Solve a system of three equations in three variables to find all fixed points of this system.

Question

Solve a system of three equations in three variables to find all fixed points of this system. \begin{align} x &= 0.7x + 0.1y + 0.3z \\ y &= 0.1x + 0.3y + 0.3z \\ z &= 0.2x + 0.6y + 0.4z \\ \end{align}

An answer provided at the back of the textbook is: $$(x, y, z) = (0.429, 0.214, 0.357)$$

I'm not sure how to get to that answer. Can someone point me in the right direction, as to how to solve this problem?

Answer 1

The equations can be rewritten as \begin{align} 3x - y - 3z &= 0 \\ x - 7y + 3z &= 0 \\ x + 3y - 3z &= 0 \end{align} These can be converted to the form $AX = B$ where $$A = \left[ \begin{matrix} 3 & -1 & -3 \\ 1 & -7 & 3 \\ 1 & 3 & -3 \end{matrix} \right], B = \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right], C = \left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right]$$ $A$ can be converted to the reduced row echelon form by performing Gaussian elimination on it such that $$A = \left[ \begin{matrix} 1 & 0 & -\dfrac65 \\ 0 & 1 & -\dfrac35 \\ 0 & 0 & 0 \end{matrix} \right]$$ From this, it follows that for any value of $z$, $$\begin{align} x - \dfrac65z &= 0 \implies x = \dfrac65z \\ y - \dfrac35z &= 0 \implies y = \dfrac35z \end{align}$$

Let $z = 0.357$, then $x = \dfrac65\cdot0.357 \approx 0.429$ and $y = \dfrac35\cdot0.357 \approx 0.214$, i.e., $(x, y, z) \approx (0.429, 0.214, 0.357)$.

Answer 2

Hint: your system is equivalent to $$-3x+y+3z=0$$ $$x-7y+3z=0$$ $$x+3y-3z=0$$ Multiplying the second equation by $3$ and adding to the first: $$-20y+12z=0$$ Doing the same with the last equation: $$10y-6z=0$$ And from these equations we get: These two equations are the same so $$y=\frac{3}{5}z$$ and $$x=\frac{6}{5}z$$

Answer 3

Your equations are linearly dependent so you really only have two equations in three unknowns. There will be a solution that can be multiplied by any constant to get another solution.

\begin{align} x &= 0.7x + 0.1y + 0.3z \\ y &= 0.1x + 0.3y + 0.3z\\ z &= 0.2x + 0.6y + 0.4z\\ 0&=-0.3x+0.1y+0.3z\\0&=0.1x-0.7y+0.3z\\0&=0.2x+0.6y-0.6z\\ y&=0.6z\\x&=1.2z \end{align} An any multiple of $(1.2,0.6,1)$ is a solution.