Change the variables in $Q(x,y,z)=(x-y+z-1)^2-2z+4$ to have $Q(f(u,v,w))=u^2+v$

Question

I have a problem with this exercise. Initially, they gave me this polynom, and I had to complete the squares:

$$Q(x,y,z)=x^2-2xy+2xz+y^2-2yz+z^2-2x+2y-4z+5.$$

I've done it, and I've checked with maple (so it's correct). We have:

$$Q(x,y,z)=(x-y+z-1)^2-2z+4$$

Now, they say that I have to find an affine transformation to change the variables to have:

$$Q(f(u,v,w))=u^2+v.$$

So I thought that I could do

$$\begin{cases} u=x-y+z-1\\v=-2z+4\\w=0 \end{cases} $$

and solve this system for $\{x,y,z\}$. But obviously it doesn't have a solution.

I'd appreciate any hint, because I don't know how to do this change of variables.

Thanks.

Answer 1

What do you means by

But obviously it doesn't have a solution

You don't need to solve this system of equations. You just find an affine transformation which is $$\begin{bmatrix}u\\v\\w\end{bmatrix}= \begin{bmatrix}1&-1&1\\0&0&-2\\0&0&0\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix}+ \begin{bmatrix}-1\\4\\0\end{bmatrix}$$ And it works.