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.
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. |
Take a system with the unknown x
3 x = a 5b x = 2The 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.
Second example
Take the system with unknowns x and y
/ 3 x + m^{2} y = 1 | - x + 4 y = 2 \ n x + y = 1This system has two parameters m and n.
the matrix of the coefficients of that system is
[3 m^{2}] [-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 m^{2} 1| |-1 4 2| = 0 | n 1 1|This condition is 2 n m^{2} - 4 n + m^{2} + 5 = 0. The unknowns x and y are not in that condition.
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. |
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 |l^{2}-m^{2}= 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. |
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 = 0Consider l en m as the unknowns
(2x+y) l + y m = 0 y l + (4x-y) m = 0We 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 x^{2} - y^{2} + xy = 0The variables l and m are eliminated.
Example 2:
Eliminate m from the system
2 / y = x + mx \ y = 2x + mConsider m as the unknown
/ x m = y - x^{2} \ m = y - 2xThe system with unknown m is not homogeneous.
x ( y - 2x ) = y - x^{2} <=> x^{2} - xy + y = 0 (2)The variable m is eliminated.
(x - 3)^{2} + y^{2} = (5 - m)^{2} (x + 3)^{2} + y^{2} = (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.
- m^{2} + 10 m + (x^{2} - 6 x + y^{2} - 16) = 0 - m^{2} - 10 m + (x^{2} + 6 x + y^{2} - 16) = 0First 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.
- m^{2} + 10 m + (x^{2} - 6 x + y^{2} - 16) = 0 20m - 12x =0Now, it is easy to calculate m.
-m^{2} + 10 m + (x^{2} - 6 x + y^{2} - 16) = 0 m = 3x/5The 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 x^{2} + 25 y^{2} - 400 = 0The variable m is eliminated from the given 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.
a^{2} + b^{2} = 1
/ a = cos(t) \ b = sin(t)
/ a = cos(t) \ b = sin(t) we write a^{2} + b^{2} = 1Example 1
Eliminate t from
/ x cos(t) + 2 sin(t) = y \ x cos(t) + sin(t) = 1First we calculate cos(t) and sin(t). We find
2 - y cos(t) = ----- en sin(t) = y - 1 xNow we write the condition
(2 - y)^{2} (y - 1)^{2}+ -------- = 1 x^{2}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 sin^{2} (t) + cos^{2} (t) = 1 <=> a 2 a y 2 (-) + (---) = 1 x b x <=> a^{2} b^{2} + a^{2} y^{2} = b^{2} x^{2} <=> x^{2} y^{2} -- - --- = 1 a^{2} b^{2}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 sin^{2} (t) + cos^{2} (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} = 2The parameter t is eliminated.