$\textbf{Question : }$ Find all $x,y,z \in \mathbb{R}$ that satisfy the following system of equations.
$$3x-4y+\frac{1}{xy}=12$$ $$4z-12x+\frac{1}{zx}=3$$ $$12y-3z+\frac{1}{yz}=4$$
$\textbf{My Attempt :}$
Let $t = \frac{1}{xyz}$ then each of our equation transforms to
$$3x-4y+tz=12$$ $$-12x+ty+4z=3$$ $$tx+12y-3z=4$$
I thought I would zero out the coefficient matrix which led to nothing :
$$\Delta = -t^3-169t=0 \implies t = 0 ,\pm13i$$ none of which is in $t$'s domain as $t \in \mathbb{R} – \{0\}$
Next I thought to write it in a form where all the constant terms are zero so I could surely zero out the coefficient matrix and find out the value of $t$. So here are those equations $$51x-4y(t+1)+z(t-16)=0$$ $$3(t+16)x+4(9-t)y-25z=0$$ $$3(1-t)x-40y+z(t+9)=0$$
as it turns out the coefficient matrix is zero for all $t \in \mathbb{R} – \{0\}$. So what to do now?
Best Answer
Solve the set of equations you have $$3x-4y+tz=12$$ $$-12x+ty+4z=3$$ $$tx+12y-3z=4$$for $x$, $y$ and $z$ in terms of $t$. It looks as if it will be horrible, but in fact you can eliminate $y$, for example, to get (multiply first equation by t and second by 4, then add) $$(3t-48)x+(t^2+16)z=12t+12$$ and (then multiply first equation by 3 and add to last) $$(t+9)x+(3t-3)z=40.$$ Now multiply the first of these equations by $(3t-3)$ and the second by $(t^2+16)$ and subtract, to get $$(9(t-1)(t-16)-(t+9)(t^2+16))x=36(t^2-1)-40(t^2+16)$$ or $$(-t^3-169t)x=-4(t^2+169),$$ so $x= \dfrac{4}{t}$ and back substitution gives $y= \dfrac{3}{t}$ and $z= \dfrac{12}{t}$.
Then $t= \dfrac{1}{xyz} = \dfrac{t^3}{144}$, so $t=\pm 12$, which completes the solution.