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!
Best Answer
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 $y, z$.