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% Title: Condensation of Determinants, Being a New and Brief Method for Computing their Arithmetical Values
%                                                                         %
% Author: Lewis Carroll (AKA Charles Lutwidge Dodgson)                    %
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% Release Date: September 8, 2011 [EBook #37354]                          %
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\begin{PGtext}
The Project Gutenberg EBook of Condensation of Determinants, Being a New
and Brief Method for Computing their Arithmetical Values, by Lewis Carroll (AKA Charles Lutwidge Dodgson)

This eBook is for the use of anyone anywhere at no cost and with
almost no restrictions whatsoever.  You may copy it, give it away or
re-use it under the terms of the Project Gutenberg License included
with this eBook or online at www.gutenberg.org


Title: Condensation of Determinants, Being a New and Brief Method for Computing their Arithmetical Values

Author: Lewis Carroll (AKA Charles Lutwidge Dodgson)

Release Date: September 8, 2011 [EBook #37354]

Language: English

Character set encoding: ISO-8859-1

*** START OF THIS PROJECT GUTENBERG EBOOK CONDENSATION OF DETERMINANTS ***
\end{PGtext}
\clearpage
\begin{PGtext}
Produced by Carlo Traverso, Anna Hall and the Online
Distributed Proofreading Team at http://www.pgdp.net (This
book was produced from scanned images of public domain
material from the Google Print project.)
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\vfill

\BookMark{0}{Transcriber's Note}
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%%%%%%%%%%%%%%%%%%%%%%%%%%% FRONT MATTER %%%%%%%%%%%%%%%%%%%%%%%%%%

\FrontMatter

%% -----File: 000.png---
\begin{center}
{\LARGE PROCEEDINGS} \\
\vfill
{\scriptsize OF THE }\\
\vfill
{\LARGE ROYAL SOCIETY OF LONDON.}
\vfill
\textit{From January~11, 1866, to May~23, 1867, inclusive.}
\vfill
VOL.~XV.
\vfill
LONDON: \\
PRINTED BY TAYLOR AND FRANCIS, \\
{\scriptsize RED LION COURT, FLEET STREET.} \\
{\footnotesize MDCCCLXVII.}
\end{center}
%% -----File: 001.png---
\MainMatter
\begin{quote}
IV. ``Condensation of Determinants, being a new and brief Method
for computing their arithmetical values.'' By the Rev.\ C.~L.
\textsc{Dodgson},~M.A., Student of Christ Church, Oxford. Communicated
by the Rev.\ \textsc{Bartholomew Price}, M.A., F.R.S\@.
Received May~15,~1866.
\end{quote}

If it be proposed to solve a set of $n$~simultaneous linear equations, not
being all homogeneous, involving $n$~unknowns, or to test their compatibility
when all are homogeneous, by the method of determinants, in these, as
well as in other cases of common occurrence, it is necessary to compute
the arithmetical values of one or more determinants---such, for example, as
\[
\left|\begin{array}{rrr}
1, &3, &-2\\
2, &1, & 4\\
3, &5, &-1
\end{array}\right|.
\]

Now the only method, so far as I am aware, that has been hitherto
employed for such a purpose, is that of multiplying each term of the first
row or column by the determinant of its complemental minor, and affecting
the products with the signs $+$~and~$-$ alternately, the determinants required
in the process being, in their turn, broken up in the same manner
until determinants are finally arrived at sufficiently small for mental computation.

This process, in the above instance, would run thus:---
\[
\begin{split}
\left|\begin{array}{rrr}
1, &3, &-2\\
2, &1, & 4\\
3, &5, &-1
\end{array}\right|
= 1 × \left|\begin{array}{rr}
1, & 4\\
5, &-1
\end{array}\right|
-2 × \left|\begin{array}{rr}
3, &-2\\
5, &-1
\end{array}\right|
+ 3 × \left|\begin{array}{rr}
3, &-2\\
1, & 4
\end{array}\right|\\
= -21 - 14 + 42 = 7.
\end{split}
\]

But such a process, when the block consists of $16$, $25$,~or more terms, is
so tedious that the old method of elimination is much to be preferred for
solving simultaneous equations; so that the new method, excepting for
equations containing $2$~or $3$~unknowns, is practically useless.

The new method of computation, which I now proceed to explain, and
for which ``Condensation'' appears to be an appropriate name, will be
found, I believe, to be far shorter and simpler than any hitherto employed.

In the following remarks I shall use the word ``Block'' to denote any
number of terms arranged in rows and columns, and ``interior of a block''
to denote the block which remains when the first and last rows and columns
are erased.

The process of ``Condensation'' is exhibited in the following rules, in
which the given block is supposed to consist of $n$~rows and $n$~columns:---

(1) Arrange the given block, if necessary, so that no ciphers occur in its
interior. This may be done either by transposing rows or columns, or by
adding to certain rows the several terms of other rows multiplied by
certain multipliers.

(2) Compute the determinant of every minor consisting of four adjacent
%% -----File: 002.png---
terms. These values will constitute a second block, consisting of $(n - 1)$
rows and $(n - 1)$~columns.

(3) Condense this second block in the same manner, dividing each term,
when found, by the corresponding term in the interior of the first block.

(4) Repeat this process as often as may be necessary (observing that in
condensing any block of the series, the $r$th for example, the terms so found
must be divided by the corresponding terms in the interior of the $(r - 1)$th
block), until the block is condensed to a single term, which will be the
required value.

As an instance of the foregoing rules, let us take the block
\[
\left|\begin{array}{rrrr}
-2 &-1 &-1 &-4\\
-1 &-2 &-1 &-6\\
-1 &-1 & 2 & 4\\
 2 & 1 &-3 &-8
\end{array}\right|.
\]

By rule~(2) this is condensed into
$\left|\begin{array}{rrr}
 3 &-1 & 2\\
-1 &-5 & 8\\
 1 & 1 &-4
\end{array}\right|
$;
this, again, by
rule~(3), is condensed into
$\left|\begin{array}{rr}
 8 &-2\\
-4 & 6
\end{array}\right|$; and this, by rule~(4), into~$-8$,
which is the required value.

The simplest method of working this rule appears to be to arrange the
series of blocks one under another, as here exhibited; it will then be found
very easy to pick out the divisors required in rules (3)~and~(4).
\[
\left|\begin{array}{rrrr}
-2 &-1 &-1 &-4\\
-1 &-2 &-1 &-6\\
-1 &-1 & 2 & 4\\
 2 & 1 &-3 &-8
\end{array}\right|
\]
\[
\left|\begin{array}{rrr}
 3 &-1 &2\\
-1 &-5 &8\\
 1 & 1 &-4
\end{array}\right|
\]
\[
\left|\begin{array}{rr}
 8 &-2\\
-4 & 6
\end{array}\right|
\]
\[
-8.
\]

This process cannot be continued when ciphers occur in the interior of
any one of the blocks, since infinite values would be introduced by employing
them as divisors. When they occur in the given block itself, it
may be rearranged as has been already mentioned; but this cannot be done
when they occur in any one of the derived blocks; in such a case the
given block must be rearranged as circumstances require, and the operation
commenced anew.

The best way of doing this is as follows:---

Suppose a cipher to occur in the $h$th~row and $k$th~column of one of the
derived blocks (reckoning both row and column from the \emph{nearest} corner
of the block); find the term in the $h$th~row and $k$th~column of the given
%% -----File: 003.png---
block (reckoning from the corresponding corner), and transpose rows or
columns cyclically until it is left in an outside row or column. When the
necessary alterations have been made in the derived blocks, it will be found
that the cipher now occurs in an outside row or column, and therefore
need no longer be used as a divisor.

The advantage of \emph{cyclical} transposition is, that most of the terms in the
new blocks will have been computed already, and need only be copied; in
no case will it be necessary to compute more than \emph{one} new row or column
for each block of the series.

In the following instance it will be seen that in the first series of blocks
a cipher occurs in the interior of the third. We therefore abandon the
process at that point and begin again, rearranging the given block by
transferring the top row to the bottom; and the cipher, when it occurs, is
now found in an exterior row. It will be observed that in each block of
the new series, there is only \emph{one} new row to be computed; the other rows
are simply copied from the work already done.
\[
\begin{gathered}[t]
\left|\begin{array}{rrrrr}
2 &-1 & 2 & 1 &-3\\
1 & 2 & 1 &-1 & 2\\
1 &-1 &-2 &-1 &-1\\
2 & 1 &-1 &-2 &-1\\
1 &-2 &-1 &-1 & 2\\
\end{array}\right|
\\
\left|\begin{array}{rrrr}
 5 &-5 &-3 &-1\\
-3 &-3 &-3 & 3\\
 3 & 3 & 3 &-1\\
-5 &-3 &-1 &-5\\
\end{array}\right|
\\
\left|\begin{array}{rrr}
\DPtypo{-30}{-15} & 6 &\DPtypo{-12}{12}\\
  0 & 0 &  6\\
  6 &-6 &  8\\
\end{array}\right|
\end{gathered}
\qquad
\begin{gathered}[t]
\left|\begin{array}{rrrrr}
1 & 2 & 1 &-1 & 2\\
1 &-1 &-2 &-1 &-1\\
2 & 1 &-1 &-2 &-1\\
1 &-2 &-1 &-1 & 2\\
2 &-1 & 2 & 1 &-3\\
\end{array}\right|
\\
\left|\begin{array}{rrrr}
-3 &-3 &-3 & 3\\
 3 & 3 & 3 &-1\\
-5 &-3 &-1 &-5\\
 3 &-5 & 1 & 1\\
\end{array}\right|
\\
\left|\begin{array}{rrr}
  0 & 0 & 6\\
  6 &-6 & 8\\
-17 & 8 &-4\\
\end{array}\right|
\\
\left|\begin{array}{rr}
 0 &12\\
18 &40\\
\end{array}\right|
\\
36.
\end{gathered}
\]

The fact that, whenever ciphers occur in the interior of a derived block,
it is necessary to recommence the operation, may be thought a great
obstacle to the use of this method; but I believe it will be found in practice
that, even though this should occur several times in the course of one
operation, the whole amount of labour will still be much less than that involved
in the old process of computation.
\tb

I now proceed to give a proof of the validity of this process, deduced
from a well-known theorem in determinants; and in doing so, I shall use
the word ``adjugate'' in the following sense:---if there be a square block,
and if a new block be formed, such that each of its terms is the determinant
of the complemental minor of the corresponding term of the first
block, the second block is said to be \emph{adjugate} to the first.
%% -----File: 004.png---

The theorem referred to is the following:---

``If the determinant of a block $= R$, the determinant of any minor of
the $m$th~degree of the adjugate block is the product of~$R^{m - 1}$ and the
coefficient which, in~$R$, multiplies the determinant of the corresponding
minor.''

Let us first take a block of $9$~terms,
\[
\left|\begin{array}{ccc}
a_{1,1} &a_{1,2} &a_{1,3}\\
a_{2,1} &a_{2,2} &a_{2,3}\\
a_{3,1} &a_{3,2} &a_{3,3}
\end{array}\right|
= R;
\]
and let $\alpha_{1,1}$~represent the determinant of the complemental minor of~$a_{1,1}$,
and so on.

If we ``condense'' this, by the method already given, we get the block
$\left\{\begin{array}{cc}
\alpha_{3,3} &\alpha_{3,1}\\
\alpha_{1,3} &\alpha_{1,1}
\end{array}\right\},$ and, by the theorem above cited, the determinant of this,
viz.\
\[
\left|\begin{array}{cc}
\alpha_{3,3} &\alpha_{3,1}\\
\alpha_{1,3} &\alpha_{1,1}
\end{array}\right|
= R × a_{2,2}.
\]

Hence
\[
R = \frac{\left|\begin{array}{cc}
\alpha_{3,3} &\alpha_{3,1}\\
\alpha_{1,3} &\alpha_{1,1}
\end{array}\right|}{a_{2,2}},
\]
which proves the rule.

Secondly, let us take a block of $16$~terms:
\[
\left|\begin{array}{ccc}
a_{1,1}&\dots &a_{1,4}\\
\vdots &      &\vdots \\
a_{4,1}&\dots &a_{4,4}
\end{array}\right|
= R.
\]
If we ``condense'' this, we get a block of $9$~terms; let us denote it by
\[
\left\{\begin{array}{ccc}
b_{1,1} &\dots &b_{1,3}\\
\vdots  &      &\vdots \\
b_{3,1} &\dots &b_{3,3}
\end{array}\right\},
\text{ in which }
b_{1,1} =
\left|\begin{array}{cc}
a_{1,1} &a_{1,2}\\
a_{2,1} &a_{2,2}
\end{array}\right|, \text{ \&c.}
\]

If we ``condense'' this block again, we get a block of $4$~terms, each
of which, by the preceding paragraph, is the determinant of $9$~terms
of the original block; that is to say, we get the block
$\left\{\begin{array}{cc}
\alpha_{4,4} &\alpha_{4,1}\\
\alpha_{1,4} &\alpha_{1,1}
\end{array}\right\}$;
but, by the theorem already quoted,
$\left|\begin{array}{cc}
\alpha_{4,4} &\alpha_{4,1}\\
\alpha_{1,4} &\alpha_{1,1}
\end{array}\right| = R × b_{2,2}$; therefore
$R = \dfrac{\left|\begin{array}{cc}
\alpha_{4,4} &\alpha_{4,1}\\
\alpha_{1,4} &\alpha_{1,1}
\end{array}\right|}{b_{2,2}}$; that is, $R$~may be obtained by ``condensing'' the block
$\left\{\begin{array}{cc}
\alpha_{4,4} &\alpha_{4,1}\\
\alpha_{1,4} &\alpha_{1,1}
\end{array}\right\}$.

This proves the rule for a block of $16$~terms; and similar proofs might
be given for larger blocks.
\tb

I shall conclude by showing how this process may be applied to the
solution of simultaneous linear equations.
%% -----File: 005.png---

If we take a block consisting of $n$~rows and $(n + 1)$~columns, and ``condense''
it, we reduce it at last to $2$~terms, the first of which is the determinant
of the first $n$~columns, the other of the last $n$~columns.

Hence, if we take the $n$~simultaneous equations,
\begin{alignat*}{3}
a_{1,1}x_{1} +{}& a_{1,2}x_{2} + \cdots\,&\cdots + a_{1,n}x_{n} &+ a_{1,n + 1} = 0,\\
\multispan{4}{\dotfill} \\
a_{n,1}x_{1} +{}&\multispan{2}{\dotfill}&+ a_{n,n + 1} = 0;
\end{alignat*}
and if we condense the whole block of coefficients and constants, viz.\
\[
\left\{\begin{array}{ccc}
a_{1,1} &\dots &a_{1,n + 1}\\
\vdots  &      &\vdots \\
a_{n,1} &\dots &a_{n,n + 1}
\end{array}\right\},
\]
we reduce it at last to $2$~terms: let us denote them by $S$,~$T$, so that
\[
S = \left|\begin{array}{ccc}
a_{1,1} &\dots &a_{1,n}\\
\vdots  &      &\vdots \\
a_{n,1} &\dots &a_{n,n}
\end{array}\right|,
\text{ and }
T = \left|\begin{array}{ccc}
a_{\DPtypo{1,1}{1,2}} &\dots &a_{1,n+1}\\
\vdots  &      &\vdots \\
a_{n,2} &\dots &a_{n,n+1}
\end{array}\right|.
\]

Now we know that $x_{1} = (-)^{n} \dfrac{T}{S}$, which may be written in the form
$(-)^{n}S · x_{1} = T$.

Hence the $2$~terms obtained by the process of condensation may be
converted into an equation for~$x_{1}$, by multiplying the first of them by~$x_{1}$,
affected with $+$~or~$-$, according as $n$~is even or odd. The latter part of
the rule may be simply expressed thus:---``place the signs $+$~and~$-$
alternately over the several columns, beginning with the last, and the sign
which occurs over the column containing~$x_{1}$ is the sign with which $x_{1}$ is to
be affected.''

When the value of~$x_{1}$ has been thus found, it may be substituted in the
first $(n - 1)$~equations, and the same operation repeated on the new block,
which will now consist of $(n - 1)$~rows and $n$~columns. But in calculating
the second series of blocks, it will be found that most of the work has been
already done; in fact, of the $2$~determinants required in the new block, one
has been already computed correctly, and the other so nearly so that it
only requires the \emph{last} column in each of the derived blocks to be corrected.

In the \hyperlink{example}{example} given opposite, after writing $+$~and~$-$ alternately over
the columns, beginning with the last, we first condense the whole block, and
thus obtain the $2$~terms $36$~and~$-72$. Observing that the $x$-column has
the sign~$-$ placed over it, we multiply the~$36$ by~$-x$, and so form the
equation $-36x = -72$, which gives~$x = 2$.

Hence the $x$-terms in the first four equations become respectively
$2$,~$2$,~$4$, and~$2$; adding these values to the constant terms in the same equations,
we obtain a block of which we need only write down the last two
columns, viz.\
\[
\left|\begin{array}{rr}
 2 & 4\\
-1 &-2\\
-1 &-2\\
 2 & 6
\end{array}\right|.
\]
%% -----File: 006.png---

We then condense these into the column~$\left|\begin{array}{r}
0\\
0\\
\DPtypo{2}{-2}
\end{array}\right|$,
and, supplying from
the second block of the first series the column~$\left|\begin{array}{r}
 3\\
-1\\
-5
\end{array}\right|$, we obtain
$\left|\begin{array}{rr}
 3 &0\\
-1 &0\\
-5 &\DPtypo{2}{-2}
\end{array}\right|$ as the last two columns of the \emph{second} block of the new series;
and proceeding thus we ultimately obtain the two terms $12$,~$12$. Observing
that the $y$-column has the sign~$+$ placed over it, we multiply the first~$12$
by~$+y$, and so form the equation $12y = 12$, which gives $y = 1$. The
values of $z$,~$u$, and~$v$ are similarly found.

% In Dodgson's two examples, the coefficient matrix of the original
% system is at upper left. Each column of the calculation, which Dodgson
% calls a "series", solves the system for one variable. Successive matrices
% in a column are obtained by condensations, reducing the size of the matrix
% until the value of one variable can be read off, with the sign adjusted
% according to the +/- over the corresponding column in the original system.
%
% Next, the solved value of the variable is substituted into the system, the
% last equation/coefficient row is dropped, the constants are "updated", and
% the process begins anew. Only condensations involving the last column need
% to be recomputed, so the second and subsequent series involve only 2 x n
% matrices. In the matrix at the top of each series, the two columns are
% (i) the penultimate column of the preceding series and (ii) the updated
% values of the constants.
It will be seen that when once the given block has been successfully
condensed, and the value of the first unknown obtained, there is no further
danger of the operation being interrupted by the occurrence of ciphers.
\hypertarget{example}
\[
\begin{array}{ccccccc}
 - &+   &-   &+   &-   &+  &   \\
 x &+2y &+ z &- u &+2v &+2 &= 0\\
 x &- y &-2z &- u &- v &-4 &= 0\\
2x &+ y &- z &-2u &- v &-6 &= 0\\
 x &-2y &- z &- u &+2v &+4 &= 0\\
2x &- y &+2z &+ u &-3v &-8 &= 0
\end{array}
\]
\[
\SquashArrays
% First series of blocks
\begin{gathered}
\left|\begin{array}{rrrrrr}
1 & 2 & 1 &-1 & 2 & 2\\
1 &-1 &-2 &-1 &-1 &-4\\
2 & 1 &-1 &-2 &-1 &-6\\
1 &-2 &-1 &-1 & 2 & 4\\
2 &-1 & 2 & 1 &-3 &-8
\end{array}\right|
\\
\left|\begin{array}{rrrrr}
-3 &-3 &-3 & 3 &-6\\
 3 & 3 & 3 &-1 & 2\\
-5 &-3 &-1 &-5 & 8\\
 3 &-5 & 1 & 1 &-4
\end{array}\right|
\\
\left|\begin{array}{rrrr}
  0 & 0 & 6 & 0\\
  6 &-6 & 8 &-2\\
-17 & 8 &-4 & 6
\end{array}\right|
\\
\left|\begin{array}{rrr}
 0 &12 &12\\
18 &40 &-8
\end{array}\right|
\\
\left|\begin{array}{rr} 36 & -72 \end{array}\right| \\
\DPtypo{\because}{\therefore} -36x = -72 \\
x = 2
\end{gathered}
\;
% Second series
\begin{gathered}
\left|\begin{array}{rr}
 2 & 4\\
-1 &-2\\
-1 &-2\\
 2 & 6
\end{array}\right|
\\
\left|\begin{array}{rr}
 3 & 0\\
-1 & 0\\
-5 &-2
\end{array}\right|
\\
\left|\begin{array}{rr}
6 & 0\\
8 &-2
\end{array}\right|
\\
\left|\begin{array}{rr} 12 & 12 \end{array}\right| \\
\therefore 12y = 12 \\
y = 1
\\
\Tall{5}
\end{gathered}
\;
% Third series
\begin{gathered}
\left|\begin{array}{rr}
 2 & 6\\
-1 &-3\\
-1 &-1
\end{array}\right|
\\
\left|\begin{array}{rr}
 3 & 0\\
-1 &-2
\end{array}\right|
\\
\left|\begin{array}{rr} 6 & 6\end{array}\right| \\
\therefore -6z = 6 \\
z = -1
\\
\Tall{5}
\\
\Tall{4}
\end{gathered}
\;
% Fourth series
\begin{gathered}
\left|\begin{array}{rr}
 2 & 5\\
-1 &-1
\end{array}\right|
\\
\left|\begin{array}{rr} 3 & 3 \end{array}\right| \\
\therefore 3u = 3 \\
u = 1
\\
\Tall{5}
\\
\Tall{4}
\\
\Tall{3}
\end{gathered}
\;
% Fifth series
\begin{gathered}
\left|\begin{array}{rr} 2 & 4 \end{array}\right| \\
\therefore -2v = 4 \\
v = -2
\\
\Tall{5}
\\
\Tall{4}
\\
\Tall{3}
\\
\Tall{2}
\end{gathered}
\]

\tb
%
\[
\begin{array}{ccccc}
-  &+   &-   &+   &   \\
5x &+2y &-3z &+ 3 &= 0\\
3x &- y &-2z &+ 7 &= 0\\
2x &+3y &+ z &-12 &= 0
\end{array}
\]
\[
% First series
\begin{gathered}
\left|\begin{array}{rrrr}
5 & 2 &-3 &  3\\
3 &-1 &-2 &  7\\
2 & 3 & 1 &-12
\end{array}\right|
\\
\left|\begin{array}{rrr}
-11 &-7 &-15\\
 11 & 5 & 17
\end{array}\right|
\\
\left|\begin{array}{cc} -22 & 22 \end{array}\right| \\
\therefore 22x =22 \\
x = 1
\end{gathered}
\quad
\begin{gathered}
\left|\begin{array}{rr}
-3 & 8\\
-2 &10
\end{array}\right|
\\
\left|\begin{array}{cc} -7 & -14 \end{array}\right| \\
\therefore -7y = -14 \\
y = 2
\\
\Tall{3}
\end{gathered}
\quad
\begin{gathered}
\left|\begin{array}{rr} -3 & 12 \end{array}\right| \\
\therefore 3z = 12 \\
z = 4
\\
\Tall{3}
\\
\Tall{2}
\end{gathered}
\]

The Society then adjourned over the Whitsuntide Recess to Thursday,
May~31.

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