Theory of Equation

Elimination

582

Solution of 𝑛 Linear Equations in 𝑛 Variables

The equations and the values of the variables are arranged below: 𝑎₁𝑥₁+𝑎₂𝑥₂+⋯+𝑎_𝑛𝑥_𝑛=𝜉₁ 𝑏₁𝑥₁+𝑏₂𝑥₂+⋯+𝑏_𝑛𝑥_𝑛=𝜉₂ ⋯ 𝑙₁𝑥₁+𝑙₂𝑥₂+⋯+𝑙_𝑛𝑥_𝑛=𝜉_𝑛 and 𝑥₁∆=𝐴₁𝜉₁+𝐵₁𝜉₂+⋯+𝐿₁𝜉_𝑛 𝑥₂∆=𝐴₂𝜉₁+𝐵₂𝜉₂+⋯+𝐿₂𝜉_𝑛 ⋯ 𝑥_𝑛∆=𝐴_𝑛𝜉₁+𝐵_𝑛𝜉₂+⋯+𝐿_𝑛𝜉_𝑛 where ∆ is the determinant annexed, and 𝐴₁, 𝐵₁, ⋯, are its first minors. 𝑎₁⋯𝑎_𝑛⋯𝑙₁⋯𝑙_𝑛 To find the value of one of the unknowns 𝑥_𝑟.

Rule

Multiply the equations respectively by the minors of the 𝑟^th column, and add the results. 𝑥_𝑟 will be equal to the fraction whose numerator is the determinant ∆, with its 𝑟^th column replaced by 𝜉₁, 𝜉₂, ⋯, 𝜉_𝑛, and whose denominator is ∆ itself. 582 If 𝜉₁, 𝜉₂, ⋯, 𝜉_𝑛, and ∆ all vanish, then 𝑥₁, 𝑥₂, ⋯, 𝑥_𝑛 are in the ratios of the minors of any row of the determinant ∆. For example, in the ratios 𝐶₁:𝐶₂:𝐶₃: ⋯:𝐶_𝑛.
The eliminant of the given equations is now ∆=0. 584

Orthogonal Transformation

If the two sets of variables in the 𝑛 equations (582) be connected by the relation 𝑥₁+𝑥²₂+⋯+𝑥²_𝑛=𝜉₁+𝜉²₂+⋯+𝜉²_𝑛1 then the changing from one set of variables to the other, by substituting the values of the 𝜉's in terms of the 𝑥's in any function of th former, or vice verssa, is called orthogonal transformation. When equation [1] is satisified, two results follow. I. The determinant ∆=±1. II. Each of the constituents of ∆ is equal to the corresponding minor, or else to minus that minor according as ∆ is positiive or negative

Proof

Substitute the values of 𝜉₁, 𝜉₂, ⋯, 𝜉_𝑛 in terms of 𝑥₁, 𝑥₂, ⋯, 𝑥_𝑛 in equation [1], and equate coefficients of the squares and products of the new variables. We get the 𝑛² equations 𝑎²₁+𝑏²₁+=1𝑎₁𝑎₂+𝑏₁𝑏₂+=0𝑎₁𝑎₃+𝑏₁𝑏₃+=0⋯⋯⋯⋯⋯𝑎₁𝑎_𝑛+𝑏₁𝑏_𝑛+=0} 𝑎₂𝑎₁+𝑏₂𝑏₁+=0𝑎²₂+𝑏²₂+=1𝑎₂𝑎₃+𝑏₂𝑏₃+=0⋯⋯⋯⋯⋯𝑎₂𝑎_𝑛+𝑏₂𝑏_𝑛+=0} 𝑎₃𝑎₁+𝑏₃𝑏₁+=0𝑎₃𝑎₂+𝑏₃𝑏₂+=0𝑎²₃+𝑏²₃+=1⋯⋯⋯⋯⋯𝑎₃𝑎_𝑛+𝑏₃𝑏_𝑛+=0} Also ∆=𝑎₁𝑏₁⋯𝑙₁𝑎₂𝑏₂⋯𝑙₂𝑎₃𝑏₃⋯𝑙₃⋯⋯⋯⋯𝑎_𝑛𝑏_𝑛⋯𝑙_𝑛 From the square of the determinant ∆ by the rule (570), and these equations show that the product is a determinant in which the only constituents that do not vanish constitute a diagonal of 'ones'. Therefore ∆²=1 and ∆=±1 Again, solving the first set of equations for 𝑎₁ (writing 𝑎²₁ as 𝑎₁𝑎₁, ⋯), the second set for 𝑎₂, the third for 𝑎₃, and so on, we have, by (582), the results annexed; which proves the second proposition. {𝑎₁∆=𝐴₁+𝐴₂0+𝐴₃0+=𝐴₁𝑎₂∆=𝐴₁0+𝐴₂+𝐴₃0+=𝐴₂𝑎₃∆=𝐴₁0+𝐴₂0+𝐴₃+=𝐴₃⋯ 585

Theorem

The 𝑛−2^th power of a determinant of the 𝑛^th order multiplied by any constituent is equal to the corresponding minor of the reciprocal determinant.

Proof

Let 𝜌 be the reciprocal determinant of ∆, and 𝛽_𝑟 the minor of 𝐵_𝑟 in 𝜌. Write the transformed equations (582) for the 𝑥's in terms of the 𝜉's, and solve them for 𝜉₂. Then equate the coefficient of 𝑥_𝑟 in the result with its coefficient in the original value of 𝜉₂.
Thus 𝜌𝜉₂=∆(𝛽₁𝑥₁+⋯+𝛽_𝑟𝑥_𝑟+⋯), and 𝜉₂=𝑏₁𝑥₁+⋯+𝑏_𝑟𝑥_𝑟+⋯; ∴ ∆𝛽_𝑟=𝜌𝑏_𝑟=∆^𝑛−1𝑏_𝑟 by (575); ∴ 𝛽_𝑟=∆^𝑛−2𝑏_𝑟 586 To eliminate 𝑥 from the two equations 𝑎𝑥^𝑚+𝑏𝑥^𝑚−1+𝑐𝑥^𝑚−2+⋯=01 𝑎′𝑥^𝑛+𝑏′𝑥^𝑛−1+𝑐′𝑥^𝑛−2+⋯=02 If it is desired that the equation should be homogeneous in 𝑥 and 𝑦; put 𝑥𝑦 instead of 𝑥, and clear of fractions. The following methods will still be applicable.

I. Bezout's Method

Suppose 𝑚>𝑛
Rule: Bring the equations to the same degree by multiplying [2] by 𝑥^𝑚−𝑛. Then multiply [1] by 𝑎′, and [2] by 𝑎, and subtract.
Again, multiply [1] by 𝑎′𝑥+𝑏′, and [2] by (𝑎𝑥+𝑏), and subtract.
Again, multiply [1] by 𝑎′𝑥²+𝑏′𝑥+𝑐′, and [2] by (𝑎𝑥²+𝑏𝑥+𝑐), and subtract, and so on until 𝑛 equations have been obtained. Each will be of the degree 𝑚−1.
Write under these the 𝑚−𝑛 equations obtained by multiplying [2] successively by 𝑥. The eliminant of the 𝑚 equations is the result required.

Example

Let the equations be {𝑎𝑥⁵+𝑏𝑥⁴+𝑐𝑥³+𝑑𝑥²+𝑒𝑥+𝑓=0𝑎′𝑥³+𝑏′𝑥²+𝑐′𝑥+𝑑′=0 The five equations obtained by the method, and their eliminant, by (583), are, writing capital letters for the functions of 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓, 𝐴₁𝑥⁴+𝐵₁𝑥³+𝐶₁𝑥²+𝐷₁𝑥+𝐸₁=0𝐴₂𝑥⁴+𝐵₂𝑥³+𝐶₂𝑥²+𝐷₂𝑥+𝐸₂=0𝐴₃𝑥⁴+𝐵₃𝑥³+𝐶₃𝑥²+𝐷₃𝑥+𝐸₃=0𝑎′𝑥⁴+𝑏′𝑥³+𝑐′𝑥²+𝑑′𝑥 =0 𝑎′𝑥³+𝑏′𝑥²+𝑐′𝑥+𝑑′=0} and 𝐴₁𝐵₁𝐶₁𝐷₁𝐸₁𝐴₂𝐵₂𝐶₂𝐷₂𝐸₂𝐴₃𝐵₃𝐶₃𝐷₃𝐸₃𝑎′𝑏′𝑐′𝑑′00𝑎′𝑏′𝑐′𝑑′=0 Should the equations be of the same degree, the eliminant will be a symmetrical determinant. 587

II. Sylvester's Dialytic Method

Rule: Multiply equation [1] successively by 𝑥, 𝑛−1 times; and equation [2] 𝑚−1 times; and eliminate 𝑥 from the 𝑚+𝑛 resulting equations.

Example

To eliminate 𝑥 from 𝑎𝑥³+𝑏𝑥²+𝑐𝑥+𝑑=0𝑝𝑥²+𝑞𝑥+𝑟=0} The 𝑚+𝑛 equations and their eliminant are 𝑝𝑥²+𝑞𝑥+𝑟=0 𝑝𝑥³+𝑞𝑥²+𝑟𝑥 =0𝑝𝑥⁴+𝑞𝑥³+𝑟𝑥² =0 𝑎𝑥³+𝑏𝑥²+𝑐𝑥+𝑑=0𝑎𝑥⁴+𝑏𝑥³+𝑐𝑥²+𝑑𝑥 =0} and 00𝑝𝑞𝑟0𝑝𝑞𝑟0𝑝𝑞𝑟000𝑎𝑏𝑐𝑑𝑎𝑏𝑐𝑑0=0 588

III. Method of elimination by Symmetrical Functions

Divide the two equations in (586) respectively by the coefficients of their first terms, thus reducing them to the forms 𝑓(𝑥)≡𝑥^𝑚+𝑝₁𝑥^𝑚−1+⋯+𝑝_𝑚=0 𝜙(𝑥)≡𝑥^𝑛+𝑞₁𝑥^𝑛−1+⋯+𝑞_𝑛=0

Rule

Let 𝑎, 𝑏, 𝑐, ⋯, represent the roots of 𝑓(𝑥). Form the equation 𝜙(𝑎)𝜙(𝑏)𝜙(𝑐)⋯=0. This will contain symmetrical functions only of the roots 𝑎, 𝑏, 𝑐, ⋯.
Express these functions in terms of 𝑝₁, 𝑝₂, ⋯ by (538), ⋯, and the equation becomes the eliminant.
Reason of the rule: The eliminant is the condition for a common root of the two equations. That root must make one of the factors 𝜙(𝑎), 𝜙(𝑏), ⋯, vanish, and therefore it makes their product vanish. 589 The eliminant expressed in terms of the roots 𝑎, 𝑏, 𝑐, ⋯, of 𝑓(𝑥), and the roots 𝛼, 𝛽, 𝛾, ⋯, of 𝜙(𝑥), will be (𝑎−𝛼)(𝑎−𝛽)(𝑎−𝛾)⋯(𝑏−𝛼)(𝑏−𝛽)(𝑏−𝛾)⋯ ⋯ being the product of all possible differences between a root of one equation and a root of another. 590 The eliminant is a homogeneous function of the coefficients of either equation, beign of the 𝑛^th degree in the coefficients of 𝑓(𝑥), and of the 𝑚^th degree in the coefficients of 𝜙(𝑥). 591 The sum of the suffixes of 𝑝 and 𝑞 in each term of the eliminant = 𝑚𝑛. Also, if 𝑝, 𝑞 contain 𝑧; if 𝑝₂, 𝑞₂ contain 𝑧²; if 𝑝₃, 𝑞₃ contain 𝑧³; and so on, the eliminant will contain 𝑧^𝑚𝑛.
Proved by the fact that 𝑝_𝑟 is a homogeneous function of 𝑟 dimensions of the roots 𝑎, 𝑏, 𝑐, ⋯, by (406). 592 If the two equations involve 𝑥 and 𝑦, the elimination may be conducted with respect to 𝑥; and 𝑦 will be contained in the coefficients 𝑝₁, 𝑝₂, ⋯, 𝑞₁, 𝑞₂, ⋯. 593

Elimination by the Method of Highest Common Factor

Let two algebraical equations in 𝑥 and 𝑦 be represented by 𝐴=0 and 𝐵=0.
It is required to eliminate 𝑥.
Arrange 𝐴 and 𝐵 according to descending powers of 𝑥, and, having rejected any factor which is a function of 𝑦 only, proceed to find the Highest Common Factor of 𝐴 and 𝐵.
The process may be exhibited as follows: 𝑐₁𝐴=𝑞₁𝐵+𝑟₁𝑅₁ 𝑐₂𝐵=𝑞₂𝑅₁+𝑟₂𝑅₂ 𝑐₃𝑅₁=𝑞₃𝑅₂+𝑟₃𝑅₃ 𝑐₄𝑅₂=𝑞₄𝑅₃+𝑟₄ } 𝑐₁, 𝑐₂, 𝑐₃, 𝑐₄ are the mulipliers required at each stage in order to avoid fractional quotients; and these must be constants or functions of 𝑦 only.
𝑞₁, 𝑞₂, 𝑞₃, 𝑞₄ are the successive quotients.
𝑟₁𝑅₁, 𝑟₂𝑅₂, 𝑟₃𝑅₃, 𝑟₄ are the successive remainders; 𝑟₁, 𝑟₂, 𝑟₃, 𝑟₄ being functions of 𝑦 only.
The process terminates as soon as a remainder is obtained which is a function of 𝑦 only; 𝑟₄ is here supposed to be such a remainder.
Now, the simplest factors having been taken for 𝑐₁, 𝑐₂, 𝑐₃, 𝑐₄, we see that 1 is the H.C.F> of 𝑐₁ and 𝑟₁ 𝑑₂ is the H.C.F> of 𝑐₁ and 𝑟₂ 𝑑₃ is the H.C.F> of 𝑐₁𝑐₂𝑑₂ and 𝑟₃ 𝑑₄ is the H.C.F> of 𝑐₁𝑐₂𝑐₃𝑑₂𝑑₃ and 𝑟₄ } The values of 𝑥 and 𝑦, which satisfy simultaneously the equations 𝐴=0 and 𝐵=0, are those obtained by the four pairs of simultaneous equations following: 𝑟₁=0 and 𝐵=01 ~~𝑟₂𝑑₂~~=0 and 𝑅₁=02 ~~𝑟₃𝑑₃~~=0 and 𝑅₂=03 ~~𝑟₄𝑑₄~~=0 and 𝑅₃=04 } The final equation in 𝑦, which gives all admissible values, is 𝑟₁𝑟₂𝑟₃𝑟₄𝑑₂𝑑₃𝑑₄=0,
If it should happen that the remainder 𝑟₄ is zero, the simultaneous equations [1], [2], [3], and [4] reduce to 𝑟₁=0, and 𝐵𝑅₃=0; 𝑟₂𝑑₂=0, and 𝑅₁𝑅₃=0; 𝑟₃𝑑₃=0, and 𝑅₂𝑅₃=0; 594 To find infinite values of 𝑥 or 𝑦 which satisfy the given equations.
Put 𝑥=1𝑧. Clear of fractions, and make 𝑧=0.
If the two resulting equations in 𝑦 have any common roots, such roots, together with 𝑥=∞, satisfy simultaneously the equations proposed.
similarly we may put 𝑦=1𝑧.