example of resultant (2)

example of resultant (2)

This example shows how

resultants

can be used to solve simultaneous

algebraic equations

in two variables. We shall compute the

intersection

of two ellipses.

Consider the system of equations

f

⁢

(

x

,

y

)

=

0

,

g

⁢

(

x

,

y

)

=

0

where

f

⁢

(

x

,

y

)

=

3

⁢

x

2

+

2

⁢

x

⁢

y

+

3

⁢

y

2

-

2

g

⁢

(

x

,

y

)

=

3

⁢

x

2

-

2

⁢

x

⁢

y

+

3

⁢

y

2

-

2

We will consider

f

and

g

as

polynomials

in

x

whose coefficients are functions of

x

. What this means can be seen by writing

f

and

g

as

(

3

)

⁢

x

2

+

(

2

⁢

y

)

⁢

x

+

(

3

⁢

y

2

-

2

)

(

3

)

⁢

x

2

+

(

-

2

⁢

y

)

⁢

x

+

(

3

⁢

y

2

-

2

)

We will now

construct

the resultant by computing Sylvester’s determinant. In the notation of the main article, the coefficients of the various powers of

x

may be notated as

a

0

=

3

a

1

=

2

⁢

y

a

2

=

3

⁢

y

2

-

2

b

0

=

3

b

1

=

-

2

⁢

y

b

2

=

3

⁢

y

2

-

2

The determinant is

|

3

2

⁢

y

3

⁢

y

2

-

2

0

0

3

2

⁢

y

3

⁢

y

2

-

2

3

-

2

⁢

y

3

⁢

y

2

-

2

0

0

3

-

2

⁢

y

3

⁢

y

2

-

2

|

This determinant evaluates to

144

⁢

y

4

-

96

⁢

y

2

. Hence, in order for the system of equations to have a solution,

y

must satisfy the equation

144

⁢

y

4

-

96

⁢

y

2

=

0

We can factor the polynomial as

144

⁢

y

4

-

96

⁢

y

2

=

144

⁢

(

y

+

2

3

)

⁢

y

2

⁢

(

y

-

2

3

)

Hence, the solutions are

y

=

-

2

3

y

=

0

y

=

+

2

3

Note that the solution

y

=

0

occurs with

multiplicity

2

. We shall see what that means shortly.

Having found the possible values of

y

, let us now find the corresponding values for

x

. Substituting the possible value

y

=

-

2

3

into the equation

f

⁢

(

x

,

y

)

=

0

, we obtain

3

⁢

x

2

-

2

⁢

2

3

⁢

x

=

0

Hence, either

x

=

0

or

x

=

+

2

⁢

2

3

. If we substitute

x

=

0

and

y

=

-

2

3

into

g

⁢

(

x

,

y

)

, we obtain zero so

x

=

0

y

=

-

2

3

is a solution of our system. However, if we substitute

x

=

+

2

⁢

2

3

and

y

=

-

2

3

into

g

⁢

(

x

,

y

)

, we obtain

16

9

, so this root does not lead to a solution of the original system of equations.

Substituting the possible value

y

=

+

2

3

into the equation

f

⁢

(

x

,

y

)

=

0

, we obtain

3

⁢

x

2

+

2

⁢

2

3

⁢

x

=

0

Hence, either

x

=

0

or

x

=

-

2

⁢

2

3

. If we substitute

x

=

0

and

y

=

+

2

3

into

g

⁢

(

x

,

y

)

, we obtain zero so

x

=

0

y

=

+

2

3

is a solution of our system. However, if we substitute

x

=

-

2

⁢

2

3

and

y

=

+

2

3

into

g

⁢

(

x

,

y

)

, we obtain

-

16

9

, so this root does not lead to a solution of the original system of equations.

Finally, let us consider the value

y

=

0

. Substituting this value into

f

⁢

(

x

,

y

)

, we obtain the equation

3

⁢

x

2

-

2

=

0

This equation has the solutions

x

=

-

2

/

3

and

x

=

+

2

/

3

. Substituting

y

=

0

and

x

=

-

2

/

3

into

g

⁢

(

x

,

y

)

, we obtain

0

, so

y

=

0

x

=

-

2

/

3

is a solution of the system of equations. Likewise, substituting

y

=

0

and

x

=

+

2

/

3

into

g

⁢

(

x

,

y

)

, we obtain

0

, so

y

=

0

x

=

+

2

/

3

is a solution of the system of equations. In this case, we obtained two solutions to the system of equations.

At this point, recall the remark that

y

=

0

was a double root of the resultant. This fact explains why both values of

x

gave rise to solutions of the system when

y

=

0

. In general, the number of solutions (counted with multiplicity) of the system of equations for a particular value of

y

equals the mutiplicity of that value of

y

as a root of the resultant.

Title

example of resultant (2)

Canonical name

ExampleOfResultant2

Date of creation

2013-03-22 14:36:36

Last modified on

2013-03-22 14:36:36

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

7

Author

rspuzio (6075)

Entry type

Example

Classification

msc 13P10

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