &cdot &cdot &cdot &cdot &cdot &cdot \
0 &cdot &cdot &-1 &alpha_{n} &cdot &cdot &kappa_{n}
end{vmatrix}
end{equation*}

end{document}


When the above file gets compiled with TeXLive on a Linux desktop computer, it produces the following ugly result:enter image description here I thought using hdotsfor{} for the horizontal line, but the spacing between the dots is too close. If there is a simple way to tell hdotsfor{} to skip every dot, I believe that would do. However, perhaps someone here knows a much better way.










share|improve this question













I need to create a determinant (vmatrix) as shown in this picture.enter image description here So, I wrote a simple LaTEX file for testing purposes as follows:



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

begin{document}

begin{equation*}
begin{vmatrix}
0 &cdot &cdot &0 &a_{1} alpha_{1}+ cdots +k_{1} alpha_{n} &cdot &cdot &a_{1} kappa_{1}+ cdots +k_{1} kappa_{n} \
cdot &cdot &cdot &cdot &cdot &cdot &cdot &cdot \
0 &cdot &cdot &0 &a_{n} alpha_{1}+ cdots +k_{n} alpha_{n} &cdot &cdot &a_{n} kappa_{1}+ cdots +k_{n} kappa_{n} \
-1 &cdot &cdot &0 &alpha_{1} &cdot &cdot &kappa_{1} \
cdot &cdot &cdot &cdot &cdot &cdot &cdot &cdot \
0 &cdot &cdot &-1 &alpha_{n} &cdot &cdot &kappa_{n}
end{vmatrix}
end{equation*}

end{document}


When the above file gets compiled with TeXLive on a Linux desktop computer, it produces the following ugly result:enter image description here I thought using hdotsfor{} for the horizontal line, but the spacing between the dots is too close. If there is a simple way to tell hdotsfor{} to skip every dot, I believe that would do. However, perhaps someone here knows a much better way.







matrices






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asked 2 days ago









user91822

1354




1354












  • hdotsfor has an optional argument which is a multiplier for the amount of space between consecutive dots, so you could try hdotsfor[2]{<number of rows>}. I'm not sure if it'd look good though, since they wouldn't align with the dots in your first row.
    – Circumscribe
    2 days ago


















  • hdotsfor has an optional argument which is a multiplier for the amount of space between consecutive dots, so you could try hdotsfor[2]{<number of rows>}. I'm not sure if it'd look good though, since they wouldn't align with the dots in your first row.
    – Circumscribe
    2 days ago
















hdotsfor has an optional argument which is a multiplier for the amount of space between consecutive dots, so you could try hdotsfor[2]{<number of rows>}. I'm not sure if it'd look good though, since they wouldn't align with the dots in your first row.
– Circumscribe
2 days ago




hdotsfor has an optional argument which is a multiplier for the amount of space between consecutive dots, so you could try hdotsfor[2]{<number of rows>}. I'm not sure if it'd look good though, since they wouldn't align with the dots in your first row.
– Circumscribe
2 days ago










4 Answers
4






active

oldest

votes

















up vote
8
down vote



accepted










This seems a good approximation:



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

newcommand{widedots}{%
.cleadershbox to 2tabcolsep{hss.hss}hfill.%
hspace*{0pt}%
}

begin{document}

begin{equation*}
newcommand{ak}[2]{%
makebox[14tabcolsep][s]{$displaystyle a_{#2}#1_1+dots+k_{#2}#1_n$}%
}
begin{vmatrix}
0 & . & . & 0 & ak{alpha}{1} & . & . & ak{kappa}{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & 0 & ak{alpha}{n} & . & . & ak{kappa}{n} \
-1 & . & . & 0 & alpha_{1} & . & . & kappa_{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & -1 & alpha_{n} & . & . & kappa_{n}
end{vmatrix}
end{equation*}

end{document}


enter image description here



The value 14tabcolsep has been determined by first looking at the entries at their natural width.



Even better than the original if we make -1 to hide its width.



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

newcommand{widedots}{%
.cleadershbox to 2tabcolsep{hss.hss}hfill.%
hspace*{0pt}%
}

begin{document}

begin{equation*}
newcommand{ak}[2]{%
makebox[14tabcolsep][s]{$displaystyle a_{#2}#1_1+dots+k_{#2}#1_n$}%
}
newcommand{?}{makebox[0pt]{$-1$}}
begin{vmatrix}mspace{8mu}
begin{matrix}
0 & . & . & 0 & ak{alpha}{1} & . & . & ak{kappa}{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & 0 & ak{alpha}{n} & . & . & ak{kappa}{n} \
? & . & . & 0 & alpha_{1} & . & . & kappa_{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & ? & alpha_{n} & . & . & kappa_{n}
end{matrix}mspace{3mu}
end{vmatrix}
end{equation*}

end{document}


enter image description here



With dots and hdotsfor:



begin{equation*}
newcommand{ak}[2]{a_{#2}#1_1+dots+k_{#2}#1_n}
begin{vmatrix}
0 & dots & 0 & ak{alpha}{1} & dots & ak{kappa}{1} \
hdotsfor{6} \
0 & dots & 0 & ak{alpha}{n} & dots & ak{kappa}{n} \
-1 & dots & 0 & alpha_{1} & dots & kappa_{1} \
hdotsfor{6} \
0 & dots & -1 & alpha_{n} & dots & kappa_{n}
end{vmatrix}
end{equation*}


enter image description here






share|improve this answer























  • Although I may not be able to decipher how your code works, it looks like the one with the capability of hiding the with of -1 fits my need. I have a feeling after perusing your code, I will understand better. Thank you.
    – user91822
    2 days ago


















up vote
4
down vote













One way is to manually put dots with some horizontal space.



enter image description here



documentclass[a4paper,10pt]{article}
usepackage{amsmath}
begin{document}

begin{equation*}
begin{vmatrix}
0 &cdot &cdot &0 &a_{1} alpha_{1}+ cdots +k_{1} alpha_{n} &cdot &cdot &a_{1} kappa_{1}+ cdots +k_{1} kappa_{n} \
cdot &cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot \
0 &cdot &cdot &0 &a_{n} alpha_{1}+ cdots +k_{n} alpha_{n} &cdot &cdot &a_{n} kappa_{1}+ cdots +k_{n} kappa_{n} \
-1 &cdot &cdot &0 &alpha_{1} &cdot &cdot &kappa_{1} \
cdot &cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot \
0 &cdot &cdot &-1 &alpha_{n} &cdot &cdot &kappa_{n}
end{vmatrix}
end{equation*}

end{document}





share|improve this answer





















  • Honestly, I had never thought of using 'hfil'. Thank you for the pointer.
    – user91822
    2 days ago


















up vote
3
down vote













Defining two newcommands twodts and fivedts for empty columns and rows can simplify the table.



documentclass[a4paper,10pt]{article}
usepackage{amsmath}

begin{document}

newcommand{twodts}{begin{tabular}{@{}cc@{}}$cdot$&$cdot$end{tabular}}
newcommand{fivedts}{begin{tabular}{@{}*5{c}@{}}$cdot$&$cdot$&$cdot$&$cdot$&$cdot$end{tabular}}

begin{equation*}
begin{array}{|cccccc|}
0 & twodts & 0 & a_1 alpha_1+ cdots +k_1 alpha_n & twodts & a_1 kappa_1+ cdots +k_1 kappa_n \
cdot & twodts & cdot & fivedts & twodts & fivedts \
0 & twodts & 0 & a_n alpha_1+ cdots +k_n alpha_n & twodts & a_n kappa_1+ cdots +k_n kappa_n \
-1 & twodts & 0 & alpha_1 & twodts & kappa_1 \
cdot & twodts & cdot & fivedts & twodts & fivedts \
0 & twodts & -1 & alpha_n & twodts & kappa_n
end{array}
end{equation*}

end{document}



enter image description here




Update:



You can easily adapt my answer to add six dots instead of five like this:



documentclass[a4paper,10pt]{article}
usepackage{amsmath,array}

begin{document}

deftwocdots{$cdot$&$cdot$}
newcommand{twodts}{begin{tabular}{@{}cc@{}}twocdotsend{tabular}}
newcommand{sixdts}{begin{tabular}{@{}*6{c}@{}}twocdots&twocdots&twocdotsend{tabular}}

begin{equation*}
begin{array}{|cccccc|}
0 & twodts & 0 & a_1 alpha_1+ cdots +k_1 alpha_n & twodts & a_1 kappa_1+ cdots +k_1 kappa_n \
cdot & twodts & cdot & sixdts & twodts & sixdts \
0 & twodts & 0 & a_n alpha_1+ cdots +k_n alpha_n & twodts & a_n kappa_1+ cdots +k_n kappa_n \
-1 & twodts & 0 & alpha_1 & twodts & kappa_1 \
cdot & twodts & cdot & sixdts & twodts & sixdts \
0 & twodts & -1 & alpha_n & twodts & kappa_n
end{array}
end{equation*}

end{document}



enter image description here







share|improve this answer























  • This looks pretty simple and good solution. Since there is still some gaps that seems to be able to put one more dot when using the `fivedts' command, I wonder if this command can be modded to accommodate the gap. I tried to modify with a "sixdots" and it does not work, unfortunately. Anyway, thank you too.
    – user91822
    2 days ago










  • Please see my update for doing this.
    – AboAmmar
    yesterday


















up vote
1
down vote













Would this be acceptable?



enter image description here



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

begin{document}

begin{equation*}
begin{vmatrix}
0 &cdots &0 &a_{1} alpha_{1}+ cdots +k_{1} alpha_{n} &cdots &a_{1} kappa_{1}+ cdots +k_{1} kappa_{n} \
vdots &ddots &vdots &vdots &ddots &vdots \
0 &cdots & 0 &a_{n} alpha_{1}+ cdots +k_{n} alpha_{n} &cdots &a_{n} kappa_{1}+ cdots +k_{n} kappa_{n} \
-1 &cdots &0 &alpha_{1} &cdots &kappa_{1} \
vdots &ddots &vdots &vdots &ddots &vdots \
0 &cdots & -1 &alpha_{n} &cdots &kappa_{n}
end{vmatrix}
end{equation*}

end{document}





share|improve this answer

















  • 1




    This is I usually do and I don't think it will do for me. Sorry and I try to stick closer to the example I posted above. Anyway, thank you.
    – user91822
    2 days ago













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4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
8
down vote



accepted










This seems a good approximation:



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

newcommand{widedots}{%
.cleadershbox to 2tabcolsep{hss.hss}hfill.%
hspace*{0pt}%
}

begin{document}

begin{equation*}
newcommand{ak}[2]{%
makebox[14tabcolsep][s]{$displaystyle a_{#2}#1_1+dots+k_{#2}#1_n$}%
}
begin{vmatrix}
0 & . & . & 0 & ak{alpha}{1} & . & . & ak{kappa}{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & 0 & ak{alpha}{n} & . & . & ak{kappa}{n} \
-1 & . & . & 0 & alpha_{1} & . & . & kappa_{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & -1 & alpha_{n} & . & . & kappa_{n}
end{vmatrix}
end{equation*}

end{document}


enter image description here



The value 14tabcolsep has been determined by first looking at the entries at their natural width.



Even better than the original if we make -1 to hide its width.



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

newcommand{widedots}{%
.cleadershbox to 2tabcolsep{hss.hss}hfill.%
hspace*{0pt}%
}

begin{document}

begin{equation*}
newcommand{ak}[2]{%
makebox[14tabcolsep][s]{$displaystyle a_{#2}#1_1+dots+k_{#2}#1_n$}%
}
newcommand{?}{makebox[0pt]{$-1$}}
begin{vmatrix}mspace{8mu}
begin{matrix}
0 & . & . & 0 & ak{alpha}{1} & . & . & ak{kappa}{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & 0 & ak{alpha}{n} & . & . & ak{kappa}{n} \
? & . & . & 0 & alpha_{1} & . & . & kappa_{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & ? & alpha_{n} & . & . & kappa_{n}
end{matrix}mspace{3mu}
end{vmatrix}
end{equation*}

end{document}


enter image description here



With dots and hdotsfor:



begin{equation*}
newcommand{ak}[2]{a_{#2}#1_1+dots+k_{#2}#1_n}
begin{vmatrix}
0 & dots & 0 & ak{alpha}{1} & dots & ak{kappa}{1} \
hdotsfor{6} \
0 & dots & 0 & ak{alpha}{n} & dots & ak{kappa}{n} \
-1 & dots & 0 & alpha_{1} & dots & kappa_{1} \
hdotsfor{6} \
0 & dots & -1 & alpha_{n} & dots & kappa_{n}
end{vmatrix}
end{equation*}


enter image description here






share|improve this answer























  • Although I may not be able to decipher how your code works, it looks like the one with the capability of hiding the with of -1 fits my need. I have a feeling after perusing your code, I will understand better. Thank you.
    – user91822
    2 days ago















up vote
8
down vote



accepted










This seems a good approximation:



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

newcommand{widedots}{%
.cleadershbox to 2tabcolsep{hss.hss}hfill.%
hspace*{0pt}%
}

begin{document}

begin{equation*}
newcommand{ak}[2]{%
makebox[14tabcolsep][s]{$displaystyle a_{#2}#1_1+dots+k_{#2}#1_n$}%
}
begin{vmatrix}
0 & . & . & 0 & ak{alpha}{1} & . & . & ak{kappa}{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & 0 & ak{alpha}{n} & . & . & ak{kappa}{n} \
-1 & . & . & 0 & alpha_{1} & . & . & kappa_{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & -1 & alpha_{n} & . & . & kappa_{n}
end{vmatrix}
end{equation*}

end{document}


enter image description here



The value 14tabcolsep has been determined by first looking at the entries at their natural width.



Even better than the original if we make -1 to hide its width.



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

newcommand{widedots}{%
.cleadershbox to 2tabcolsep{hss.hss}hfill.%
hspace*{0pt}%
}

begin{document}

begin{equation*}
newcommand{ak}[2]{%
makebox[14tabcolsep][s]{$displaystyle a_{#2}#1_1+dots+k_{#2}#1_n$}%
}
newcommand{?}{makebox[0pt]{$-1$}}
begin{vmatrix}mspace{8mu}
begin{matrix}
0 & . & . & 0 & ak{alpha}{1} & . & . & ak{kappa}{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & 0 & ak{alpha}{n} & . & . & ak{kappa}{n} \
? & . & . & 0 & alpha_{1} & . & . & kappa_{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & ? & alpha_{n} & . & . & kappa_{n}
end{matrix}mspace{3mu}
end{vmatrix}
end{equation*}

end{document}


enter image description here



With dots and hdotsfor:



begin{equation*}
newcommand{ak}[2]{a_{#2}#1_1+dots+k_{#2}#1_n}
begin{vmatrix}
0 & dots & 0 & ak{alpha}{1} & dots & ak{kappa}{1} \
hdotsfor{6} \
0 & dots & 0 & ak{alpha}{n} & dots & ak{kappa}{n} \
-1 & dots & 0 & alpha_{1} & dots & kappa_{1} \
hdotsfor{6} \
0 & dots & -1 & alpha_{n} & dots & kappa_{n}
end{vmatrix}
end{equation*}


enter image description here






share|improve this answer























  • Although I may not be able to decipher how your code works, it looks like the one with the capability of hiding the with of -1 fits my need. I have a feeling after perusing your code, I will understand better. Thank you.
    – user91822
    2 days ago













up vote
8
down vote



accepted







up vote
8
down vote



accepted






This seems a good approximation:



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

newcommand{widedots}{%
.cleadershbox to 2tabcolsep{hss.hss}hfill.%
hspace*{0pt}%
}

begin{document}

begin{equation*}
newcommand{ak}[2]{%
makebox[14tabcolsep][s]{$displaystyle a_{#2}#1_1+dots+k_{#2}#1_n$}%
}
begin{vmatrix}
0 & . & . & 0 & ak{alpha}{1} & . & . & ak{kappa}{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & 0 & ak{alpha}{n} & . & . & ak{kappa}{n} \
-1 & . & . & 0 & alpha_{1} & . & . & kappa_{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & -1 & alpha_{n} & . & . & kappa_{n}
end{vmatrix}
end{equation*}

end{document}


enter image description here



The value 14tabcolsep has been determined by first looking at the entries at their natural width.



Even better than the original if we make -1 to hide its width.



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

newcommand{widedots}{%
.cleadershbox to 2tabcolsep{hss.hss}hfill.%
hspace*{0pt}%
}

begin{document}

begin{equation*}
newcommand{ak}[2]{%
makebox[14tabcolsep][s]{$displaystyle a_{#2}#1_1+dots+k_{#2}#1_n$}%
}
newcommand{?}{makebox[0pt]{$-1$}}
begin{vmatrix}mspace{8mu}
begin{matrix}
0 & . & . & 0 & ak{alpha}{1} & . & . & ak{kappa}{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & 0 & ak{alpha}{n} & . & . & ak{kappa}{n} \
? & . & . & 0 & alpha_{1} & . & . & kappa_{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & ? & alpha_{n} & . & . & kappa_{n}
end{matrix}mspace{3mu}
end{vmatrix}
end{equation*}

end{document}


enter image description here



With dots and hdotsfor:



begin{equation*}
newcommand{ak}[2]{a_{#2}#1_1+dots+k_{#2}#1_n}
begin{vmatrix}
0 & dots & 0 & ak{alpha}{1} & dots & ak{kappa}{1} \
hdotsfor{6} \
0 & dots & 0 & ak{alpha}{n} & dots & ak{kappa}{n} \
-1 & dots & 0 & alpha_{1} & dots & kappa_{1} \
hdotsfor{6} \
0 & dots & -1 & alpha_{n} & dots & kappa_{n}
end{vmatrix}
end{equation*}


enter image description here






share|improve this answer














This seems a good approximation:



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

newcommand{widedots}{%
.cleadershbox to 2tabcolsep{hss.hss}hfill.%
hspace*{0pt}%
}

begin{document}

begin{equation*}
newcommand{ak}[2]{%
makebox[14tabcolsep][s]{$displaystyle a_{#2}#1_1+dots+k_{#2}#1_n$}%
}
begin{vmatrix}
0 & . & . & 0 & ak{alpha}{1} & . & . & ak{kappa}{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & 0 & ak{alpha}{n} & . & . & ak{kappa}{n} \
-1 & . & . & 0 & alpha_{1} & . & . & kappa_{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & -1 & alpha_{n} & . & . & kappa_{n}
end{vmatrix}
end{equation*}

end{document}


enter image description here



The value 14tabcolsep has been determined by first looking at the entries at their natural width.



Even better than the original if we make -1 to hide its width.



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

newcommand{widedots}{%
.cleadershbox to 2tabcolsep{hss.hss}hfill.%
hspace*{0pt}%
}

begin{document}

begin{equation*}
newcommand{ak}[2]{%
makebox[14tabcolsep][s]{$displaystyle a_{#2}#1_1+dots+k_{#2}#1_n$}%
}
newcommand{?}{makebox[0pt]{$-1$}}
begin{vmatrix}mspace{8mu}
begin{matrix}
0 & . & . & 0 & ak{alpha}{1} & . & . & ak{kappa}{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & 0 & ak{alpha}{n} & . & . & ak{kappa}{n} \
? & . & . & 0 & alpha_{1} & . & . & kappa_{1} \
. & . & . & . & widedots & . & . & widedots \
0 & . & . & ? & alpha_{n} & . & . & kappa_{n}
end{matrix}mspace{3mu}
end{vmatrix}
end{equation*}

end{document}


enter image description here



With dots and hdotsfor:



begin{equation*}
newcommand{ak}[2]{a_{#2}#1_1+dots+k_{#2}#1_n}
begin{vmatrix}
0 & dots & 0 & ak{alpha}{1} & dots & ak{kappa}{1} \
hdotsfor{6} \
0 & dots & 0 & ak{alpha}{n} & dots & ak{kappa}{n} \
-1 & dots & 0 & alpha_{1} & dots & kappa_{1} \
hdotsfor{6} \
0 & dots & -1 & alpha_{n} & dots & kappa_{n}
end{vmatrix}
end{equation*}


enter image description here







share|improve this answer














share|improve this answer



share|improve this answer








edited 2 days ago

























answered 2 days ago









egreg

703k8618753154




703k8618753154












  • Although I may not be able to decipher how your code works, it looks like the one with the capability of hiding the with of -1 fits my need. I have a feeling after perusing your code, I will understand better. Thank you.
    – user91822
    2 days ago


















  • Although I may not be able to decipher how your code works, it looks like the one with the capability of hiding the with of -1 fits my need. I have a feeling after perusing your code, I will understand better. Thank you.
    – user91822
    2 days ago
















Although I may not be able to decipher how your code works, it looks like the one with the capability of hiding the with of -1 fits my need. I have a feeling after perusing your code, I will understand better. Thank you.
– user91822
2 days ago




Although I may not be able to decipher how your code works, it looks like the one with the capability of hiding the with of -1 fits my need. I have a feeling after perusing your code, I will understand better. Thank you.
– user91822
2 days ago










up vote
4
down vote













One way is to manually put dots with some horizontal space.



enter image description here



documentclass[a4paper,10pt]{article}
usepackage{amsmath}
begin{document}

begin{equation*}
begin{vmatrix}
0 &cdot &cdot &0 &a_{1} alpha_{1}+ cdots +k_{1} alpha_{n} &cdot &cdot &a_{1} kappa_{1}+ cdots +k_{1} kappa_{n} \
cdot &cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot \
0 &cdot &cdot &0 &a_{n} alpha_{1}+ cdots +k_{n} alpha_{n} &cdot &cdot &a_{n} kappa_{1}+ cdots +k_{n} kappa_{n} \
-1 &cdot &cdot &0 &alpha_{1} &cdot &cdot &kappa_{1} \
cdot &cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot \
0 &cdot &cdot &-1 &alpha_{n} &cdot &cdot &kappa_{n}
end{vmatrix}
end{equation*}

end{document}





share|improve this answer





















  • Honestly, I had never thought of using 'hfil'. Thank you for the pointer.
    – user91822
    2 days ago















up vote
4
down vote













One way is to manually put dots with some horizontal space.



enter image description here



documentclass[a4paper,10pt]{article}
usepackage{amsmath}
begin{document}

begin{equation*}
begin{vmatrix}
0 &cdot &cdot &0 &a_{1} alpha_{1}+ cdots +k_{1} alpha_{n} &cdot &cdot &a_{1} kappa_{1}+ cdots +k_{1} kappa_{n} \
cdot &cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot \
0 &cdot &cdot &0 &a_{n} alpha_{1}+ cdots +k_{n} alpha_{n} &cdot &cdot &a_{n} kappa_{1}+ cdots +k_{n} kappa_{n} \
-1 &cdot &cdot &0 &alpha_{1} &cdot &cdot &kappa_{1} \
cdot &cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot \
0 &cdot &cdot &-1 &alpha_{n} &cdot &cdot &kappa_{n}
end{vmatrix}
end{equation*}

end{document}





share|improve this answer





















  • Honestly, I had never thought of using 'hfil'. Thank you for the pointer.
    – user91822
    2 days ago













up vote
4
down vote










up vote
4
down vote









One way is to manually put dots with some horizontal space.



enter image description here



documentclass[a4paper,10pt]{article}
usepackage{amsmath}
begin{document}

begin{equation*}
begin{vmatrix}
0 &cdot &cdot &0 &a_{1} alpha_{1}+ cdots +k_{1} alpha_{n} &cdot &cdot &a_{1} kappa_{1}+ cdots +k_{1} kappa_{n} \
cdot &cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot \
0 &cdot &cdot &0 &a_{n} alpha_{1}+ cdots +k_{n} alpha_{n} &cdot &cdot &a_{n} kappa_{1}+ cdots +k_{n} kappa_{n} \
-1 &cdot &cdot &0 &alpha_{1} &cdot &cdot &kappa_{1} \
cdot &cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot \
0 &cdot &cdot &-1 &alpha_{n} &cdot &cdot &kappa_{n}
end{vmatrix}
end{equation*}

end{document}





share|improve this answer












One way is to manually put dots with some horizontal space.



enter image description here



documentclass[a4paper,10pt]{article}
usepackage{amsmath}
begin{document}

begin{equation*}
begin{vmatrix}
0 &cdot &cdot &0 &a_{1} alpha_{1}+ cdots +k_{1} alpha_{n} &cdot &cdot &a_{1} kappa_{1}+ cdots +k_{1} kappa_{n} \
cdot &cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot \
0 &cdot &cdot &0 &a_{n} alpha_{1}+ cdots +k_{n} alpha_{n} &cdot &cdot &a_{n} kappa_{1}+ cdots +k_{n} kappa_{n} \
-1 &cdot &cdot &0 &alpha_{1} &cdot &cdot &kappa_{1} \
cdot &cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot &cdot &cdot &cdot hfil hfil cdot hfil cdot hfil cdot hfil cdot \
0 &cdot &cdot &-1 &alpha_{n} &cdot &cdot &kappa_{n}
end{vmatrix}
end{equation*}

end{document}






share|improve this answer












share|improve this answer



share|improve this answer










answered 2 days ago









nidhin

2,714926




2,714926












  • Honestly, I had never thought of using 'hfil'. Thank you for the pointer.
    – user91822
    2 days ago


















  • Honestly, I had never thought of using 'hfil'. Thank you for the pointer.
    – user91822
    2 days ago
















Honestly, I had never thought of using 'hfil'. Thank you for the pointer.
– user91822
2 days ago




Honestly, I had never thought of using 'hfil'. Thank you for the pointer.
– user91822
2 days ago










up vote
3
down vote













Defining two newcommands twodts and fivedts for empty columns and rows can simplify the table.



documentclass[a4paper,10pt]{article}
usepackage{amsmath}

begin{document}

newcommand{twodts}{begin{tabular}{@{}cc@{}}$cdot$&$cdot$end{tabular}}
newcommand{fivedts}{begin{tabular}{@{}*5{c}@{}}$cdot$&$cdot$&$cdot$&$cdot$&$cdot$end{tabular}}

begin{equation*}
begin{array}{|cccccc|}
0 & twodts & 0 & a_1 alpha_1+ cdots +k_1 alpha_n & twodts & a_1 kappa_1+ cdots +k_1 kappa_n \
cdot & twodts & cdot & fivedts & twodts & fivedts \
0 & twodts & 0 & a_n alpha_1+ cdots +k_n alpha_n & twodts & a_n kappa_1+ cdots +k_n kappa_n \
-1 & twodts & 0 & alpha_1 & twodts & kappa_1 \
cdot & twodts & cdot & fivedts & twodts & fivedts \
0 & twodts & -1 & alpha_n & twodts & kappa_n
end{array}
end{equation*}

end{document}



enter image description here




Update:



You can easily adapt my answer to add six dots instead of five like this:



documentclass[a4paper,10pt]{article}
usepackage{amsmath,array}

begin{document}

deftwocdots{$cdot$&$cdot$}
newcommand{twodts}{begin{tabular}{@{}cc@{}}twocdotsend{tabular}}
newcommand{sixdts}{begin{tabular}{@{}*6{c}@{}}twocdots&twocdots&twocdotsend{tabular}}

begin{equation*}
begin{array}{|cccccc|}
0 & twodts & 0 & a_1 alpha_1+ cdots +k_1 alpha_n & twodts & a_1 kappa_1+ cdots +k_1 kappa_n \
cdot & twodts & cdot & sixdts & twodts & sixdts \
0 & twodts & 0 & a_n alpha_1+ cdots +k_n alpha_n & twodts & a_n kappa_1+ cdots +k_n kappa_n \
-1 & twodts & 0 & alpha_1 & twodts & kappa_1 \
cdot & twodts & cdot & sixdts & twodts & sixdts \
0 & twodts & -1 & alpha_n & twodts & kappa_n
end{array}
end{equation*}

end{document}



enter image description here







share|improve this answer























  • This looks pretty simple and good solution. Since there is still some gaps that seems to be able to put one more dot when using the `fivedts' command, I wonder if this command can be modded to accommodate the gap. I tried to modify with a "sixdots" and it does not work, unfortunately. Anyway, thank you too.
    – user91822
    2 days ago










  • Please see my update for doing this.
    – AboAmmar
    yesterday















up vote
3
down vote













Defining two newcommands twodts and fivedts for empty columns and rows can simplify the table.



documentclass[a4paper,10pt]{article}
usepackage{amsmath}

begin{document}

newcommand{twodts}{begin{tabular}{@{}cc@{}}$cdot$&$cdot$end{tabular}}
newcommand{fivedts}{begin{tabular}{@{}*5{c}@{}}$cdot$&$cdot$&$cdot$&$cdot$&$cdot$end{tabular}}

begin{equation*}
begin{array}{|cccccc|}
0 & twodts & 0 & a_1 alpha_1+ cdots +k_1 alpha_n & twodts & a_1 kappa_1+ cdots +k_1 kappa_n \
cdot & twodts & cdot & fivedts & twodts & fivedts \
0 & twodts & 0 & a_n alpha_1+ cdots +k_n alpha_n & twodts & a_n kappa_1+ cdots +k_n kappa_n \
-1 & twodts & 0 & alpha_1 & twodts & kappa_1 \
cdot & twodts & cdot & fivedts & twodts & fivedts \
0 & twodts & -1 & alpha_n & twodts & kappa_n
end{array}
end{equation*}

end{document}



enter image description here




Update:



You can easily adapt my answer to add six dots instead of five like this:



documentclass[a4paper,10pt]{article}
usepackage{amsmath,array}

begin{document}

deftwocdots{$cdot$&$cdot$}
newcommand{twodts}{begin{tabular}{@{}cc@{}}twocdotsend{tabular}}
newcommand{sixdts}{begin{tabular}{@{}*6{c}@{}}twocdots&twocdots&twocdotsend{tabular}}

begin{equation*}
begin{array}{|cccccc|}
0 & twodts & 0 & a_1 alpha_1+ cdots +k_1 alpha_n & twodts & a_1 kappa_1+ cdots +k_1 kappa_n \
cdot & twodts & cdot & sixdts & twodts & sixdts \
0 & twodts & 0 & a_n alpha_1+ cdots +k_n alpha_n & twodts & a_n kappa_1+ cdots +k_n kappa_n \
-1 & twodts & 0 & alpha_1 & twodts & kappa_1 \
cdot & twodts & cdot & sixdts & twodts & sixdts \
0 & twodts & -1 & alpha_n & twodts & kappa_n
end{array}
end{equation*}

end{document}



enter image description here







share|improve this answer























  • This looks pretty simple and good solution. Since there is still some gaps that seems to be able to put one more dot when using the `fivedts' command, I wonder if this command can be modded to accommodate the gap. I tried to modify with a "sixdots" and it does not work, unfortunately. Anyway, thank you too.
    – user91822
    2 days ago










  • Please see my update for doing this.
    – AboAmmar
    yesterday













up vote
3
down vote










up vote
3
down vote









Defining two newcommands twodts and fivedts for empty columns and rows can simplify the table.



documentclass[a4paper,10pt]{article}
usepackage{amsmath}

begin{document}

newcommand{twodts}{begin{tabular}{@{}cc@{}}$cdot$&$cdot$end{tabular}}
newcommand{fivedts}{begin{tabular}{@{}*5{c}@{}}$cdot$&$cdot$&$cdot$&$cdot$&$cdot$end{tabular}}

begin{equation*}
begin{array}{|cccccc|}
0 & twodts & 0 & a_1 alpha_1+ cdots +k_1 alpha_n & twodts & a_1 kappa_1+ cdots +k_1 kappa_n \
cdot & twodts & cdot & fivedts & twodts & fivedts \
0 & twodts & 0 & a_n alpha_1+ cdots +k_n alpha_n & twodts & a_n kappa_1+ cdots +k_n kappa_n \
-1 & twodts & 0 & alpha_1 & twodts & kappa_1 \
cdot & twodts & cdot & fivedts & twodts & fivedts \
0 & twodts & -1 & alpha_n & twodts & kappa_n
end{array}
end{equation*}

end{document}



enter image description here




Update:



You can easily adapt my answer to add six dots instead of five like this:



documentclass[a4paper,10pt]{article}
usepackage{amsmath,array}

begin{document}

deftwocdots{$cdot$&$cdot$}
newcommand{twodts}{begin{tabular}{@{}cc@{}}twocdotsend{tabular}}
newcommand{sixdts}{begin{tabular}{@{}*6{c}@{}}twocdots&twocdots&twocdotsend{tabular}}

begin{equation*}
begin{array}{|cccccc|}
0 & twodts & 0 & a_1 alpha_1+ cdots +k_1 alpha_n & twodts & a_1 kappa_1+ cdots +k_1 kappa_n \
cdot & twodts & cdot & sixdts & twodts & sixdts \
0 & twodts & 0 & a_n alpha_1+ cdots +k_n alpha_n & twodts & a_n kappa_1+ cdots +k_n kappa_n \
-1 & twodts & 0 & alpha_1 & twodts & kappa_1 \
cdot & twodts & cdot & sixdts & twodts & sixdts \
0 & twodts & -1 & alpha_n & twodts & kappa_n
end{array}
end{equation*}

end{document}



enter image description here







share|improve this answer














Defining two newcommands twodts and fivedts for empty columns and rows can simplify the table.



documentclass[a4paper,10pt]{article}
usepackage{amsmath}

begin{document}

newcommand{twodts}{begin{tabular}{@{}cc@{}}$cdot$&$cdot$end{tabular}}
newcommand{fivedts}{begin{tabular}{@{}*5{c}@{}}$cdot$&$cdot$&$cdot$&$cdot$&$cdot$end{tabular}}

begin{equation*}
begin{array}{|cccccc|}
0 & twodts & 0 & a_1 alpha_1+ cdots +k_1 alpha_n & twodts & a_1 kappa_1+ cdots +k_1 kappa_n \
cdot & twodts & cdot & fivedts & twodts & fivedts \
0 & twodts & 0 & a_n alpha_1+ cdots +k_n alpha_n & twodts & a_n kappa_1+ cdots +k_n kappa_n \
-1 & twodts & 0 & alpha_1 & twodts & kappa_1 \
cdot & twodts & cdot & fivedts & twodts & fivedts \
0 & twodts & -1 & alpha_n & twodts & kappa_n
end{array}
end{equation*}

end{document}



enter image description here




Update:



You can easily adapt my answer to add six dots instead of five like this:



documentclass[a4paper,10pt]{article}
usepackage{amsmath,array}

begin{document}

deftwocdots{$cdot$&$cdot$}
newcommand{twodts}{begin{tabular}{@{}cc@{}}twocdotsend{tabular}}
newcommand{sixdts}{begin{tabular}{@{}*6{c}@{}}twocdots&twocdots&twocdotsend{tabular}}

begin{equation*}
begin{array}{|cccccc|}
0 & twodts & 0 & a_1 alpha_1+ cdots +k_1 alpha_n & twodts & a_1 kappa_1+ cdots +k_1 kappa_n \
cdot & twodts & cdot & sixdts & twodts & sixdts \
0 & twodts & 0 & a_n alpha_1+ cdots +k_n alpha_n & twodts & a_n kappa_1+ cdots +k_n kappa_n \
-1 & twodts & 0 & alpha_1 & twodts & kappa_1 \
cdot & twodts & cdot & sixdts & twodts & sixdts \
0 & twodts & -1 & alpha_n & twodts & kappa_n
end{array}
end{equation*}

end{document}



enter image description here








share|improve this answer














share|improve this answer



share|improve this answer








edited yesterday

























answered 2 days ago









AboAmmar

31.8k22781




31.8k22781












  • This looks pretty simple and good solution. Since there is still some gaps that seems to be able to put one more dot when using the `fivedts' command, I wonder if this command can be modded to accommodate the gap. I tried to modify with a "sixdots" and it does not work, unfortunately. Anyway, thank you too.
    – user91822
    2 days ago










  • Please see my update for doing this.
    – AboAmmar
    yesterday


















  • This looks pretty simple and good solution. Since there is still some gaps that seems to be able to put one more dot when using the `fivedts' command, I wonder if this command can be modded to accommodate the gap. I tried to modify with a "sixdots" and it does not work, unfortunately. Anyway, thank you too.
    – user91822
    2 days ago










  • Please see my update for doing this.
    – AboAmmar
    yesterday
















This looks pretty simple and good solution. Since there is still some gaps that seems to be able to put one more dot when using the `fivedts' command, I wonder if this command can be modded to accommodate the gap. I tried to modify with a "sixdots" and it does not work, unfortunately. Anyway, thank you too.
– user91822
2 days ago




This looks pretty simple and good solution. Since there is still some gaps that seems to be able to put one more dot when using the `fivedts' command, I wonder if this command can be modded to accommodate the gap. I tried to modify with a "sixdots" and it does not work, unfortunately. Anyway, thank you too.
– user91822
2 days ago












Please see my update for doing this.
– AboAmmar
yesterday




Please see my update for doing this.
– AboAmmar
yesterday










up vote
1
down vote













Would this be acceptable?



enter image description here



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

begin{document}

begin{equation*}
begin{vmatrix}
0 &cdots &0 &a_{1} alpha_{1}+ cdots +k_{1} alpha_{n} &cdots &a_{1} kappa_{1}+ cdots +k_{1} kappa_{n} \
vdots &ddots &vdots &vdots &ddots &vdots \
0 &cdots & 0 &a_{n} alpha_{1}+ cdots +k_{n} alpha_{n} &cdots &a_{n} kappa_{1}+ cdots +k_{n} kappa_{n} \
-1 &cdots &0 &alpha_{1} &cdots &kappa_{1} \
vdots &ddots &vdots &vdots &ddots &vdots \
0 &cdots & -1 &alpha_{n} &cdots &kappa_{n}
end{vmatrix}
end{equation*}

end{document}





share|improve this answer

















  • 1




    This is I usually do and I don't think it will do for me. Sorry and I try to stick closer to the example I posted above. Anyway, thank you.
    – user91822
    2 days ago

















up vote
1
down vote













Would this be acceptable?



enter image description here



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

begin{document}

begin{equation*}
begin{vmatrix}
0 &cdots &0 &a_{1} alpha_{1}+ cdots +k_{1} alpha_{n} &cdots &a_{1} kappa_{1}+ cdots +k_{1} kappa_{n} \
vdots &ddots &vdots &vdots &ddots &vdots \
0 &cdots & 0 &a_{n} alpha_{1}+ cdots +k_{n} alpha_{n} &cdots &a_{n} kappa_{1}+ cdots +k_{n} kappa_{n} \
-1 &cdots &0 &alpha_{1} &cdots &kappa_{1} \
vdots &ddots &vdots &vdots &ddots &vdots \
0 &cdots & -1 &alpha_{n} &cdots &kappa_{n}
end{vmatrix}
end{equation*}

end{document}





share|improve this answer

















  • 1




    This is I usually do and I don't think it will do for me. Sorry and I try to stick closer to the example I posted above. Anyway, thank you.
    – user91822
    2 days ago















up vote
1
down vote










up vote
1
down vote









Would this be acceptable?



enter image description here



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

begin{document}

begin{equation*}
begin{vmatrix}
0 &cdots &0 &a_{1} alpha_{1}+ cdots +k_{1} alpha_{n} &cdots &a_{1} kappa_{1}+ cdots +k_{1} kappa_{n} \
vdots &ddots &vdots &vdots &ddots &vdots \
0 &cdots & 0 &a_{n} alpha_{1}+ cdots +k_{n} alpha_{n} &cdots &a_{n} kappa_{1}+ cdots +k_{n} kappa_{n} \
-1 &cdots &0 &alpha_{1} &cdots &kappa_{1} \
vdots &ddots &vdots &vdots &ddots &vdots \
0 &cdots & -1 &alpha_{n} &cdots &kappa_{n}
end{vmatrix}
end{equation*}

end{document}





share|improve this answer












Would this be acceptable?



enter image description here



documentclass[a4paper,10pt]{article}

usepackage{amsmath}

begin{document}

begin{equation*}
begin{vmatrix}
0 &cdots &0 &a_{1} alpha_{1}+ cdots +k_{1} alpha_{n} &cdots &a_{1} kappa_{1}+ cdots +k_{1} kappa_{n} \
vdots &ddots &vdots &vdots &ddots &vdots \
0 &cdots & 0 &a_{n} alpha_{1}+ cdots +k_{n} alpha_{n} &cdots &a_{n} kappa_{1}+ cdots +k_{n} kappa_{n} \
-1 &cdots &0 &alpha_{1} &cdots &kappa_{1} \
vdots &ddots &vdots &vdots &ddots &vdots \
0 &cdots & -1 &alpha_{n} &cdots &kappa_{n}
end{vmatrix}
end{equation*}

end{document}






share|improve this answer












share|improve this answer



share|improve this answer










answered 2 days ago









prt13463

765




765








  • 1




    This is I usually do and I don't think it will do for me. Sorry and I try to stick closer to the example I posted above. Anyway, thank you.
    – user91822
    2 days ago
















  • 1




    This is I usually do and I don't think it will do for me. Sorry and I try to stick closer to the example I posted above. Anyway, thank you.
    – user91822
    2 days ago










1




1




This is I usually do and I don't think it will do for me. Sorry and I try to stick closer to the example I posted above. Anyway, thank you.
– user91822
2 days ago






This is I usually do and I don't think it will do for me. Sorry and I try to stick closer to the example I posted above. Anyway, thank you.
– user91822
2 days ago




















draft saved

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