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<body>
<font size=3>The document to which you referred me was excellent.
Everything now looks great.<br><br>
Thanks very much.<br><br>
<br><br>
\documentclass[reqno]{amsart}<br>
\usepackage{amssymb,latexsym}<br>
\usepackage{pstricks}<br>
\usepackage{pst-node}<br><br>
<br>
\begin{document}<br><br>
\subsection{Submatrices and partitions}<br><br>
Before discussing the computation of determinants using cofactors a<br>
few definitions concerning matrices and submatrices will be
useful.<br><br>
\subsubsection{Submatrix}<br><br>
A submatrix is a matrix formed from a matrix A by taking a subset<br>
consisting of j rows with column elements from a set k of the<br>
columns. For example consider A(\{1,3\},\{2,3\}) below<br><br>
<br>
\begin{equation}<br>
\begin{aligned}<br>
A~=&~\left (<br>
\begin{matrix}3 & 4&7 \\<br>
2&5& 2 \\<br>
1&0&4<br>
\end{matrix} \right )\\[6pt]<br>
A(\{1,3\},\{2,3\})~=&~\left(\raisebox{-4.2ex}{<br>
\begin{psmatrix}[rowsep=3pt,colsep=10pt]<br>
3 &4&7 \\<br>
2&5& 2 \\<br>
1&0&4<br>
\end{psmatrix}} \right)<br>
\psset{nodesep=.1,boxsize=.25,linearc=.3,linestyle=dashed,linecolor=blue}<br>
\ncbox[nodesep=.2,linestyle=solid,linecolor=red]{1,2}{3,2}<br>
\ncbox[nodesep=.2,linestyle=solid,linecolor=red]{1,3}{3,3}<br>
\ncbox{1,1}{1,3} \ncbox{3,1}{3,3}<br>
\\[6pt]<br>
=&~\left (<br>
\begin{matrix}<br>
4 &7 \\<br>
0&4<br>
\end{matrix}<br>
\right )\\[4pt]<br>
\end{aligned}<br>
\end{equation}<br><br>
The notation A(\{1,3\},\{2,3\}) means that we take the first and<br>
third rows of A and include the second and third elements of each<br>
row.<br><br>
\subsubsection{Principal submatrix}<br><br>
A principal submatrix is a matrix formed from a square matrix A
by<br>
taking a subset consisting of n rows and column elements from the<br>
same numbered columns. For example consider A(\{1,3\},\{1,3\})<br>
below<br><br>
\begin{equation}<br>
\begin{aligned}<br>
A~=&~\left (<br>
\begin{matrix}3 & 4&7 \\<br>
2&5& 2 \\<br>
1&0&4<br>
\end{matrix} \right )\\[6pt]<br>
A(\{1,3\},\{1,3\})~=&~\left(\raisebox{-4.2ex}{<br>
\begin{psmatrix}[rowsep=3pt,colsep=10pt]<br>
3 &4&7 \\<br>
2&5& 2 \\<br>
1&0&4<br>
\end{psmatrix}} \right)<br>
\psset{nodesep=.1,boxsize=.25,linearc=.3,linestyle=dashed,linecolor=blue}<br>
\ncbox[nodesep=.2,linestyle=solid,linecolor=red]{1,1}{3,1}<br>
\ncbox[nodesep=.2,linestyle=solid,linecolor=red]{1,3}{3,3}<br>
\ncbox{1,1}{1,3} \ncbox{3,1}{3,3}<br>
\\[6pt]<br>
=&~\left (<br>
\begin{matrix}<br>
3 &7 \\<br>
1&4<br>
\end{matrix}<br>
\right )\\[4pt]<br>
\end{aligned}<br>
\end{equation}<br><br>
\subsubsection{Minor}<br><br>
A minor is the determinant of a square submatrix of the matrix A.<br>
For example consider $|\, A(\{2,3\},\{1,3\})\, |$.<br><br>
\begin{equation}<br>
\begin{aligned}<br>
A~=&~\left (<br>
\begin{matrix}3 & 4&7 \\<br>
2&5& 2 \\<br>
1&0&4<br>
\end{matrix} \right )\\[4pt]<br>
A(\{2,3\},\{1,3\})~=&~\left(\raisebox{-4.1ex}{<br>
\begin{psmatrix}[rowsep=3pt,colsep=10pt]<br>
3 &4&7 \\<br>
2&5& 2 \\<br>
1&0&4<br>
\end{psmatrix}} \right)<br>
\psset{nodesep=.1,boxsize=.25,linearc=.3,linestyle=dashed,linecolor=blue}<br>
\ncbox[nodesep=.2,linestyle=solid,linecolor=red]{1,1}{3,1}<br>
\ncbox[nodesep=.2,linestyle=solid,linecolor=red]{1,3}{3,3}<br>
\ncbox{2,1}{2,3} \ncbox{3,1}{3,3}<br>
\\[6pt]<br>
=&~\left (<br>
\begin{matrix}<br>
2 &2 \\<br>
1&4<br>
\end{matrix}<br>
\right )\\[4pt]<br>
|A\,(\{2,3\},\{1,3\})\, | ~=&~6<br>
\end{aligned}<br>
\end{equation}<br><br>
\subsubsection{Principal minor}<br><br>
A principal minor is the determinant of a principal submatrix of A.<br>
For example consider $|\,A(\{1,2\},\{1,2\})\,|$.<br><br>
\begin{equation}<br>
\begin{aligned}<br>
A~=&~\left (<br>
\begin{matrix}3 & 4&7 \\<br>
2&5& 2 \\<br>
1&0&4<br>
\end{matrix} \right )\\[4pt]<br>
A(\{1,2\},\{1,2\})~=&~\left(\raisebox{-4.1ex}{<br>
\begin{psmatrix}[rowsep=3pt,colsep=10pt]<br>
3 &4&7 \\<br>
2&5& 2 \\<br>
1&0&4<br>
\end{psmatrix}} \right)<br>
\psset{nodesep=.1,boxsize=.25,linearc=.3,linestyle=dashed,linecolor=blue}<br>
\ncbox[nodesep=.2,linestyle=solid,linecolor=red]{1,1}{3,1}<br>
\ncbox[nodesep=.2,linestyle=solid,linecolor=red]{1,2}{3,2}<br>
\ncbox{1,1}{1,3} \ncbox{2,1}{2,3}<br>
\\[6pt]<br>
=&~\left (<br>
\begin{matrix}<br>
3 &4 \\<br>
2&5<br>
\end{matrix}<br>
\right )\\[4pt]<br>
|A\,(\{1,2\},\{1,2\})\, | ~=&~7<br>
\end{aligned}<br>
\end{equation}<br><br>
\subsubsection{Leading principal minor}<br><br>
Let A =(a$_{ij}$) be any n $\times$ n . The {\bf leading
principle<br>
minors} of A are the n determinants:<br><br>
\begin{equation}<br>
D_k~=~\left[\,<br>
\begin{matrix}<br>
a_{11}&a_{12}&\dots&a_{1k}\\<br>
a_{21}&a_{22}&\dots&a_{2k}\\<br>
\vdots&\vdots&\ddots&\vdots\\<br>
a_{k1}&a_{k2}&\dots&a_{kk}<br>
\end{matrix}\,<br>
\right], \quad \quad k~=~1, 2, \dots, n<br>
\end{equation}<br><br>
D$_k$ is obtained by crossing out the last n-k columns and n-k rows<br>
of the matrix. Thus for k = 1, 2, 3, $\dots$ n, the leading<br>
principle minors are, respectively<br><br>
\begin{equation}<br>
a_{11}, \ \left|\,<br>
\begin{matrix}<br>
a_{11}&a_{12}\\<br>
a_{21}&a_{22}<br>
\end{matrix}\,<br>
\right|,\ \left|\,<br>
\begin{matrix}<br>
a_{11}&a_{12}&a_{13}\\<br>
a_{21}&a_{22}&a_{23}\\<br>
a_{31}&a_{32}&a_{33}<br>
\end{matrix}\,<br>
\right|,\ \ \dots\,, \quad \left|\,<br>
\begin{matrix}<br>
a_{11}&a_{12}&\dots&a_{1n}\\<br>
a_{21}&a_{22}&\dots&a_{2n}\\<br>
\vdots&\vdots&\ddots&\vdots\\<br>
a_{n1}&a_{n2}&\dots&a_{nn}<br>
\end{matrix}\,<br>
\right|<br>
\end{equation}<br><br>
<br>
\subsubsection{Cofactor}<br><br>
The cofactor (denoted A$_{ij}$) of the element a$_{ij}$ of any<br>
square matrix A is (-1)$^{i+j}$ times the minor of A that is<br>
obtained by including all but the i$^{\rm th}$ row and the j$^{\rm<br>
th}$ column, or alternatively the minor that is obtained by deleting<br>
the i$^{\rm th}$ row and the j$^{\rm th}$ column. For example<br>
consider the matrix A.<br><br>
\begin{equation}<br>
\begin{aligned}<br>
A=& \left (<br>
\begin{matrix}<br>
a_{11} &a_{12} &a_{13} \\<br>
a_{21} &a_{22} &a_{23} \\<br>
a_{31} &a_{32} &a_{33}<br>
\end{matrix}<br>
\right )~ ~=<br>
\left (<br>
\begin{matrix}<br>
3 & 4&7 \\<br>
2&5& 2 \\<br>
1&0&4<br>
\end{matrix}<br>
\right )<br>
\end{aligned}<br>
\end{equation}<br><br>
To find the cofactor of of a$_{12}$ we first find the submatrix that<br>
includes all rows and columns except the first and second.<br><br>
\bigskip<br><br>
\begin{equation}<br>
A(\{2,3\},\{1,3\})~=~\left(\raisebox{-3.0ex}{ $<br>
\begin{psmatrix}[rowsep=6pt,colsep=6pt]%<br>
3 & 4&7 \\<br>
2&5& 2 \\<br>
1&0&4<br>
\end{psmatrix}<br>
$ } \right) \ncline[linecolor=red,nodesep=-0.5em]{1,1}{1,3}<br>
\ncline[linecolor=red,nodesep=-0.5em]{1,2}{3,2}~=~\left (<br>
\begin{matrix}<br>
2& 2\\<br>
1&4<br>
\end{matrix}<br>
\right )<br>
\end{equation}<br><br>
We then multiply the determinant of this matrix by (-1)$^{i+j}$. For<br>
the example this gives<br><br>
\begin{equation}<br>
\begin{aligned}<br>
cofactor(a_{12})~=&~A_{12} =(-1)^{3} ~ | A(\{2,3\},\{1,3\}|<br>
~=~(-1)^{3} ~\left | \,<br>
\begin{matrix}<br>
2&2 \\<br>
1&4<br>
\end{matrix}<br>
\, \right | \\[3pt]<br>
~=&~(-1)^{3} ~(6)~=~-6<br>
\end{aligned}<br>
\end{equation}<br><br>
<br>
\subsection{Computing determinants using cofactors}<br><br>
\subsubsection{Definition of a cofactor expansion}<br><br>
The determinant of a square matrix A can be found inductively using<br>
the following formula<br><br>
\begin{equation}<br>
\det\,\, A~=~|\, A\, | ~=~\sum_{j=1}^{n} ~a_{ij}\, A_{ij}<br>
\end{equation}<br><br>
where i denotes the i$^{\rm th}$ row of the matrix A. This is<br>
called an expansion of $\mid$A$\mid$ by column i of A. The
result<br>
is the same for any other row. This can also be done for
columns<br>
letting the sum range over i instead of j.<br><br>
\subsubsection{Examples}<br><br>
\begin{description}<br><br>
\item[1]<br><br>
Consider as an example the following 2x2 matrix and expand using the<br>
first row<br><br>
\begin{displaymath}<br>
\begin{aligned}<br>
B~=& \left (\begin{matrix}<br>
4&3 \\<br>
1& 2<br>
\end{matrix}<br>
\right ) \\<br>
~&~ \\<br>
| B |~=&~4*(-1)^{2} *(2)~+~3*(-1)^{3} *(1)\,\, =\,\, 8\, -\,
3\,\,<br>
=\,\, 5<br>
\end{aligned}<br>
\end{displaymath}<br><br>
<br>
where the cofactor of 4 is (-1)$^{{(1+1)}}$ times the submatrix that<br>
remains when we delete row and column 1 from the matrix B while the<br>
cofactor of 3 is (-1)$^{{ (1+2)}}$ times the submatrix that remains<br>
when we delete row 1 and column 2 from the matrix B. In this
case<br>
each the principle<br>
submatrices are just single numbers so there is no need to formally
compute a determinant. \\<br><br>
\item[2]<br><br>
Now consider a 3x3 matrix B.<br><br>
\begin{displaymath}<br>
B= \left (<br>
\begin{matrix}1 & 2&3 \\<br>
0&5& 2 \\<br>
1&0&4<br>
\end{matrix}<br>
\right )<br>
\end{displaymath}<br><br>
If we expand using the first row of the matrix, the relevant minors<br>
are<br><br>
\begin{equation}<br>
\begin{aligned}<br>
cofactor(b_{11})~=&~B\left(\{2,3\},\{2,3\}\right)~=~\left|\raisebox{-3.0ex}{<br>
$<br>
\begin{psmatrix}[rowsep=6pt,colsep=6pt]%<br>
1 &2&3 \\<br>
0&5& 2 \\<br>
1&0&4<br>
\end{psmatrix}<br>
$ } \right| \ncline[linecolor=red,nodesep=-0.5em]{1,1}{1,3}<br>
\ncline[linecolor=red,nodesep=-0.5em]{1,1}{3,1}\\[4pt]<br>
cofactor(b_{12})~=&~B\left(\{2,3\},\{1,3\}\right)~=~\left|\raisebox{-3.0ex}{<br>
$<br>
\begin{psmatrix}[rowsep=6pt,colsep=6pt]%<br>
1 &2&3 \\<br>
0&5& 2 \\<br>
1&0&4<br>
\end{psmatrix}<br>
$ } \right| \ncline[linecolor=red,nodesep=-0.5em]{1,1}{1,3}<br>
\ncline[linecolor=red,nodesep=-0.5em]{1,2}{3,2}\\[4pt]<br>
cofactor(b_{13})~=&~B\left(\{2,3\},\{1,2\}\right)~=~\left|\raisebox{-3.0ex}{<br>
$<br>
\begin{psmatrix}[rowsep=6pt,colsep=6pt]%<br>
1 &2&3 \\<br>
0&5& 2 \\<br>
1&0&4<br>
\end{psmatrix}<br>
$ } \right| \ncline[linecolor=red,nodesep=-0.5em]{1,1}{1,3}<br>
\ncline[linecolor=red,nodesep=-0.5em]{1,3}{3,3}<br>
\end{aligned}<br>
\end{equation}<br><br>
The determinant is then<br><br>
\begin{displaymath}<br>
\begin{aligned}<br>
| B | ~=&~1*(-1)^{2}\,\, \left |<br>
\begin{matrix}5&2 \\<br>
0&4<br>
\end{matrix} \right |<br>
~+~2*(-1)^{3}\,\, \left |<br>
\begin{matrix}<br>
0&2 \\<br>
1&4\end{matrix} \, \right | ~+~ 3*(-1)^{4} ~\left |<br>
\begin{matrix}<br>
0&5 \\<br>
1&0\end{matrix}<br>
\, \right | \\[4pt]<br>
=&~(1)*(20)\, +\, (-2)*(-2)\, +\, 3*(-5)~=~9<br>
\end{aligned}<br>
\end{displaymath}<br><br>
We can also compute it using the third row of the matrix<br><br>
\begin{displaymath}<br>
\begin{aligned}<br>
B=& \left (<br>
\begin{matrix}<br>
1 & 2&3 \\<br>
0&5&2 \\<br>
1&0&4<br>
\end{matrix}<br>
\right ) \\<br>
~&~ \\<br>
| B | ~=&~1*(-1)^{4}\,\, \left |<br>
\begin{matrix}<br>
2&3 \\<br>
5&2<br>
\end{matrix}<br>
\right | ~+~0*(-1)^{5}\,\, \left |<br>
\begin{matrix}<br>
1&3 \\<br>
0&2<br>
\end{matrix}\, \right |<br>
~+~ 4*(-1)^{6} ~\left | \begin{matrix}<br>
1&2 \\<br>
0&5\end{matrix}<br>
\, \right | \\<br>
=&~(1)*(-11)\, +\, (0)*(2)\, +\, (4)*(5)~=~9<br>
\end{aligned}<br>
\end{displaymath}<br><br>
or the first column<br><br>
\begin{displaymath}<br>
\begin{aligned}<br>
B=& \left (<br>
\begin{matrix}<br>
1 & 2&3 \\<br>
0&5& 2 \\<br>
1&0&4<br>
\end{matrix}<br>
\right ) \\<br>
~&~ \\<br>
| B |~=&~1*(-1)^{2}\,\, \left |<br>
\begin{matrix}5&2 \\<br>
0&4\end{matrix} \right |<br>
~+~0*(-1)^{3}\,\, \left |<br>
\begin{matrix}2&3 \\<br>
0&4\end{matrix}\, \right |<br>
~+~ 1*(-1)^{4}<br>
~\left | \begin{matrix}2&3 \\<br>
5&2\end{matrix}\, \right | \\<br>
=&~(1)*(20)\, +\, (0)*(-8)\, +\, (1)*(-11)~=~9<br>
\end{aligned}<br>
\end{displaymath}<br><br>
\item[3]<br><br>
Consider the following 3x3 matrix<br><br>
\begin{equation}<br>
A=~ \left(<br>
\begin{matrix}<br>
-1&2&4 \\<br>
2&1&-3 \\<br>
-1&2&0<br>
\end{matrix}<br>
\right )<br>
\end{equation}<br><br>
The determinant of A computed using the cofactor expansion along the<br>
third row is given by<br><br>
\begin{displaymath}<br>
\begin{aligned}<br>
| A | ~=&~-1*(-1)^{4}\,\, \left |<br>
\begin{matrix}<br>
2&4 \\<br>
1&-3<br>
\end{matrix}<br>
\right | ~+~2*(-1)^{5}\,\, \left |<br>
\begin{matrix}<br>
-1&4 \\<br>
2&-3<br>
\end{matrix}\, \right |<br>
~+~ 0*(-1)^{6} ~\left | \begin{matrix}<br>
-1&2 \\<br>
2&1\end{matrix}<br>
\, \right | \\<br>
=&~(-1)*(-10)\, +\, (-2)*(-5)\, +\, (0)*(-5)~=~20<br>
\end{aligned}<br>
\end{displaymath}<br><br>
\end{description}<br>
\end{document}<br><br>
<br><br>
<br>
At 11:43 PM 3/6/2006, you wrote:<br>
<blockquote type=cite class=cite cite="">Arne Hallam wrote:<br>
<blockquote type=cite class=cite cite="">I did and could not find any in
the material on the web.<br>
I am not interesting in drawing a circlular line from one node to
another but a box that is not rectangular from one node to another.<br>
Maybe I am missing a part of the documentation.</blockquote><br>
<a href="http://tex.loria.fr/graph-pack/pstricks/pst-doc1.pdf" eudora="autourl">
http://tex.loria.fr/graph-pack/pstricks/pst-doc1.pdf</a><br><br>
Herbert<br><br>
<br>
_______________________________________________<br>
pstricks mailing list<br>
pstricks@tug.org<br>
<a href="http://tug.org/mailman/listinfo/pstricks" eudora="autourl">
http://tug.org/mailman/listinfo/pstricks</a><br>
</font></blockquote>
<x-sigsep><p></x-sigsep>
<font face="arial" size=2>with all thy getting get understanding --
Proverbs 4:7.<br><br>
In the wildness of speculation it has been suggested (of course more in
jest than in earnest), that Europe ought to grow its corn in America, and
devote itself solely to manufactures and commerce, as the best sort of
division of the labour of the globe -- Thomas Malthus, <i>An Essay on the
Principle of Population Book III, Chapter XII.<br><br>
</i></font><font size=3>Arne Hallam<br>
Department of Economics<br>
266 Heady Hall<br>
Iowa State University<br>
Ames, IA 50011<br><br>
Work<x-tab> </x-tab> 515-294-5861<br>
FAX:<x-tab> </x-tab> 515-294-0221<br>
Home: 515-292-8739<br><br>
<br>
</font></body>
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