How can I show that there are only finitely many solutions for the following system?

${x}^{2}+yz=x$

${y}^{2}+zx=y$

${z}^{2}+xy=z$

${x}^{2}+yz=x$

${y}^{2}+zx=y$

${z}^{2}+xy=z$

lilmoore11p8
2022-07-16
Answered

How can I show that there are only finitely many solutions for the following system?

${x}^{2}+yz=x$

${y}^{2}+zx=y$

${z}^{2}+xy=z$

${x}^{2}+yz=x$

${y}^{2}+zx=y$

${z}^{2}+xy=z$

You can still ask an expert for help

persstemc1

Answered 2022-07-17
Author has **18** answers

First assume $x=y=z$. That will give you two solutions.

Otherwise, one of the three unknowns differs from both others. If we assume $x\ne y$ and $x\ne z$, use lab bhatteacharjee's hint to obtain two linear equations in $x,y,z$. You will notice that $y=z$ and $x=1$ follows from these and then there is a unique solution of the original equations (plus two others, by symmetry)

Otherwise, one of the three unknowns differs from both others. If we assume $x\ne y$ and $x\ne z$, use lab bhatteacharjee's hint to obtain two linear equations in $x,y,z$. You will notice that $y=z$ and $x=1$ follows from these and then there is a unique solution of the original equations (plus two others, by symmetry)

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 2022-07-09

Suppose we define a summation graph $G$ as follows:

Each vertex $v\in G$ has a unique but unknown value ascribed to it. Each edge $e\in G$ is labelled with the sum of the values of the two vertices it joins.

This construction corresponds to a system of equations where each equation is of the form ${v}_{a}+{v}_{b}={e}_{ab}$ where ${v}_{1}$ and ${v}_{2}$ correspond to the unknown values of two vertices $\u0444$ and $b$, and ${e}_{ab}$ the value of the edge joining the two.

Now, if there is any odd cycle, it's known that there will always be a unique solution for just that cycle. But with an even cycle, there are either an infinite number of possible values, or an inherent contradiction in the values of the edges that make a solution impossible.

So my question is this: is it possible that there is some bipartite graph (which has no odd cycles, but can have a bunch of even cycles joined together) that has a configuration of values with a unique solution?

Each vertex $v\in G$ has a unique but unknown value ascribed to it. Each edge $e\in G$ is labelled with the sum of the values of the two vertices it joins.

This construction corresponds to a system of equations where each equation is of the form ${v}_{a}+{v}_{b}={e}_{ab}$ where ${v}_{1}$ and ${v}_{2}$ correspond to the unknown values of two vertices $\u0444$ and $b$, and ${e}_{ab}$ the value of the edge joining the two.

Now, if there is any odd cycle, it's known that there will always be a unique solution for just that cycle. But with an even cycle, there are either an infinite number of possible values, or an inherent contradiction in the values of the edges that make a solution impossible.

So my question is this: is it possible that there is some bipartite graph (which has no odd cycles, but can have a bunch of even cycles joined together) that has a configuration of values with a unique solution?

asked 2022-07-03

System of linear equations dependent on $a$:

$x+y+z=1$

$-2x+2y+az=3$

$-ax+y+2z=2$

I'm transforming this to a matrix.

${M}_{1}=\left[\begin{array}{cccc}1& 1& 1& 1\\ -2& 2& a& 3\\ -a& 1& 2& 2\end{array}\right]$

I'm switching $x$ with $y$ column and sort.

${M}_{2}=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& -a& 2& 3\\ 2& -2& a& 2\end{array}\right]$

Solving this:

${M}_{3}=\left[\begin{array}{cccc}1& 1& 1& 1\\ 0& -a-1& 1& 2\\ 0& 0& a+2& 5\end{array}\right]$

Therefore I'm getting

$a=3$

Putting it into the $M1$ yields no result, as I end up with.

$M=\left[\begin{array}{cccc}1& 1& 1& 1\\ 0& 4& 5& 1\\ 0& 4& 5& 5\end{array}\right]$

Is there an error in my calculation?

$x+y+z=1$

$-2x+2y+az=3$

$-ax+y+2z=2$

I'm transforming this to a matrix.

${M}_{1}=\left[\begin{array}{cccc}1& 1& 1& 1\\ -2& 2& a& 3\\ -a& 1& 2& 2\end{array}\right]$

I'm switching $x$ with $y$ column and sort.

${M}_{2}=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& -a& 2& 3\\ 2& -2& a& 2\end{array}\right]$

Solving this:

${M}_{3}=\left[\begin{array}{cccc}1& 1& 1& 1\\ 0& -a-1& 1& 2\\ 0& 0& a+2& 5\end{array}\right]$

Therefore I'm getting

$a=3$

Putting it into the $M1$ yields no result, as I end up with.

$M=\left[\begin{array}{cccc}1& 1& 1& 1\\ 0& 4& 5& 1\\ 0& 4& 5& 5\end{array}\right]$

Is there an error in my calculation?

asked 2022-06-22

Given the system of differential equations ${x}^{\prime}=2x+{y}^{3}$ and ${y}^{\prime}=-y$ i found the flow

${\varphi}_{t}(x,y)=(({x}_{0}+1/5{y}_{0}^{3}){e}^{2t}-1/5{y}_{0}^{3}{e}^{-3t},{y}_{0}{e}^{-t})$

${\varphi}_{t}(x,y)=(({x}_{0}+1/5{y}_{0}^{3}){e}^{2t}-1/5{y}_{0}^{3}{e}^{-3t},{y}_{0}{e}^{-t})$

asked 2022-11-08

You have three numbers. The sum of these numbers are 7.2. The second number is twice as large as the first one. The third number is three times as large as the first one.

This is easily solveable by setting up an equation system such as:

$$a+b+c=7.2$$

$$b=2a$$

$$c=3a$$

This is easily solveable by setting up an equation system such as:

$$a+b+c=7.2$$

$$b=2a$$

$$c=3a$$

asked 2022-07-03

How to solve system equation

$\{\begin{array}{cc}{x}^{3}+{y}^{3}+{z}^{3}=x+y+z& \\ {x}^{2}+{y}^{2}+{z}^{2}=xyz& \end{array}$

$x,y,z\in \mathbb{R}$

$\{\begin{array}{cc}{x}^{3}+{y}^{3}+{z}^{3}=x+y+z& \\ {x}^{2}+{y}^{2}+{z}^{2}=xyz& \end{array}$

$x,y,z\in \mathbb{R}$

asked 2022-09-05

For a square matrix $$A$$, $$A$$ is invertible if and only if $A{A}^{T}$ is