Gaaljh

Answered

2022-06-27

system of three equations

$-sx+sy=0$

$rx-y-xz=0$

$xy-bz=0$

where $s,r,$ and b are given parameters and $x,y,z$ are my three variables. Trouble is the "$xz$" and the "$xy$" terms since they are coupled between two different variables. Ignoring those two terms, my matrix equation would read:

$\begin{array}{r}\left[\begin{array}{ccc}-s& s& 0\\ r& -1& ?\\ 0& ?& -b\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\end{array}$

Where "?" represents the coupled terms

$-sx+sy=0$

$rx-y-xz=0$

$xy-bz=0$

where $s,r,$ and b are given parameters and $x,y,z$ are my three variables. Trouble is the "$xz$" and the "$xy$" terms since they are coupled between two different variables. Ignoring those two terms, my matrix equation would read:

$\begin{array}{r}\left[\begin{array}{ccc}-s& s& 0\\ r& -1& ?\\ 0& ?& -b\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\end{array}$

Where "?" represents the coupled terms

Answer & Explanation

jarakapak7

Expert

2022-06-28Added 14 answers

(1) $-sx+sy=0$

(2) $rx-y-xz=0$

(3) $xy-bz=0$

You can try something like this:

(1) $\to x=y$

(3) $\to z=\frac{{x}^{2}}{b}$

(2) $\to xz=rx-x$

And continue in the similar manner.

(2) $rx-y-xz=0$

(3) $xy-bz=0$

You can try something like this:

(1) $\to x=y$

(3) $\to z=\frac{{x}^{2}}{b}$

(2) $\to xz=rx-x$

And continue in the similar manner.

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