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!

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 $(*1)$.
$$(x,y,z) = (x,xu,xv) = 
\left(\pm\frac{5}{3\sqrt{2}},
\mp\frac{1}{3\sqrt{2}},
\mp\frac{7}{3\sqrt{2}}
\right)
$$
