$y+x{y}^{2}-{x}^{3}+2x{z}^{4}=0\phantom{\rule{0ex}{0ex}}-x-{y}^{3}-3{x}^{2}y+3y{z}^{4}=0\phantom{\rule{0ex}{0ex}}-\frac{5}{2}{y}^{2}{z}^{3}-2{x}^{2}{z}^{3}-\frac{{z}^{7}}{2}=0$

there is one real and twenty complex solutions: $(0,0,0)$

slijmigrd
2022-07-05
Answered

Why does the following nonlinear system have 21 solutions?

$y+x{y}^{2}-{x}^{3}+2x{z}^{4}=0\phantom{\rule{0ex}{0ex}}-x-{y}^{3}-3{x}^{2}y+3y{z}^{4}=0\phantom{\rule{0ex}{0ex}}-\frac{5}{2}{y}^{2}{z}^{3}-2{x}^{2}{z}^{3}-\frac{{z}^{7}}{2}=0$

there is one real and twenty complex solutions: $(0,0,0)$

$y+x{y}^{2}-{x}^{3}+2x{z}^{4}=0\phantom{\rule{0ex}{0ex}}-x-{y}^{3}-3{x}^{2}y+3y{z}^{4}=0\phantom{\rule{0ex}{0ex}}-\frac{5}{2}{y}^{2}{z}^{3}-2{x}^{2}{z}^{3}-\frac{{z}^{7}}{2}=0$

there is one real and twenty complex solutions: $(0,0,0)$

You can still ask an expert for help

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 $3\cdot 3=9$ solutions.Then z only occurs as ${z}^{4}$. Multiply the first equation by $x$, the second by $y$ and express all equations in $a={x}^{2}$, $b={y}^{2}$, $c=xy$ and $d={z}^{4}$, adding the equation $ab-{c}^{2}=0$. Then the system reads as

$\begin{array}{rl}c+{c}^{2}-{a}^{2}+2ad& =0\\ -a-bc-3ac+3cd& =0\\ -5b-4a-d& =0\\ ab-{c}^{2}& =0\end{array}$

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 $5\cdot 5\cdot 7=175$ or $5\cdot 5\cdot 4+3\cdot 3=109$ for the Bezout bound of the (reduced) original system.

$\begin{array}{rl}c+{c}^{2}-{a}^{2}+2ad& =0\\ -a-bc-3ac+3cd& =0\\ -5b-4a-d& =0\\ ab-{c}^{2}& =0\end{array}$

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 $5\cdot 5\cdot 7=175$ or $5\cdot 5\cdot 4+3\cdot 3=109$ for the Bezout bound of the (reduced) original system.

asked 2021-09-09

Consider the system of equations described by

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

$\frac{d{x}_{1}}{dx}(t)={x}_{2}(t)$

$\frac{d{x}_{2}}{dx}(t)=\frac{-k}{m}{x}_{1}(t)+\frac{f(t)}{m}$

Then we find the equilibrium point by setting

$0={x}_{2}(t)$

$0=\frac{-k}{m}{x}_{1}(t)+\frac{f(t)}{m}$

Which gives as result

${x}_{eq}=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=\left[\begin{array}{c}f/k\\ 0\end{array}\right]$

Finally, the book defines the increment with respect to the equilibrium point $\mathrm{\Delta}x(t)=x(t)-{x}_{eq}$. Substracting equation $(1,3)$ and $(2,4)$ the result is

$\frac{d\mathrm{\Delta}{x}_{1}}{dx}(t)=\mathrm{\Delta}{x}_{2}(t)$

$\frac{d\mathrm{\Delta}{x}_{2}}{dx}(t)=\frac{-k}{m}\mathrm{\Delta}{x}_{1}(t)$

So far so good, but for the next step they get "the general solution, parametrized by the initial state" as

$\mathrm{\Delta}{x}_{1}(t)=\mathrm{\Delta}{x}_{1}(0)cos(\omega t)+\frac{\mathrm{\Delta}{x}_{2}(0)}{\omega}sin(\omega t)$

$\mathrm{\Delta}{x}_{2}(t)=-\mathrm{\Delta}{x}_{1}(0)\omega sin(\omega t)+\mathrm{\Delta}{x}_{2}(0)cos(\omega t)$

This result I don't understand where does it comes from, could someone give me a hint?

$\frac{d{x}_{1}}{dx}(t)={x}_{2}(t)$

$\frac{d{x}_{2}}{dx}(t)=\frac{-k}{m}{x}_{1}(t)+\frac{f(t)}{m}$

Then we find the equilibrium point by setting

$0={x}_{2}(t)$

$0=\frac{-k}{m}{x}_{1}(t)+\frac{f(t)}{m}$

Which gives as result

${x}_{eq}=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=\left[\begin{array}{c}f/k\\ 0\end{array}\right]$

Finally, the book defines the increment with respect to the equilibrium point $\mathrm{\Delta}x(t)=x(t)-{x}_{eq}$. Substracting equation $(1,3)$ and $(2,4)$ the result is

$\frac{d\mathrm{\Delta}{x}_{1}}{dx}(t)=\mathrm{\Delta}{x}_{2}(t)$

$\frac{d\mathrm{\Delta}{x}_{2}}{dx}(t)=\frac{-k}{m}\mathrm{\Delta}{x}_{1}(t)$

So far so good, but for the next step they get "the general solution, parametrized by the initial state" as

$\mathrm{\Delta}{x}_{1}(t)=\mathrm{\Delta}{x}_{1}(0)cos(\omega t)+\frac{\mathrm{\Delta}{x}_{2}(0)}{\omega}sin(\omega t)$

$\mathrm{\Delta}{x}_{2}(t)=-\mathrm{\Delta}{x}_{1}(0)\omega sin(\omega t)+\mathrm{\Delta}{x}_{2}(0)cos(\omega t)$

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

$2x+2y+z=2$

$-x+2y-z=-5$

$x-3y+2z=8$

Form an augmented matrix, then reduce this matrix to reduced row echelon form and solve the system.

My answer/working:

Given:

$2x+2y+z=2$

$-x+2y-z=-5$

$x-3y+2z=8$

Matrix form:

$\left(\begin{array}{cccc}2& 2& 1& 2\\ -1& 2& -1& -5\\ 1& -3& 2& 8\end{array}\right)$

$\left(\begin{array}{cccc}2& 0& 0& 2\\ 0& 3& 0& -3\\ 0& 0& \frac{5}{6}& \frac{5}{3}\end{array}\right)$

Solution: $x=1;y=-1;z=2;$

Consider the following system of equations

$2x+2y+z=2$

$-x+2y-z=-5$

$x-3y+2z=8$

Form an augmented matrix, then reduce this matrix to reduced row echelon form and solve the system.

My answer/working:

Given:

$2x+2y+z=2$

$-x+2y-z=-5$

$x-3y+2z=8$

Matrix form:

$\left(\begin{array}{cccc}2& 2& 1& 2\\ -1& 2& -1& -5\\ 1& -3& 2& 8\end{array}\right)$

$\left(\begin{array}{cccc}2& 0& 0& 2\\ 0& 3& 0& -3\\ 0& 0& \frac{5}{6}& \frac{5}{3}\end{array}\right)$

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 $\Rightarrow x+y=24$ 𝑥+𝑦=24 [⋯(1)]

Area=135 m${}^{2}$.

As we know, Area=𝑙𝑏

Therefore, 135 m${}^{2}$ =𝑥𝑦 [⋯(2)]

Let the length be 𝑥 m and the breadth be 𝑦 m

As we know, Perimeter =2(𝑥+𝑦)=48 $\Rightarrow x+y=24$ 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?

$\{\begin{array}{l}y+2kz=0\\ x+2y+6z=2\\ kx+2z=1\end{array}$

I have $\left[\begin{array}{cccc}1& 2& 6& 2\\ 0& 1& 2k& 0\\ k& 0& 2& 1\end{array}\right]$

When I row reduce, I get:

$\left[\begin{array}{cccc}1& 0& 6-4k& 2\\ 0& 1& 2k& 0\\ 0& 0& 2-6k-4{k}^{2}& 1-2k\end{array}\right]$

$\{\begin{array}{l}y+2kz=0\\ x+2y+6z=2\\ kx+2z=1\end{array}$

I have $\left[\begin{array}{cccc}1& 2& 6& 2\\ 0& 1& 2k& 0\\ k& 0& 2& 1\end{array}\right]$

When I row reduce, I get:

$\left[\begin{array}{cccc}1& 0& 6-4k& 2\\ 0& 1& 2k& 0\\ 0& 0& 2-6k-4{k}^{2}& 1-2k\end{array}\right]$

asked 2022-07-02

Let $A$ and $B$ be two $2\times 2$-matrices, and considering the equation

$AX=B$

Where $X$ is an unknown $2\times 2$-matrix. What I want to do is explain why the equation is equivalent to solving two $2\times 2$-systems of equations simultaneously and determine how this generalizes to $n\times n$.

And, on the corollary, let $A,B,C$ be three $2\times 2$-matrices and consider the equation

$AX+XB=C$

What I want to do here explain why you can not solve this as two $2\times 2$-systems of equations simultaneously

$AX=B$

Where $X$ is an unknown $2\times 2$-matrix. What I want to do is explain why the equation is equivalent to solving two $2\times 2$-systems of equations simultaneously and determine how this generalizes to $n\times n$.

And, on the corollary, let $A,B,C$ be three $2\times 2$-matrices and consider the equation

$AX+XB=C$

What I want to do here explain why you can not solve this as two $2\times 2$-systems of equations simultaneously

asked 2022-05-15

System of equations with multiplication

I want to find all $x,y,z\in \mathbb{R}$ such that

$(x+1)yz=12,(y+1)zx=4,(z+1)xy=4$.

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

$\frac{(x+1)y}{(y+1)x}}=4$, which simplifies to $3xy+4x-y=0$

None of these seems to help. How can I solve the equations?

I want to find all $x,y,z\in \mathbb{R}$ such that

$(x+1)yz=12,(y+1)zx=4,(z+1)xy=4$.

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

$\frac{(x+1)y}{(y+1)x}}=4$, which simplifies to $3xy+4x-y=0$

None of these seems to help. How can I solve the equations?