1
$\begingroup$

I want to find a matrix $\mathbf A$ such that $x^2 + 4y^2 + 9z^2 + 4xy - 6xz = 1$ can be written as

$$\begin{bmatrix}x&y&z\end{bmatrix}\mathbf A\begin{bmatrix}x&y&z\end{bmatrix}^\top = 1$$

Is there a quick way to do this? I can tell $\mathbf A$ is

$$\begin{bmatrix} 1 & x_{12} & x_{13}\\ x_{21} &4 & x_{23}\\ x_{31} & x_{32} & 9 \end{bmatrix}$$

But I need to guess the other elements. Is there a quick way to solve this?

$\endgroup$
  • 1
    $\begingroup$ watch this : youtube.com/watch?v=0yEiCV-xEWQ $\endgroup$ – rapidracim Mar 28 '18 at 20:34
  • $\begingroup$ Multiply it out once, and compare terms. No need to guess. $\endgroup$ – jgon Mar 28 '18 at 20:35
  • $\begingroup$ You can choose your matrix to be symmetric. In that case, divide coefficients of $xy,xz$ and $yz$ terms by $2$ to obtain the $x_{12},x_{13}$ and $x_{23}$ entries respectively. $\endgroup$ – StubbornAtom Mar 28 '18 at 20:39
3
$\begingroup$

Let consider

  • $x_{12}=x_{21}=2$
  • $x_{13}=x_{31}=-3$
  • $x_{23}=x_{32}=0$

and in general

$$ax^2 + by^2 + cz^2 + 2dxy+2exz+2fyz\implies A=\begin{bmatrix} a & d & e\\ d & b & f\\ e & f & c \end{bmatrix}$$

$\endgroup$
2
$\begingroup$

It' rather simple: it is the symmetric matrix $$\begin{bmatrix} 1 & 2 & -3\\ 2 & 4 & 0\\ -3 & 0 & 9 \end{bmatrix},$$ where the coefficients $a_{ij}$ are half the coefficients of $x_ix_j$ in the quadratic form.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.