<s>[INST] 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!
 [/INST] 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 $y, z$.
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