Empty elements as dots in matrix
up vote
6
down vote
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I need to create a determinant (vmatrix) as shown in this picture. 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: 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
add a comment |
up vote
6
down vote
favorite
I need to create a determinant (vmatrix) as shown in this picture. 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: 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
hdotsfor
has an optional argument which is a multiplier for the amount of space between consecutive dots, so you could tryhdotsfor[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
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I need to create a determinant (vmatrix) as shown in this picture. 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: 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
I need to create a determinant (vmatrix) as shown in this picture. 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: 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
matrices
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 tryhdotsfor[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
add a comment |
hdotsfor
has an optional argument which is a multiplier for the amount of space between consecutive dots, so you could tryhdotsfor[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
add a comment |
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}
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}
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*}
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
add a comment |
up vote
4
down vote
One way is to manually put dots with some horizontal space.
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}
Honestly, I had never thought of using 'hfil'. Thank you for the pointer.
– user91822
2 days ago
add a comment |
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}
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}
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
add a comment |
up vote
1
down vote
Would this be acceptable?
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}
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
add a comment |
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}
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}
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*}
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
add a comment |
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}
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}
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*}
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
add a comment |
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}
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}
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*}
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}
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}
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*}
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
add a comment |
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
add a comment |
up vote
4
down vote
One way is to manually put dots with some horizontal space.
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}
Honestly, I had never thought of using 'hfil'. Thank you for the pointer.
– user91822
2 days ago
add a comment |
up vote
4
down vote
One way is to manually put dots with some horizontal space.
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}
Honestly, I had never thought of using 'hfil'. Thank you for the pointer.
– user91822
2 days ago
add a comment |
up vote
4
down vote
up vote
4
down vote
One way is to manually put dots with some horizontal space.
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}
One way is to manually put dots with some horizontal space.
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}
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
add a comment |
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
add a comment |
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}
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}
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
add a comment |
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}
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}
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
add a comment |
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}
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}
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}
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}
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
add a comment |
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
add a comment |
up vote
1
down vote
Would this be acceptable?
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}
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
add a comment |
up vote
1
down vote
Would this be acceptable?
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}
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
add a comment |
up vote
1
down vote
up vote
1
down vote
Would this be acceptable?
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}
Would this be acceptable?
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}
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
add a comment |
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
add a comment |
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hdotsfor
has an optional argument which is a multiplier for the amount of space between consecutive dots, so you could tryhdotsfor[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