Why does the following nonlinear system have 21 solutions? y + x y 2 </msup>

slijmigrd 2022-07-05 Answered
Why does the following nonlinear system have 21 solutions?
y + x y 2 x 3 + 2 x z 4 = 0 x y 3 3 x 2 y + 3 y z 4 = 0 5 2 y 2 z 3 2 x 2 z 3 z 7 2 = 0
there is one real and twenty complex solutions: ( 0 , 0 , 0 )
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Gornil2
Answered 2022-07-06 Author has 20 answers
Mark the solution ( 0 , 0 , 0 ) and remove z 3 from the last equation. The reduced systems for z = 0 has two cubic equations, so at most 3 3 = 9 solutions.Then z only occurs as z 4 . Multiply the first equation by x, the second by y and express all equations in a = x 2 , b = y 2 , c = x y and d = z 4 , adding the equation a b c 2 = 0. Then the system reads as
c + c 2 a 2 + 2 a d = 0 a b c 3 a c + 3 c d = 0 5 b 4 a d = 0 a b c 2 = 0
This system has the Bezout bound 8 for the number of solutions for ( a , b , c , d ), each of these will have at most 8 solutions for ( x , y , z ). This reduces the estimate to 64 + 9 + 1 = 74 solutions from the 5 5 7 = 175 or 5 5 4 + 3 3 = 109 for the Bezout bound of the (reduced) original system.
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2021-09-09

Consider the system of equations described by
{x1=2x13x2x2=4x15x2
1. Write down the system of equations in matrix form.
2. Find the eigenvalues of the system of equations.
3. Find the associated eigenvectors.

asked 2022-05-22
Dynamical system equilibrium point increment
d x 1 d x ( t ) = x 2 ( t )
d x 2 d x ( t ) = k m x 1 ( t ) + f ( t ) m
Then we find the equilibrium point by setting
0 = x 2 ( t )
0 = k m x 1 ( t ) + f ( t ) m
Which gives as result
x e q = [ x 1 x 2 ] = [ f / k 0 ]
Finally, the book defines the increment with respect to the equilibrium point Δ x ( t ) = x ( t ) x e q . Substracting equation ( 1 , 3 ) and ( 2 , 4 ) the result is
d Δ x 1 d x ( t ) = Δ x 2 ( t )
d Δ x 2 d x ( t ) = k m Δ x 1 ( t )
So far so good, but for the next step they get "the general solution, parametrized by the initial state" as
Δ x 1 ( t ) = Δ x 1 ( 0 ) c o s ( ω t ) + Δ x 2 ( 0 ) ω s i n ( ω t )
Δ x 2 ( t ) = Δ x 1 ( 0 ) ω s i n ( ω t ) + Δ x 2 ( 0 ) c o s ( ω t )
This result I don't understand where does it comes from, could someone give me a hint?
asked 2022-06-21
Solving a system of three linear equations with three unknowns
Consider the following system of equations
2 x + 2 y + z = 2
x + 2 y z = 5
x 3 y + 2 z = 8
Form an augmented matrix, then reduce this matrix to reduced row echelon form and solve the system.
My answer/working:
Given:
2 x + 2 y + z = 2
x + 2 y z = 5
x 3 y + 2 z = 8
Matrix form:
( 2 2 1 2 1 2 1 5 1 3 2 8 )
( 2 0 0 2 0 3 0 3 0 0 5 6 5 3 )
Solution: x = 1 ; y = 1 ; z = 2 ;
asked 2022-09-26
Perimeter =48 m
Let the length be 𝑥 m and the breadth be 𝑦 m
As we know, Perimeter =2(𝑥+𝑦)=48 x + y = 24 𝑥+𝑦=24 [⋯(1)]
Area=135 m 2 .
As we know, Area=𝑙𝑏
Therefore, 135 m 2 =𝑥𝑦 [⋯(2)]
asked 2022-06-07
For what values of k does this system of equations have a unique solution?
{ y + 2 k z = 0 x + 2 y + 6 z = 2 k x + 2 z = 1
I have [ 1 2 6 2 0 1 2 k 0 k 0 2 1 ]
When I row reduce, I get:
[ 1 0 6 4 k 2 0 1 2 k 0 0 0 2 6 k 4 k 2 1 2 k ]
asked 2022-07-02
Let A and B be two 2 × 2-matrices, and considering the equation
A X = B
Where X is an unknown 2 × 2-matrix. What I want to do is explain why the equation is equivalent to solving two 2 × 2-systems of equations simultaneously and determine how this generalizes to n × n.
And, on the corollary, let A , B , C be three 2 × 2-matrices and consider the equation
A X + X B = C
What I want to do here explain why you can not solve this as two 2 × 2-systems of equations simultaneously
asked 2022-05-15
System of equations with multiplication
I want to find all x , y , z R such that
( x + 1 ) y z = 12 , ( y + 1 ) z x = 4 , ( z + 1 ) x y = 4.
I can multiply all three equations to get
( x + 1 ) ( y + 1 ) ( z + 1 ) x 2 y 2 z 2 = 192
I can divide the first equation by the second to get
( x + 1 ) y ( y + 1 ) x = 4, which simplifies to 3 x y + 4 x y = 0
None of these seems to help. How can I solve the equations?