Theory of Equation

Product of Two Determinants of the 𝑛² Order

570 𝑎1𝑎2⋯𝑎𝑛𝑏1𝑏2⋯𝑏𝑛⋯⋯⋯⋯𝑙1𝑙2⋯𝑙𝑛 (𝑃) 𝛼1𝛼2⋯𝛼𝑛𝛽1𝛽2⋯𝛽𝑛⋯⋯⋯⋯𝜆1𝜆2⋯𝜆𝑛 (𝑄) =𝐴1𝐴2⋯𝐴𝑛𝐵1𝐵2⋯𝐵𝑛⋯⋯⋯⋯𝐿1𝐿2⋯𝐿𝑛 (𝑆) The values of 𝐴₁, 𝐵₁, ⋯, 𝐿₁ in the first column of 𝑆 are annexed. For the second column write 𝑏's in the place of 𝑎's. For the third column write 𝑐's and so on. {𝐴₁=𝑎₁𝛼₁+𝑎₂𝛼₂+⋯+𝑎_𝑛𝛼_𝑛 𝐵₁=𝑎₁𝛽₁+𝑎₂𝛽₂+⋯+𝑎_𝑛𝛽_𝑛 ⋯ 𝐿₁=𝑎₁𝜆₁+𝑎₂𝜆₂+⋯+𝑎_𝑛𝜆_𝑛 For proof substitute the values of 𝐴₁, 𝐵₁, ⋯ in the determinant 𝑆, and then resolve 𝑆 into the sum of a number of determinants by (562), and note the determinants which vanish through having identical columns.
Rule: To form the determinant 𝑆, which is the product of two determinants 𝑃 and 𝑄. First connect by plus signs the constituents in the rows of both the determinants 𝑃 and 𝑄.
Now place the first row of 𝑃 upon each row of 𝑄 in turn, and let each two constituents as they touch become products. This is the first column of 𝑆.
Perform the same operation upon 𝑄 with the second row of 𝑃 to obtain the second column of 𝑆; and again with the third row of 𝑃 to obtain the third column of 𝑆, and so on. 571 If the number of columns, both in 𝑃 and 𝑄, be 𝑛, and the number of rows, 𝑟, and if 𝑟 be >𝑟, then the determinant 𝑆, found in the same way from 𝑃 and 𝑄, is equal to the sum of the 𝐶(𝑛,𝑟) products of pairs of determinants obtained by taking any 𝑟 columns out of 𝑃, and the corresponding 𝑟 columns out of 𝑄.
But if 𝑛 be <𝑟 the determinant 𝑆 vanishes.
For in that case, in every one of the component determinants, there will be two columns identical. 572 The product of the determinants 𝑃 and 𝑄 may be formed in four ways by changing the rows into columns in either or both 𝑃 and 𝑄. 573 Let the following system of 𝑛 equations in 𝑥₁𝑥₂⋯𝑥_𝑛 be transformed by substituting the accompanying values of variables, 𝑎₁𝑥₁+𝑎₂𝑥₂+⋯+𝑎_𝑛𝑥_𝑛=0 𝑏₁𝑥₁+𝑏₂𝑥₂+⋯+𝑏_𝑛𝑥_𝑛=0 ⋯ 𝑙₁𝑥₁+𝑙₂𝑥₂+⋯+𝑙_𝑛𝑥_𝑛=0 that is 𝑥₁=𝛼₁𝜉₁+𝛼₂𝜉₂+⋯+𝛼_𝑛𝜉_𝑛 𝑥₂=𝛽₁𝜉₁+𝛽₂𝜉₂+⋯+𝛽_𝑛𝜉_𝑛 ⋯ 𝑥_𝑛=𝜆₁𝜉₁+𝜆₂𝜉₂+⋯+𝜆_𝑛𝜉_𝑛 The eliminant of the resulting equations in 𝜉₁𝜉₂⋯𝜉_𝑛 is the determinant 𝑆 in (570), and is therefore equal to the product of the determinants 𝑃 and 𝑄. The determinant 𝑄 is then termed the modulus of transformation. 574 A Symmetrical determinant is symmetrical about the leading diagonal. If the 𝑅's form the 𝑟^th row, and the 𝐾's the 𝑘^th row; then 𝑅_𝑘=𝐾_𝑟 throughout a symmetrical determinant.
The square of a determinant is a symmetrical determinant. 575 A Reciprocal determinant has for its constituents the first minors of the original determinant, and is equal to its 𝑛−1^th power; that is, 𝐴₁⋯𝐴_𝑛 ⋯⋯⋯ ⋯⋯⋯ 𝐿₁⋯𝐿_𝑛 =𝑎₁⋯𝑎_𝑛 ⋯⋯⋯ ⋯⋯⋯ 𝑙₁⋯𝑙_𝑛 ^𝑛−1 Proof: Multiply both sides of the equation by the original determinant (555). The constituents on the left side all vanish except the diagonal of ∆'s. 576

Partial and complementary Determinants

If 𝑟 rows and the same number of columns be selected from a determinant, and if the rows be brought to the top, and the columns to the left side, without changing their order, then the elements common to the selected rows and columns form a Partial determinant of the order 𝑟, and the elements not found in any of those rows and columns form the Complementary determinant, its order being 𝑛−𝑟.

Example

Let the selected rows from the determinant (𝑎₁𝑏₂𝑐₃𝑑₄𝑐₅) be the second, third, and fifth; and the selected columns be the third, fourth, and fifth. The original and the transformed determinants will be 𝛼₁𝛼₂𝛼₃𝛼₄𝛼₅𝑏₁𝑏₂𝑏₃𝑏₄𝑏₅𝑐₁𝑐₂𝑐₃𝑐₄𝑐₅𝑑₁𝑑₂𝑑₃𝑑₄𝑑₅𝑒₁𝑒₂𝑒₃𝑒₄𝑒₅ and 𝑏₃𝑏₄𝑏₅𝑏₁𝑏₂𝑐₃𝑐₄𝑐₅𝑐₁𝑐₂𝑒₃𝑒₄𝑒₅𝑒₁𝑒₂𝛼₃𝛼₄𝛼₅𝛼₁𝛼₂𝑑₃𝑑₄𝑑₅𝑑₁𝑑₂ The partial determinant of the third order is (𝑏₃𝑐₄𝑒₅), and its complementary of the second order is (𝛼₁𝑑₂).
The complete altered determinant is plus or minus, according as the permuations of the rows and columns are of the same or of different class. In the example they are of the same class, for there have been four transpositions of rows, and six of columns. Thus (−1)¹⁰=+1 gives the sign of the altered determinant. 577

Theorem

A partial reciprocal determinant of the 𝑟^th order is equal to the product of the 𝑟−1^th power of the original determinant, and the complementary of its corresponding partial determinant.
Take the last determinant for an example. Here 𝑛=5, 𝑟=3; and by the theorem, 𝐵₃𝐵₄𝐵₅𝐶₃𝐶₄𝐶₅𝐸₃𝐸₄𝐸₅=∆²𝛼₁𝛼₂𝑑₁𝑑₂ where 𝐵, 𝐶, 𝐸 are the respective minors.

Proof

Raise the Partial Reciprocal to the original order five without altering its value, by (564); and multiply it by ∆, with the rows and columns changed to correspond as in Ex. (576); thus, by (570), we have 𝐵₃𝐵₄𝐵₅𝐵₁𝐵₂𝐶₃𝐶₄𝐶₅𝐶₁𝐶₂𝐸₃𝐸₄𝐸₅𝐸₁𝐸₂0001000001𝑏₃𝑏₄𝑏₅𝑏₁𝑏₂𝑐₃𝑐₄𝑐₅𝑐₁𝑐₂𝑒₃𝑒₄𝑒₅𝑒₁𝑒₂𝑎₃𝑎₄𝑎₅𝑎₁𝑎₂𝑑₃𝑑₄𝑑₅𝑑₁𝑑₂=∆00𝑏₁𝑏₂0∆0𝑐₁𝑐₂00∆𝑒₁𝑒₂000𝑎₁𝑎₂000𝑑₁𝑑₂=∆³𝑎₁𝑎₂𝑑₁𝑑₂ 578 The product of the differences between every pair of 𝑛 quantities 𝑎₁, 𝑎₂, ⋯, 𝑎_𝑛, (𝑎₁−𝑎₂)(𝑎₁−𝑎₃)(𝑎₁−𝑎₄)⋯(𝑎₁−𝑎_𝑛)×(𝑎₂−𝑎₃)(𝑎₂−𝑎₄)⋯(𝑎₂−𝑎_𝑛)×(𝑎₃−𝑎₄)⋯(𝑎₃−𝑎_𝑛)⋯×(𝑎_𝑛−1−𝑎_𝑛)}=111⋯1𝑎₁𝑎₂𝑎₃⋯𝑎_𝑛𝑎²₁𝑎²₂𝑎²₃⋯𝑎²_𝑛⋯⋯⋯⋯⋯𝑎^𝑛−1₁𝑎^𝑛−1₂𝑎^𝑛−1₃⋯𝑎^𝑛−1_𝑛

Proof

The determinant vanishes when any two of the quantities are equal. Therefore it is divisible by each of the factors on the left; therefore by their product. And the quotient is seen to be unity, for both sides of the equation are of the same degree; viz., 12𝑛(𝑛−1). 579 The product of the squares of the differences of the same 𝑛 quantities= 𝑠₀𝑠₁⋯𝑠_𝑛−1𝑠₁𝑠₂⋯𝑠_𝑛⋯⋯⋯⋯𝑠_𝑛−1𝑠_𝑛⋯𝑠_2𝑛−2 580 With teh same meaning for 𝑠₁, 𝑠₂, ⋯, the same determinant taken of an order 𝑟, less than 𝑛, is equal to the sum of the products of the squares of the differences of 𝑟 of the 𝑛 quantities taken in every possible way; that is, in 𝐶(𝑛, 𝑟) ways.

Example

𝑠₀𝑠₁𝑠₁𝑠₂=(𝑎₁−𝑎₂)²+(𝑎₁−𝑎₃)²+⋯≡∑(𝑎₁−𝑎₂)² 𝑠₀𝑠₁𝑠₂𝑠₁𝑠₂𝑠₃𝑠₂𝑠₃𝑠₃=≡∑(𝑎₁−𝑎₂)²(𝑎₁−𝑎₃)²(𝑎₂−𝑎₃)² The next determinant in order =∑(𝑎₁−𝑎₂)²(𝑎₁−𝑎₃)²𝑎₁−𝑎₄)²(𝑎₂−𝑎₃)²(𝑎₂−𝑎₄)²(𝑎₃−𝑎₄)² And so on until the equation (579) is reached.
Proved by substituting the values of 𝑠₁, 𝑠₂, ⋯, and resolving the determinant into its partial determinants by (571). 581 The quotient of 𝑎₀𝑥^𝑚+𝑎₁𝑥^𝑚−1+⋯+𝑎_𝑟𝑥^𝑚−𝑟+⋯𝑏₀𝑥^𝑛+𝑏₁𝑥^𝑛−1+⋯+𝑏_𝑟𝑥^𝑛−𝑟+⋯ is given by the formula 𝑞₀𝑥^𝑚−𝑛+𝑞₁𝑥^{𝑚−𝑛−1}+⋯+𝑞_𝑟𝑥^{𝑚−𝑛−𝑟}+⋯ where 𝑞_𝑟=1𝑏^𝑟+1₀𝑏₀00⋯𝑎₀𝑏₁𝑏₀0⋯𝑎₁𝑏₂𝑏₁𝑏₀⋯𝑎₂⋯⋯⋯⋯⋯𝑏_𝑟𝑏_𝑟−1𝑏_𝑟−2⋯𝑏₁𝑎_𝑟 Proved by Induction.

Sources and References

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