Matrix associated to this quadratic form

Question

I have the quadratic form

$$q(x, y) = x^2 + 6xy + 9y^2 - 6x$$

How can I find the associated matrix? The $6x$ term destroys all.

I tried to write it as

$$q(x, y) = (x+3y)^2 - 6x$$

But then?

OK UNDERSTOOD PART 1

The professor told us: "treat it as a second degree without the translational term $-6x$, and then after having found the matrix and bla bla, draw and find the whole new figure".

Ok, honestly I have done how she asked us. I found the $A$ matrix of the quadratic form without $-6x$

$$\begin{pmatrix} 1 & 3 \\ 3 & 9 \end{pmatrix}$$

Eigenvalues are $\lambda = 0, 10$ and eigenvectors are (non normalized) $\{-3, 1\}$ and $\{1, 3\}$

Now there is a problem: the matrix of the bad change has NEGATIVE determinant.

So there is also a reflection, in addition to a rotation.

Anyway how can I proceed from here?

Answer 1

This isn't a quadratic form precisely due to the presence of the $-6x$ term. When $q$ is a quadratic form, you always have $q(\lambda x,\lambda y)=\lambda^2q(x,y)$. That's clearly not the case here.

Answer 2

Consider $A=\begin{pmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{pmatrix}$, then you get $$ (x,y)A\begin{pmatrix}x\\y\end{pmatrix}=(x,y)\begin{pmatrix}a_{11}x+a_{12}y\\a_{21}x+a_{22}y\end{pmatrix}=a_{11}x^2+(a_{12}+a_{21})xy+a_{22}y^2. $$ So you see, that your function can't be a quadratic form, since each quadratic form has to have the form $q(x,y)=\alpha x^2+\beta xy+\gamma y^2$.