Solving a system of non-linear equations Let
$$(\star)\begin{cases}
 \begin{vmatrix}
 x&y\\
 z&x\\
 \end{vmatrix}=1,   \\ 
 \begin{vmatrix}
 y&z\\
 x&y\\
 \end{vmatrix}=2,  \\ 
 \begin{vmatrix}
 z&x\\
 y&z\\
 \end{vmatrix}=3.  
\end{cases}$$ 
Solving the above system of three non-linear equations with three unknowns.

I have a try. 
Let$$A=\begin{bmatrix}
 1&  1/2& -1/2\\ 
 1/2&  1& -1/2\\ 
 -1/2& -1/2& -1 
\end{bmatrix}$$
We have $$(x,y,z)A\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix}=0.$$
There must be a orthogonal matrix $T$,such that $T^{-1}A T=diag \begin{Bmatrix}
\frac{1}{2},\frac{\sqrt{33}+1}{4},-\frac{\sqrt{33}-1}{4}
\end{Bmatrix}.$
$$\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix}=T\begin{pmatrix}
x^{'}\\ 
y^{'}\\ 
z^{'}
\end{pmatrix}\Longrightarrow\frac{1}{2} {x'}^{2}+\frac{\sqrt{33}+1}{4} {y'}^{2}-\frac{\sqrt{33}-1}{4}{z'}^{2}=0.$$
But even if we find a $\begin{pmatrix}
x_0^{'}\\ 
y_0^{'}\\ 
z_0^{'}
\end{pmatrix} $ satisfying $\frac{1}{2} {x_0'}^{2}+\frac{\sqrt{33}+1}{4} {y_0'}^{2}-\frac{\sqrt{33}-1}{4}{z_0'}^{2}=0,\begin{pmatrix}
x_0\\ 
y_0\\ 
z_0
\end{pmatrix}=T\begin{pmatrix}
x_0^{'}\\ 
y_0^{'}\\ 
z_0^{'}
\end{pmatrix}$ may not be the solution of $(\star)$

If you have some  good ideas,please give me some hints. Any help would be appreciated!
 A: Given $x^2-yz = 1, \quad y^2-xz = 2, \quad z^2-xy = 3$, we can sum all of these to get
$$(x-y)^2+(y-z)^2+(z-x)^2 = 12 \tag{1}$$
OTOH, subtracting gives $(y^2-x^2)+z(y-x)=1 \implies (x+y+z)(y-x) = 1$
and similarly $(x+y+z)(z-y) = 1$, so we must have $y-x = z - y = a$, say.  Using this in $(1)$, 
$$a^2+a^2+4a^2=12 \implies a = \pm \sqrt2$$
So we have $y = x \pm \sqrt2, \quad z = x \pm 2\sqrt2$.  Using these in say the first equation, you should be able to solve for $x$ and then $y, z$.
A: Let's rewrite the system as $$\mathbf r \times \mathbf P\mathbf{r} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\x&y&z\\z&x&y\end{vmatrix} = 2\mathbf{i} + 3\mathbf{j} + 1\mathbf{k} \equiv \mathbf f$$
where $\mathbf{r} = (x,y,z)^\top$ and
$$
\mathbf P = \begin{pmatrix}
0 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0
\end{pmatrix}
$$
So we want to solve
$$
\mathbf{r} \times \mathbf{Pr} = \mathbf{f}.
$$
Let's dot multiply both sides with $\mathbf{r}$
$$
0 = \mathbf{r}^\top \mathbf{f}
$$
also when multiplying by $\mathbf{Pr}$:
$$
0 = \mathbf{r}^\top \mathbf P^\top \mathbf{f}
$$
That's a pair of linear equations in $x,y,z$. Since the equations are homogeneous, the solution will be of the form $\mathbf r = \alpha \mathbf{h}, \alpha \in \mathbb{R}$.
$$
2x + 3y + z = 0\\
2z + 3x + y = 0
$$
Row reducing the system we get
$$
\mathbf h = \begin{pmatrix}
-5\\1\\7
\end{pmatrix}.
$$
Since $\mathbf h \times \mathbf {Ph} = 18 \mathbf f$ we deduce that
$$
\alpha^2 = \frac{1}{18}\qquad\alpha = \pm\frac{1}{3\sqrt{2}}.
$$
Finally
$$
x = \mp \frac{5}{3\sqrt{2}}\\
y = \pm \frac{1}{3\sqrt{2}}\\
z = \pm \frac{7}{3\sqrt{2}}
$$
A: The resultant of $x^2-yz-1$ and $y^2-xz-2$ with respect to $z$ is $x^3-y^3-x+2y$.  The resultant of $x^2 - yz - 1$ and $z^2 - xy - 3$ with respect to $z$ is $x^4-x y^3-2 x^2-3 y^2+1$.  The resultant of $x^3-y^3-x+2y$ and 
$x^4-x y^3-2 x^2-3 y^2+1$ with respect to $x$ is $-y^4(18 y^2-1)$.  So either
$y = 0$ or $y = \pm 1/\sqrt{18}$.
With $y=0$ we get $x^2 - 1 = 0$, $-xz-2 = 0$ and $z^2 - 3 = 0$, which clearly will not work.
With $y = \pm 1/\sqrt{18} = \sqrt{2}/6$ we do get 
solutions: $x = \mp 5 \sqrt{2}/6$, $z = \pm 7 \sqrt{2}/6$.
A: Note: This is a slightly clumsy but systematic approach. On the plus side, this allow you solving similar equations of the form
$$\begin{cases}
x^2 - Ayz &= D\\
y^2 - Bxz &= E\\
z^2 - Cxy &= F
\end{cases}$$
without knowing how to complete the squares. On the minus side, you need to factor a quartic polynomial in the middle of the process.

Notice the LHS of given set of equations are all homogenous with degree $2$,
$$\begin{cases}
x^2 - yz &= 1\\
y^2 - xz &= 2\\
z^2 - xy &= 3
\end{cases}\tag{*1}$$
we can simplify it by looking at the ratios first. i.e. let $u = \frac{y}{x}$ and $v = \frac{z}{x}$, we have
$$
\begin{cases}
\frac{1-uv}{1} &= \frac{u^2 - v}{2}\\
\frac{1-uv}{1} &= \frac{v^2 - u}{3}
\end{cases}
\iff
\begin{cases}
2(1-uv) &= u^2 - v\\
3(1-uv) &= v^2 - u
\end{cases}
\implies
\begin{cases}
v &= \frac{2-u^2}{2u-1} &(*2a)\\
u &= \frac{3-v^2}{3v-1} &(*2b)
\end{cases}
$$
Substitute $(*2a)$ into $(*2b)$, we get
$$u = \frac{3 - \left(\frac{2-u^2}{2u-1}\right)^2}{3\left(\frac{2-u^2}{2u-1}\right) - 1}
\iff
\frac{5u^4+u^3-5u-1}{6u^3+u^2-16u+7} = 
\frac{(u-1)(5u+1)(u^2+u+1)}{(2u-1)(3u^2+2u-7)}
= 0\\
$$
Since $u^2 + u + 1 = (u+\frac12)^2 + \frac34 > 0$ for all real $u$, there are
only two choices for $u$:


*

*Case 1 : $u = 1 \implies v = \frac{2 - 1^2}{2-1} = 1 \implies 1 - uv = 0$.
However, this contradicts with the requirement $x^2 - yz = x^2(1-uv) = 1$, 
this doesn't lead to any solution for $(*1)$.

*Case 2 : $u = -\frac15 \implies v = \frac{2 - \left(-\frac15\right)^2}{2\left(-\frac15\right)-1} = -\frac75 \implies 1 - uv = \frac{18}{25}$.
We have $x = \frac{1}{\sqrt{1-uv}} = \pm\frac{5}{3\sqrt{2}}$ now. This
leads to two real solutions for $(*1)$.
$$(x,y,z) = (x,xu,xv) = 
\left(\pm\frac{5}{3\sqrt{2}},
\mp\frac{1}{3\sqrt{2}},
\mp\frac{7}{3\sqrt{2}}
\right)
$$
