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Theory of Equation
 Determinants
 General Theory
  Rule
   Example
 Analysis of a Determinant
  Rule
  Example
 Synthesis of a determinant
  Example
 Sources and References

Theory of Equation

Determinants

554 Definitions: The determinant π‘Ž1π‘Ž2𝑏1𝑏2 is equivalent to π‘Ž1𝑏2βˆ’π‘Ž2𝑏1, and is called a determinant of the second order.
A determinant of the third order is π‘Ž1π‘Ž2π‘Ž3𝑏1𝑏2𝑏3𝑐1𝑐2𝑐3β‰‘π‘Ž1(𝑏2𝑐3βˆ’π‘3𝑐2)+π‘Ž2(𝑏3𝑐1βˆ’π‘1𝑐3)+π‘Ž3(𝑏1𝑐2βˆ’π‘2𝑐1) Another notation is βˆ‘Β±π‘Ž1𝑏2𝑐3, or simply (π‘Ž1𝑏2𝑐3).
The letters are named constituents, and the terms are called elements. The determinant is composed of all the elements obtained by permutations of the suffixes 1, 2, 3.
The coefficients of the constituents are determinants of the next lower order, and are termed minors of the original determinant. Thus, the first determinant above is the minor of 𝑐3 in the second determinant. It is denoted by 𝐢3. So the minor of π‘Ž1 is denoted by 𝐴1, and so on. 555 A determinant of the 𝑛th order may be written in either of the forms below π‘Ž1π‘Ž2β‹―π‘Žπ‘Ÿβ‹―π‘Žπ‘›π‘1𝑏2β‹―π‘π‘Ÿβ‹―π‘π‘›β‹―β‹―β‹―β‹―β‹―β‹―π‘™1𝑙2β‹―π‘™π‘Ÿβ‹―π‘™π‘› or π‘Ž11π‘Ž12β‹―π‘Ž1π‘Ÿβ‹―π‘Ž1π‘›π‘Ž21π‘Ž22β‹―π‘Ž2π‘Ÿβ‹―π‘Ž2π‘›β‹―β‹―β‹―β‹―β‹―β‹―π‘Žπ‘›1π‘Žπ‘›2β‹―π‘Žπ‘›π‘Ÿβ‹―π‘Žπ‘›π‘› In the latter, or double suffix notation, the first suffix indicates the row, and the second the column. The former notation will be adopted in these pages.
A Composite determinant is one in which the number of columns exceeds the number of rows, and it is written as in the annexed example. π‘Ž1π‘Ž2π‘Ž3𝑏1𝑏2𝑏3 Its value is the sum of all the determinants obtaied by taking a number of rows in every possible way.
A Simple determinant has single terms for its constituents.
A Compound determinant has more than one term in some or all of its constituents. See (570) for an example.
For the definitions of Symmetrical, Reciprocal, artial, and Complementary determinants; see (574), (575), and (576).

General Theory

556 The number of constituents is 𝑛2. The number of elements in the complete determinant is |n. 557 The first or leading element is π‘Ž1𝑏2𝑐3⋯𝑙𝑛. Any element may be derived from the first by permutation of the suffixes.
The sign of an element is + or βˆ’ according as it has been obtained from the diagonal element by an even or odd number of permutations of the suffixes.
Hence the following rule for determining the sign of an element.

Rule

Take the suffixes in order, and put them back to their places in the first element. Let π‘š be the whole number of places passed over; then (βˆ’1)π‘š will give the sign required.

Example

To find the sign of the element π‘Ž4𝑏3𝑐5𝑑1𝑒2 of the determinant (π‘Ž1𝑏2𝑐3𝑑4𝑒5).  π‘Ž4𝑏3𝑐5𝑑1𝑒2 Move the suffix 1, three places β‹―14352 Move the suffix 2, three places β‹―12435 Move the suffix 3, one place β‹―12345 In all, seven places; therefore (βˆ’1)7=βˆ’1 gives the sign required. 558 If two suffixes in any element be transposed, the sign of the element is changed.
Half of the elements are plus, and half are minus. 559 The elements are not altered by changing the rows into columns.
If two rows or columns are transposed, the sign of the determinant is changed. Because each element changes its sign.
If two rows or columns are identical, the determinant vanishes. 560 If all the constituents but one in a row or column vanish, the determinant becomes the product of that constituent and a determinant of the next lower order. 561 A cyclical interchange is effected by π‘›βˆ’1 successive transpositions of adjacent rows or columns, until the top row has been brought to the bottom, or the left column to the right side. Hence
A cyclical interchange changes the sign of a determinant of an even order only.
The π‘Ÿth row may be brought to the top by π‘Ÿβˆ’1 cyclical interchanges. 562 If each constituent in a row or column be multiplied by the same factor, the determinant becomes multiplied by it.
If each constituent of a row or column is the sum of π‘š terms, the compound determinant becomes the sum of π‘š simple determinants of the same order.
Also, if every constituent of the determinant consists of π‘š terms, the compound determinant is resolvable into the sum of π‘š2 simple determinants. 563 To express the minor of the π‘Ÿth row and π‘˜th column as a determinant of the π‘›βˆ’1th order.
Put all the constituents in the π‘Ÿth row and π‘˜th column equal to 0, and then make π‘Ÿβˆ’1 cyclical interchanges in the rows and π‘˜βˆ’1 in the columns, and multiply by (βˆ’1)(π‘Ÿ+π‘˜)(π‘›βˆ’1). [∡ =(βˆ’1)(π‘Ÿβˆ’1+π‘˜βˆ’1)(π‘›βˆ’1). 564 To express a determinant as a determinant of a higher order.
continue the diagonal with constituents of "ones", and fill up with zeros on one side, and with any quantities whatever (𝛼, 𝛽, 𝛾, β‹―) on the other. 10000 𝛼1000 π›½πœ–π‘Žβ„Žπ‘” π›Ύπœβ„Žπ‘π‘“ π›Ώπœ‚π‘”π‘“π‘ 565 The sum of the products of each constituent of a column by the corresponding minor in another given column is zero. And the same is true if we read 'row' instead of 'column'. Thus, referring to the determinant in (555), Taking the 𝑝th and π‘žth columns, π‘Žπ‘π΄π‘ž+π‘π‘π΅π‘ž+β‹―+π‘™π‘πΏπ‘ž=0 Taking the π‘Ž and 𝑐 rows, π‘Ž1𝐢1+π‘Ž2𝐢2+β‹―+π‘Žn𝐢n=0 For in each case we have a determinant with two columns identical. 566 In any row or column the sum of the products of each constituent by its minor is the determinant itself. That is, Taking the 𝑝th column, π‘Žπ‘π΄π‘+𝑏𝑝𝐡𝑝+β‹―+𝑙𝑝𝐿𝑝=βˆ† Or taking the 𝑐 row, 𝑐1𝐢1+𝑐2𝐢2+β‹―+𝑐n𝐢n=βˆ† 567 The last equation may be expressed by βˆ‘π‘π‘πΆπ‘=βˆ†. Also, if (π‘Žπ‘πΆπ‘ž) express the determinant π‘Žπ‘π‘Žπ‘žπ‘π‘π‘π‘ž; then βˆ‘(π‘Žπ‘π‘π‘ž) will represent the sum of all the determinants of the second order which can be formed by taking any two columns out of the π‘Ž and 𝑐 rows. The minor of (π‘Žπ‘, π‘π‘ž) may be written (𝐴𝑝, πΆπ‘ž), and signifies the determinant obtained by suppressing the two rows and two columns of π‘Žπ‘ and π‘π‘ž. Thus βˆ†=βˆ‘(π‘Žπ‘, π‘π‘ž)(𝐴𝑝, πΆπ‘ž). And a similar notation when three or more rows and columns are selected. 568

Analysis of a Determinant

Rule

To resolve into its elements a determinant of the 𝑛th order. Express it as the sum of 𝑛 determinants of the (π‘›βˆ’1)th order by (560), and repeat the process with each of the new determinants.

Example

π‘Ž1π‘Ž2π‘Ž3π‘Ž4 𝑏1𝑏2𝑏3𝑏4 𝑐1𝑐2𝑐3𝑐4 𝑑1𝑑2𝑑3𝑑4 =π‘Ž1 𝑏2𝑏3𝑏4 𝑐2𝑐3𝑐4 𝑑2𝑑3𝑑4 βˆ’π‘Ž2 𝑏3𝑏4𝑏1 𝑐3𝑐4𝑐1 𝑑3𝑑4𝑑1 +π‘Ž3 𝑏4𝑏1𝑏2 𝑐4𝑐1𝑐2 𝑑4𝑑1𝑑2 βˆ’π‘Ž4 𝑏1𝑏2𝑏3 𝑐1𝑐2𝑐3 𝑑1𝑑2𝑑3 Again, 𝑏1𝑏2𝑏3 𝑐1𝑐2𝑐3 𝑑1𝑑2𝑑3 =𝑏1 𝑐2𝑐3 𝑑2𝑑3 +𝑏2 𝑐3𝑐1 𝑑3𝑑1 +𝑏3 𝑐1𝑐2 𝑑1𝑑2 and so on. In the first series the derminants have alternately plus and minus signs, by the rule for cyclical interchanges (561), the order being even. 569

Synthesis of a determinant

The process is facilitated by making us of two evident rules. Those constituents which belong to the row and column of a given constituent π‘Ž, will be designated "π‘Ž's constituents". Also , two pairs of constituents such as π‘Žπ‘, π‘π‘ž and π‘Žπ‘ž, 𝑐𝑝, forming the corners of a rectangle, will be said to be "conjugate: to each other. Rule I. No constituent will be found in the same term with one of its own constituents. Rule II> The conjugates of any two constituents π‘Ž and 𝑏 will be common to π‘Ž's and 𝑏's constituents.

Example

To write the following terms in the form of a determinant: π‘Žπ‘π‘π‘‘+𝑏𝑓𝑔𝑙+𝑓2β„Ž2+1𝑒𝑑𝑓+π‘π‘”β„Žπ‘+1π‘Žβ„Žπ‘Ÿ+π‘’π‘™π‘π‘Ÿβˆ’π‘“β„Žπ‘π‘Ÿβˆ’π‘Žπ‘π‘™π‘Ÿβˆ’π‘Žπ‘β„Ž2βˆ’1π‘“β„Žπ‘”βˆ’π‘π‘‘π‘“2βˆ’π‘’π‘“β„Žπ‘™βˆ’π‘π‘’π‘‘π‘ The determinant will be of the fourth order; and since every term must conatin four constituents, and the constituent 1 is supplied to make up the number in some of the terms. select any term, as π‘Žπ‘π‘π‘‘. for the leading diagonal.
Now apply Rule I., π‘Ž is not found with 𝑒, 𝑓, 𝑓, 𝑔, 𝑝, 0, β‹―1 𝑏 is not found with 𝑒, β„Ž, β„Ž, 𝑝, 1, 0, β‹―2 𝑐 is not found with 𝑓, 𝑓, 𝑙, π‘Ÿ, 1, 0, β‹―3 𝑑 is not found with 𝑔, β„Ž, β„Ž, 𝑙, π‘Ÿ, 0, β‹―4 Each constituent has 2(π‘›βˆ’1), that is, 6 constituents belonging to it, since 𝑛=4. Assuming, therefore, that the above letters are the constituents of π‘Ž, 𝑏, 𝑐, and 𝑑, and that there are no more, we supply a sixth zero constituent in each case.
Now apply Rulle II. The constituents common to π‘Ž and 𝑏 are 𝑒, 𝑝; to π‘Ž and 𝑐 are 𝑓, 𝑓; to π‘Ž and 𝑑 are 𝑔, 0; to 𝑏 and 𝑐 are 1, 0 to 𝑏 and d are β„Ž, β„Ž, 0; to 𝑐 and d are 𝑙, π‘Ÿ, 0; The determinant may now be formed. The diagonal being π‘Žπ‘π‘π‘‘; place 𝑒, 𝑝, the conjugates of π‘Ž and 𝑏, either as in the diagram or transposed.
Then 𝑓, and 𝑓, the conjugates of π‘Ž and 𝑐, may be written.
1 and 0, the conjugates of 𝑏 and 𝑐, must be placed as indicated, because 1 is one of 𝑝's constituents, since it is not found in any term with 𝑝, and must therefore be in the second row.
Similarly the places of 𝑔 and 0, and 𝑙, and π‘Ÿ, are assigned.
In the case of 𝑏 and d we have β„Ž, β„Ž, 0 from which to choose the two conjugates, but we see that 0 is not one of them because that would assign two zero constituents to 𝑏, whereas 𝑏 has but one, which is already placed.
By similar reasoning the ambiguity in selecting the conjugates 𝑙, π‘Ÿ is removed.
π‘Žπ‘’π‘“π‘” 𝑝𝑏1β„Ž 𝑓0π‘π‘Ÿ 0β„Žπ‘™d The foregoing method is rigid in the case of a complete determinant having different constituents. It becomes uncertain when the zero constituents increase in number, and when several constituents are identical. But even then, in the majority of cases, it will soon afford a clue to the required arrangement.

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive

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References

  1. B. Joseph, 1978, University Mathematics: A Textbook for Students of Science & Engineering
  2. Wheatstone, C., 1854, On the Formation of Powers from Arithmetical Progressions
  3. Stroud, K.A., 2001, Engineering Mathematics
  4. Coolidge, J.L., 1949, The Story of The Binomial Theorem
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