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I am trying to get my equation to look like the following:

enter image description here

I cannot seem to get anywhere close (is there a way to get labels in \bordermatrix to go at the bottom?). Here is my code so far (without the dotted line and column totals)

\begin{eqnarray} 
P^{\ast}_{t+1} &=& P_{t+1} + D_{t,t+1} - B_{t,t+1} \\ \nonumber
&=&
\left(
\begin{array}{llll}
1110 &  110 &    10 &   10  \\
  45 &  440 &    45 &    0  \\
  70 &   75 &   770 &   70  \\
  35 &   30 &    30 &  330  \\
\end{array}
\right) + 
\left(
\begin{array}{llll}
44.05 &   3.36 &   0.53 &   0.24  \\
 1.79 &  13.44 &   2.37 &   0.00  \\
 2.78 &   2.29 &  40.53 &   1.71  \\
 1.39 &   0.92 &   1.58 &   8.05  \\
\end{array}
\right) - 
\left(
\begin{array}{llll}
80 &     0 &     0 &     0 \\
 0 &    25 &     0 &     0 \\
 0 &     0 &    40 &     0 \\
 0 &     0 &     0 &    40 \\
\end{array}
\right)
\end{eqnarray}
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1  
eqnarray is depreciated, and most folks recommend something from the amsmath package- have a look at eqnarray-vs-align –  cmhughes May 21 '12 at 15:54
    
The columns don't sum up to the shown total. Apart from that, do you need that the computation is performed automatically? I'll hope not. –  egreg May 21 '12 at 15:57
    
uploaded the wrong image...my apologies. –  gjabel May 21 '12 at 16:01
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3 Answers

up vote 9 down vote accepted

Here is one way of obtained the desired output:

enter image description here

\documentclass{article}
\usepackage{amsmath}% http://ctan.org/pkg/amsmath
\usepackage{arydshln}% http://ctan.org/pkg/arydshln
\newcommand{\addsum}[1]{%
  \mathstrut\smash{\begin{array}[t]{@{}r@{}}#1\end{array}}%
}
\begin{document}
\begin{align*} 
  P^{\ast}_{t+1} &= P_{t+1} + D_{t,t+1} - B_{t,t+1} \\
    &=
    \left(
      \begin{array}{rrrr}
        1110 &  110 &    10 &   10 \\
          45 &  440 &    45 &    0 \\
          70 &   75 &   770 &   70 \\
          35 &   30 &    30 &  330
      \end{array}
     \right) + 
     \left(
       \begin{array}{rrrr}
         44.05 &   3.36 &   0.53 &   0.24 \\
          1.79 &  13.44 &   2.37 &   0.00 \\
          2.78 &   2.29 &  40.53 &   1.71 \\
          \addsum{1.39\\50.01} & 
          \addsum{0.92\\20.01} & 
          \addsum{1.58\\45.01} & 
          \addsum{8.05\\10.00} \\ \hdashline
       \end{array}
      \right) - 
      \left(
        \begin{array}{rrrr}
          80 &     0 &     0 &     0 \\
           0 &    25 &     0 &     0 \\
           0 &     0 &    40 &     0 \\
           0 &     0 &     0 &    40
        \end{array}
      \right)
\end{align*}
\end{document}

I modified the alignment to be right for the four columns. With some more work, it would be possible to obtain a centred sum.

The totals are added on a per-column basis using \addsum where you specify the information contained in the last and "sum" row. The arydshln package provides the dashed rule. Alternatively, using \hline within \addsum instead of \hdashline yields the following output:

enter image description here

%...
 \left(
   \begin{array}{rrrr}
     44.05 &   3.36 &   0.53 &   0.24 \\
      1.79 &  13.44 &   2.37 &   0.00 \\
      2.78 &   2.29 &  40.53 &   1.71 \\
      \addsum{1.39\\\hline 50.01} & 
      \addsum{0.92\\\hline 20.01} & 
      \addsum{1.58\\\hline 45.01} & 
      \addsum{8.05\\\hline 10.00}
   \end{array}
  \right) - 
%...

In both instances, \smash removes all vertical height/depth of its contents. As such, be careful when adding contents below this expression. Adding a blank row in the form \\& would mostly suffice.

share|improve this answer
    
+1 although if you put \lipsum[1] immediately before and after this, it seems that the smash distorts the vertical spacing at the end of the environment. Presumably this can be fixed with some manual adjustment? –  cmhughes May 21 '12 at 16:09
    
That's very similar to my solution; however I'd use a \hline in the inner arrays, rather than \hdashline. I'd also add an empty line \\ after the row of matrices. –  egreg May 21 '12 at 16:09
    
@cmhughes: I thought about that, and forgot to mention it. Will add, thanks. –  Werner May 21 '12 at 16:10
    
@egreg: Neat suggestion; I'll add. –  Werner May 21 '12 at 16:10
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Here is a solution based on TikZ. Basic hints are:

  • change matrix separator to not have problems with the align environment;
  • fix the baseline position and inner sep to adapt dimension.

The code:

\documentclass{article}
\usepackage{amsmath,amssymb}
\usepackage{tikz}
\usetikzlibrary{matrix,calc}
\begin{document}
\begin{align*}
P^{\ast}_{t+1} &= P_{t+1} + D_{t,t+1} - B_{t,t+1} \\
&=
\left(
\begin{array}{llll}
1110 &  110 &    10 &   10  \\
  45 &  440 &    45 &    0  \\
  70 &   75 &   770 &   70  \\
  35 &   30 &    30 &  330  \\
\end{array}
\right)+
\begin{tikzpicture}[baseline=-0.1cm]
%\let\&=\pgfmatrixnextcell % <= one choice to change col separator
\matrix[matrix of math nodes,left delimiter=(,right delimiter=),inner sep=2.5pt, 
 ampersand replacement=\&] % <= to change col separator for align env
{
\node {44.05}; \& \node{3.36};\& \node {0.53}; \& \node{0.24}; \\
\node {1.79}; \& \node{13.44};\& \node {2.37}; \& \node{0.00};  \\
\node {2.78}; \& \node{2.29}; \&\node {40.53}; \& \node{1.71};  \\
\node(a) {1.39}; \& \node(b){0.92};\& \node(c) {1.58}; \& \node(d){8.05};  \\
};
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\foreach \x/\xtext in {a/50.01,b/20.01,c/45.01,d/10.00}{
% line below last row
\draw ($(\x.south)-(0.3,0.1)$) -- ($(\x.south)+(0.3,-0.1)$);
% sums
\node[shift=(\x.south),yshift=-0.45cm](0,0) {\xtext} ;
}
\end{tikzpicture}
- 
\left(
\begin{array}{llll}
80 &     0 &     0 &     0 \\
 0 &    25 &     0 &     0 \\
 0 &     0 &    40 &     0 \\
 0 &     0 &     0 &    40 \\
\end{array}
\right)
\end{align*} 
\end{document}

Graphical result:

enter image description here

IMPROVMENT

It is also possible change the code of the picture into (with the same final output):

\begin{tikzpicture}[baseline=-0.1cm]
\matrix(m)[matrix of math nodes,left delimiter=(,right delimiter=),inner sep=2.5pt,ampersand replacement=\&]
{
44.05 \&   3.36 \&   0.53 \&   0.24  \\
 1.79 \&  13.44 \&   2.37 \&   0.00  \\
 2.78 \&   2.29 \&  40.53 \&   1.71  \\
 1.39 \&   0.92 \&   1.58 \&   8.05  \\
};
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\foreach \x/\xtext in {m-4-1/50.01,m-4-2/20.01,m-4-3/45.01,m-4-4/10.00}{
% line below last row
\draw ($(\x.south)-(0.3,0.1)$) -- ($(\x.south)+(0.3,-0.1)$);
% sums
\node[shift=(\x.south),yshift=-0.45cm](0,0) {\xtext} ;
}
\end{tikzpicture}
share|improve this answer
1  
The option ampersand replacement=\& is more convenient to alter the col seperator. Also you don't need to put the \node explicitly since they are already math nodes :) Then you can access the bottom nodes as (matrix node name - row # - col #) in this example (m-4-1) would be the lower left node of the matrix (m) –  percusse May 21 '12 at 16:49
    
@percusse: I didn't know this option; I'll try it immediately, thanks :) –  Claudio Fiandrino May 21 '12 at 16:50
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I was trying out Martin Scharrer's interesting work in collcell package and I managed to get the sums of the columns automatically. It's not tuned for the result since other answers are great anyway.

\documentclass{article}
\usepackage{amsmath,collcell,tikz}
\def\columnone#1{#1\expandafter\pgfmathparse\expandafter{\cone + #1}\xdef\cone{\pgfmathresult}}
\def\columntwo#1{#1\expandafter\pgfmathparse\expandafter{\ctwo + #1}\xdef\ctwo{\pgfmathresult}}
\def\columnthree#1{#1\expandafter\pgfmathparse\expandafter{\cthree + #1}\xdef\cthree{\pgfmathresult}}
\def\columnfour#1{#1\expandafter\pgfmathparse\expandafter{\cfour + #1}\xdef\cfour{\pgfmathresult}}

\newcolumntype{X}{>{\collectcell\columnone}c<{\endcollectcell}}
\newcolumntype{Y}{>{\collectcell\columntwo}c<{\endcollectcell}}
\newcolumntype{W}{>{\collectcell\columnthree}c<{\endcollectcell}}
\newcolumntype{Z}{>{\collectcell\columnfour}c<{\endcollectcell}}

\begin{document}
\begin{align*} 
  P^{\ast}_{t+1} &= P_{t+1} + D_{t,t+1} - B_{t,t+1} \\
    &=
    \left(
      \begin{array}{rrrr}
        1110 &  110 &    10 &   10 \\
          45 &  440 &    45 &    0 \\
          70 &   75 &   770 &   70 \\
          35 &   30 &    30 &  330
      \end{array}
     \right) + \left[
\begin{tikzpicture}[baseline=(current bounding box.center)]
\def\cone{0} \def\ctwo{0} \def\cthree{0} \def\cfour{0}
\node[inner sep=0] (tabnode) {
\begin{tabular}{XYWZ}
44.05 &   3.36 &   0.53 &   0.24  \\
 1.79 &  13.44 &   2.37 &   0.00  \\
 2.78 &   2.29 &  40.53 &   1.71  \\
 1.39 &   0.92 &   1.58 &   8.05  \\
\end{tabular}
};
\useasboundingbox (tabnode.north east) -- (tabnode.south west);
\draw[dashed] (tabnode.south west) -- (tabnode.south east);
\node[anchor=north] (s1) at (tabnode.205) {\cone};
\node[anchor=north] (s2) at (tabnode.237) {\ctwo};
\node[anchor=north] (s3) at (tabnode.307) {\cthree};
\node[anchor=north] (s4) at (tabnode.338) {\cfour};
\end{tikzpicture}
\right]
- 
      \left(
        \begin{array}{rrrr}
          80 &     0 &     0 &     0 \\
           0 &    25 &     0 &     0 \\
           0 &     0 &    40 &     0 \\
           0 &     0 &     0 &    40
        \end{array}
      \right)
\end{align*}
\end{document}

Basically there is a number accumulation going on in the respective macro of each column and then we put that number inside a TikZ node. There are a couple of things among many : First, the node placement is not joyful and requires an automated solution by obtaining column widths and aligning them at the decimal point etc. Second, it's best to reset those registers after the command is executed for successive uses. The number formatting is relatively easy but I skipped it.

enter image description here

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