Elimination




Homogeneous and non-homogeneous systems

A system is called homogeneous if and only if each multiple of a solution is a solution too.

Example
/ 3x+ 5y - 8z = 0
\ x + y - 2z = 0

This system has the solution (1,1,1) and each multiple of that solution is a solution too. For example (7,7,7) is a solution too.

The systems, who don't have this property, are the non-homogeneous systems.

Coexistence of equations in a non-homogeneous system.

The equations in a non-homogeneous system are coexistent if and only if this system has at least 1 solution for the unknowns.

Example:

De equations in the next system are not coexistent.
/ 2x + 4 = 0
\ 3x = 4

Generally the case is more difficult. The equations can contain 1 or more parameters. We are looking for the condition which expresses that the equations are coexistent.

Example: We take a system with x as an unknown value.

/ m x + 4 x -3 = 0
\ 3 x = m + 2

We are looking for the condition such that the equations are coexistent. The first equation has the solution 3/(m+4) and the second one has the solution (m+2)/3.
The equations are coexistent if and only if 3/(m+4) = (m+2)/3.
This condition is called the coexistence-condition for the equations of the system. If that condition is accomplished, the system has at least 1 solution for the unknowns.

Take a non-homogeneous system with one or more parameters. The necessary and sufficient condition, for the parameters, such that the system has at least one solution for the unknowns, is called the coexistence-condition.
This condition contains parameters but no unknowns.

The notion elimination for a non-homogeneous system

Simple example

Take a system with the unknown x

 
        3 x = a
        5b x = 2
The system has two parameters a en b. The solution of the first equation is a/3 and the solution of the second equation is 2/5b.
The coexistence-condition for the equations of the system is a/3 = 2/5b or 5ab - 6 = 0.
This condition does not contain an unknown. We say that 5ab -6 = 0 is obtained by elimination of x from the system.

Second example

Take the system with unknowns x and y

 
/ 3 x + m2 y  = 1
| - x +   4 y  = 2
\ n x +    y   = 1
This system has two parameters m and n.
We look for the coexistence-condition. This is the condition such that the system has at least one solution for x en y. For that purpose we rely on the theory of systems.

the matrix of the coefficients of that system is

 
[3  m2]
[-1  4]
[ n  1]
The rank of the matrix is 2 because the determinant of the matrix formed by the first two rows is not 0.

We take the the first two equations as main equations and the third equation is the side equation. The condition, so that the system has at least one solution, is that the characteristic determinant is zero.

 
| 3    m2   1|
|-1    4    2| = 0
| n    1    1|
This condition is 2 n m2 - 4 n + m2 + 5 = 0. The unknowns x and y are not in that condition.
We say that 2 n m2 - 4 n + m2 + 5 = 0 results from eliminating x and y from the given system.

To eliminate the unknowns from a non-homogeneous system, we compose the necessary and sufficient condition so that the system has a solution. This condition is the coexistence-condition for the equations from the system.

Coexistence and equations in a homogeneous system.

Definition
The equations in a homogeneous system are coexistent if and only if this system has a solution different from the zero-solution.

Example:

 
        / l x + m y = 0
        \ m x + l y = 0                 (1)
The system obviously has the solution x=0 en y=0. This is the zero-solution. We rely on the theory of systems . Previous system has a solution different from the zero-solution if and only if the following determinant is zero.
 
| l  m |
| m  l |
l2-m2= 0 is the coexistence-condition for the equations in the homogeneous system. We say that the unknowns are eliminated.

To eliminate the unknowns from a homogeneous system, we have to find the necessary and sufficient condition, so that the system has a solution different from the zero-solution.
This condition is the coexistence-condition for the equations in the system.

Elimination of some variables from a system

It occurs that we have to eliminate some variables from a system. Then we consider these variables as the unknowns and we apply to the theory above.
To eliminate some variables from a system, we consider these variables as the unknowns and we apply to the theory above.

Example:

Eliminate l and m from the system

 
        2lx + (l+m)y = 0
        4mx + (l-m)y = 0
Consider l en m as the unknowns
 
        (2x+y) l + y m = 0
        y l + (4x-y) m = 0
We see that we have a homogeneous system with unknowns l and m.

De condition for a solution different from the zero-solution is

 
        | 2x+y    y |
        |           | = 0
        | y     4x-y|

<=>     ...

<=>     4 x2  - y2  + xy = 0
The variables l and m are eliminated.

Example 2:

Eliminate m from the system

 
                2
        /  y = x  + mx
        \  y = 2x + m
Consider m as the unknown
 

        / x m = y - x2
        \   m  = y - 2x
The system with unknown m is not homogeneous.
We look for the condition , such that the system has a solution for m. The value of m in the second equation must be a solution for the first equation.
 
  x ( y - 2x ) = y - x2


<=>     x2  - xy + y = 0                       (2)
The variable m is eliminated.

Elimination of a variable who occurs quadratically.

Consider the following equations with parameter m. m occurs quadratically.
 
        (x - 3)2  + y2  = (5 - m)2

        (x + 3)2  + y2  = (5 + m)2
We want to eliminate m. This means: we want the condition for x and y such that the system has a solution for m.

m is the unknown. So we rewrite the system.

 
        - m2  + 10 m + (x2 - 6 x + y2 - 16) = 0

        - m2  - 10 m + (x2 + 6 x + y2 - 16) = 0
First we make sure that m does not occur quadratically in one of the equations. For that, we replace the second equation with the difference of the two equations.
 
        - m2  + 10 m + (x2 - 6 x + y2 - 16) = 0

             20m - 12x =0
Now, it is easy to calculate m.
 

        -m2  + 10 m + (x2 - 6 x + y2 - 16) = 0
             m = 3x/5
The system must have a solution for m. So we bring the m-value in the first equation and the coexistence-condition occurs. After simplification, we find :
 
        16 x2  + 25 y2  - 400 = 0
The variable m is eliminated from the given system.

Elimination and trigonometry

Take the system
 
        / a = cos(t)
        \ b = sin(t)
We want to eliminate t.

To eliminate t we search for the necessary and sufficient condition in order that the given system has a solution for t.

Conclusion : To eliminate t from
 
        / a = cos(t)
        \ b = sin(t)

we write

        a2  + b2  = 1
Example 1

Eliminate t from

 
        /  x cos(t) + 2 sin(t) = y
        \  x cos(t) + sin(t) = 1
First we calculate cos(t) and sin(t). We find
 
                 2 - y
        cos(t) = -----   en   sin(t) = y - 1
                   x
Now we write the condition
 

                   (2 - y)2
        (y - 1)2+ -------- = 1
                      x2
The parameter t is eliminated.

Example 2

Eliminate t from

 
     /  x = a sec(t)
     \  y = b tan(t)


We calculate cos(t) and  sin(t)

    /         a
   |    x = ------
   |        cos(t)
   |
   |        b sin(t)
   |    y = -------
    \        cos(t)


    /           a
   |   cos(t) = -
   |            x
   |
   |            a y
   |   sin(t) = ---
    \           b x


The system has a solution for t if and only if



        sin2 (t) + cos2 (t) = 1
<=>
         a 2    a y 2
        (-)  + (---)  = 1
         x      b x
<=>

        a2  b2  + a2  y2  = b2  x2
<=>
        x2   y2
        -- - --- = 1
        a2   b2
The parameter t is eliminated.

Example 3

Eliminate t from

 
  / (x/a) cos(t) + (y/b) sin(t) = 1
  |
  \ (y/b) sin(t) - (x/a) sin(t) = 1

We calculate cos(t) and  sin(t) from the system. We find:

            (x/a) + (y/b)
  cos(t) = ----------------
           (x/a)2 + (y/b)2

           (y/b) - (x/a)
  sin(t) = -----------------
           (x/a)2 + (y/b)2


The system has a solution for t if and only if

        sin2 (t) + cos2 (t) = 1

<=>
        (x/a + y/b)2  + (y/b - x/a)2
      ----------------------------------- = 1
             ((x/a)2 + (y/b)2)2

<=>
            (x/a)2 + (y/b)2
      2  --------------------------- = 1
            ((x/a)2 + (y/b)2)2

<=>
       (x/a)2 + (y/b)2 = 2
The parameter t is eliminated.


Topics and Problems

MATH-abundance home page - tutorial

MATH-tutorial Index

The tutorial address is http://home.scarlet.be/math/

Copying Conditions

Send all suggestions, remarks and reports on errors to Johan.Claeys@ping.be     The subject of the mail must contain the flemish word 'wiskunde' because other mails are filtered to Trash