Talk:PlanetPhysics/Determinant

Original TeX Content from PlanetPhysics Archive

%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: determinant %%% Primary Category Code: 02. %%% Filename: Determinant.tex %%% Version: 2 %%% Owner: bloftin %%% Author(s): bloftin %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

\setlength{\topmargin}{0.00in} \setlength{\headsep}{0.00in} \setlength{\headheight}{0.00in} \setlength{\evensidemargin}{0.00in} \setlength{\oddsidemargin}{0.00in} \setlength{\textwidth}{6.5in} \setlength{\textheight}{9.00in} \setlength{\voffset}{0.00in} \setlength{\hoffset}{0.00in} \setlength{\marginparwidth}{0.00in} \setlength{\marginparsep}{0.00in} \setlength{\parindent}{0.00in} \setlength{\parskip}{0.15in}

\usepackage{html}

% this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners.

% almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts}

% used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic}

% there are many more packages, add them here as you need them

% define commands here

\begin{document}

In attempting to solve a linear \htmladdnormallink{system}{http://planetphysics.us/encyclopedia/SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence.html} of equations for $x^1$, $x^2$ and $x^3$

\begin{equation} \begin{array}{ccc} a_1x^1 + b_1x^2 +c_1x^3 & = & d_1 \\ a_2x^1 + b_2x^2 + c_2x^3 & = & d_2 \\ a_3x^1 + b_3x^2 + c_3x^3 & = & d_3 \end{array} \end{equation}

one is led in a very natural way to consider the \htmladdnormallink{square}{http://planetphysics.us/encyclopedia/PiecewiseLinear.html} array

\begin{equation} \left | \begin{array}{ccc} a_1^1 & a_2^1 & a_3^1 \\ a_1^2 & a_2^2 & a_3^2 \\ a_1^3 & a_2^3 & a_3^3 \end{array} \right | \end{equation}

We have written $a_{ij} = a_j^i, i, j = 1, 2, 3$. The solution of (1) requires that we attach a numerical value to the \htmladdnormallink{matrix}{http://planetphysics.us/encyclopedia/Matrix.html} of elements (2). We do this in the following way: We attach $3^3 = 27$ numerical values to a set of $\epsilon^{ijk}, i, j, k = 1, 2, 3$. If at least two of the superscripts in $\epsilon^{ijk}$ are the same, the value of $\epsilon^{ijk}$ is zero. Thus $\epsilon^{223} = \epsilon^{131} = \epsilon^{333} = 0$, etc. If the $i$, $j$, $k$ are all different, the value of $\epsilon^{ijk}$ is to be $+1$ or $-1$ according to whether it takes an even or odd number of permutations to rearrange the $ijk$ into the natural order $123$. Let us lood at $\epsilon^{321}$ and hence at the arrangement $321$. Permuting the integers $2$ and $3$ permutes $321$ into $231$, then permuting $3$ and $1$ permutes $231$ into $213$, and finally $213$ permutes into $123$ if we interchange the integers $2$ and $1$. Three (an odd number) permutations were required to permute $321$ into $123$. Thus $\epsilon^{123} = -1$. We have

$$\epsilon^{123} = \epsilon^{312} = \epsilon^{231} = +1$$ $$\epsilon^{213} = \epsilon^{321} = \epsilon^{132} = -1$$

We now define

\begin{equation} \left | \begin{array}{ccc} a_1^1 & a_2^1 & a_3^1 \\ a_1^2 & a_2^2 & a_3^2 \\ a_1^3 & a_2^3 & a_3^3 \end{array} \right | \equiv \epsilon^{ijk}a_i^1 a_j^2 a_k^3 \end{equation}

The letters $i$, $j$, $k$ are indices of summation. Equation (3) defines the determinant of the \htmladdnormallink{square matrix}{http://planetphysics.us/encyclopedia/Matrix.html} of elements (2) Its numerical value is given by the right-hand side of (3). It consists, in general, of $3! = 3 \cdot 2 \cdot 1 = 6$ terms, each term a product of three elements, one element from each row and column of (3). Expand (3) to get

$$\epsilon^{ijk}a_i^1 a_j^2 a_k^3 = ( a_1^1 a_2^2 a_3^3 + a_3^1 a_1^2 a_2^3 + a_2^1 a_3^2 a_1^3) - (a_2^1 a_1^2 a_3^3 + a_3^1 a_2^2a_1^3 + a_1^1 a_3^2 a_2^3)$$

Only $3! = 6$ terms occur in the exapansion of (3) since there are $3!$ permutations of $123$. All other values of $\epsilon_{ijk}$ are zero.

We can define $\epsilon_{ijk}$ in exactly the same manner in which the $\epsilon^{ijk}$ were defined. We leave it to the reader to show that

$$ \epsilon^{ijk}a_i^1 a_j^2 a_k^3 = \epsilon_{ijk}a_1^i a_2^j a_3^k $$

The generalization of second and third order determinants (the order of a determinant is the number of rows or columns of the determinant) to the $n$th order determinants is simple. We define the $\epsilon^{i_1i_2 \cdots i_n}$ to have the following numerical values: $\epsilon^{i_1i_2 \cdots i_n} = 0$ if at least two of the superscripts are the same. The values of the superscripts range from $1$ to $n$. If the $i_1, i_2, \dots, i_n$ are distinct, the value of $\epsilon^{i_1i_2 \cdots i_n}$ is to be $+1$ or $-1$ depending on whether an even or odd number of permutations is required to rearrange $i_1, i_2, \dots, i_n$ into the natural order $123 \cdots n$. The numerical value (determinant) of the square array of elements $\left \| a_j^i \right \|$, $i, j = 1,2,\dots,n$, is defined as

$$\left | a_j^i \right | = \left | \begin{array}{cccc} a_1^1 & a_2^1 & \dots & a_n^1 \\ a_1^2 & a_2^2 & \dots & a_n^2 \\ \dots & \dots & \dots & \dots \\ a_1^n & a_2^n & \dots & a_n^n \end{array} \right | $$

$$\left | a_j^i \right | = \epsilon^{i_1i_2 \cdots i_n}a_{i_1}^1 a_{i_2}^2 \cdots a_{i_n}^n$$

\begin{equation} \left | a_j^i \right | = \epsilon_{i_1i_2 \cdots i_n}a_1^{i_1} a_2^{i_2} \cdots a_n^{i_n} \end{equation}

where the $\epsilon_{i_1i_2 \cdots i_n}$ are defined in precisely the same manner in which the $\epsilon^{i_1i_2 \cdots i_n}$ are defined. In general, (4) consists of $n!$ terms, each term a product of elements, one element fom each row and column of $\left | a_j^i \right |$.

To facilitate writing, we shall deal with third order determinants, but it will be obvious to the reader that any \htmladdnormallink{theorem}{http://planetphysics.us/encyclopedia/Formula.html} derived for third order determinants will apply to determinants of any finite order. Le us consider

\begin{equation} \Delta = \left | \begin{array}{ccc} a_1^1 & a_2^1 & a_3^1 \\ a_1^2 & a_2^2 & a_3^2 \\ a_1^3 & a_2^3 & a_3^3 \end{array} \right | = \epsilon^{ijk}a_i^1 a_j^2 a_k^3 = \epsilon_{ijk}a_1^i a_2^j a_3^k \end{equation}

We can obtain a new third order determinant by interchanging the first and third row of $\Delta$. This yields

\begin{equation} \Delta^{\prime} = \left | \begin{array}{ccc} a_1^3 & a_2^3 & a_3^3 \\ a_1^2 & a_2^2 & a_3^2 \\ a_1^1 & a_2^1 & a_3^1 \end{array} \right | = \epsilon^{ijk}a_i^3 a_j^2 a_k^1 \end{equation}

But $\epsilon^{ijk}a_i^3 a_j^2 a_k^1 = \epsilon^{ijk}a_k^1 a_j^2 a_i^1 = \epsilon^{kji}a_i^1 a_j^2 a_k^3$, since $i$ amd $k$ are dummy indices. We see that every term of (6) is the same as every term in (5) with the exception that $\epsilon^{ijk}$ is replaced by $\epsilon^{kji}$. Since $\epsilon^{ijk} = - \epsilon^{kji}$, we conclude that $\Delta = -\Delta^{\prime}$. we thus obtain the following theorems:

\textbf{THEOREM 1.1}. Interchanging two rows (or columns) of a determinant changes the sign of the determinant.

\textbf{THEOREM 1.1}. If two rows (or columns) of a determinant are the same, the value of the determinant is zero.

We note that

\begin{equation} \Delta^{\prime \prime} = \left | \begin{array}{ccc} l a_1^1 & l a_2^1 & l a_3^1 \\ a_1^2 & a_2^2 & a_3^2 \\ a_1^3 & a_2^3 & a_3^3 \end{array} \right | = \epsilon^{ijk}(la_i^1) a_j^2 a_k^3 = l \Delta \end{equation}

\textbf{THEOREM 1.3}. If a row (or column) of a determinant is multiplied by a factor $l$, the value of the determinant is thereby multiplied by $l$.

Let us now investigate the determinant

$$ \Delta^{\prime \prime \prime} = \left | \begin{array}{ccc} a_1^1 + l a_1^3 & a_2^1 + l a_2^3 & a_3^l + l a_3^3 \\ a_1^2 & a_2^2 & a_3^2 \\ a_1^3 & a_2^3 & a_3^3 \end{array} \right | $$

$$ \Delta^{\prime \prime \prime} = \epsilon^{ijk}(a_i^1 + l a_i^3) a_j^2 a_k^3 $$

$$ \Delta^{\prime \prime \prime} = \epsilon^{ijk} a_i^1 a_j^2 a_k^3 + l \epsilon^{ijk} a_i^3 a_j^2 a_k^3$$

\begin{equation} \Delta^{\prime \prime \prime} = \Delta \end{equation}

since $\epsilon^{ijk} a_i^3 a_j^2 a_k^3 = 0$ from Theorem (1.2). Hence we obtain the following theorem:

\textbf{THEOREM 1.4}. The value of a determinant remains unchanged if to the elements of any row (or column) is added a \htmladdnormallink{scalar}{http://planetphysics.us/encyclopedia/Vectors.html} multiple of the corresponding elements of another row (or column).

The theorems derived above are very useful in evaluating a determinant.

\section{References}

[1] Lass, Harry. "Elements of pure and applied mathematics" New York: McGraw-Hill Companies, 1957.

This entry is a derivative of the Public domain work [1].

\end{document}