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I am trying to solve the following system of equations:

$$x = 0.09 x + 0.6p y + (1.3p - p^2) z,$$ $$y= 0.49 x + 0.16 y + 0.7 z,$$ $$z = 0.42 x + (0.84 - 0.6p) y + (p^2 - 2p + 1) z,$$ $$x + y + z = 1$$

When I attempt to use WolframAlpha to solve this, the system starts giving me values of $p$ for some reason. I just want the equations $x$, $y$, and $z$ in terms of (unknown) $p$.

I would greatly appreciate it if someone would please take the time to help me with this.

The Pointer
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  • Try: Take first 3 equation from your system, and check is such homogeneous system (with first 3 equation) having nontrivial solutions. We know that homogeneous system $\left( AX=0 \right)$ is having nontrivial solutions iff determinant of matrix $A$ is zero, i.e. $\det(A)=0$. Finally, if you will have one-parametric solution(s), put them into 4-th equation of the original system to find its solution... – Anton Vrdoljak May 04 '20 at 05:47

1 Answers1

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A simplistic approach

From $(1)$ extract $z$ as a function of $x$ and $y$.

Plug the result in $(2)$ and solve for $y$ as a function of $x$. These two steps give $$y=\frac{7 \left(10 p^2-13 p-13\right) x}{24 p (5 p-9)}\qquad \qquad z=\frac{7 (5 p-13) x}{20 p (5 p-9)}$$

Plug in $(3)$ to get $$\frac{49 (p-1) (5 p-13) x}{200 p (5 p-9)}=0$$ You cannot have $x=0$ since this would make $x=y=z=0$ which cannot satisfy $(4)$.

So, the only possibilities are $p=1$ or $p=\frac {13} 5$ Now, consider each case and use $(4)$ to solve.

Claude Leibovici
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