We can read here quadric wikipedia that the cone $x^2+y^2-z^2=0$ is a degenerate quadric but they don't define what is a degenerate quadric.
I know hat a quadric is non degenerate if the matrix associated to its quadratic form is invertible.
For the cone $x^2+y^2-z^2=0$, its matrix is the diagonal matrix whose diagonal is $(1,1,-1)$ which is an invertible matrix.
Is there another definition of non degenerate quadric in the sense of wikipedia ?
EDIT The relation between quadric and quadratic form could be also seen using the projective space of $R^3$. Let $(x,y,z)\in P(R^4)$ and $(K,X,Y,Z)\in R^4$ homogeneous coordinates : $x = X/K, y=Y/K ...$. In that case a quadric of $R^3$ is a quadratic form on $P(R^3)$.
Actually, the matrix of this quadric is $\operatorname{diag}(1,1,-1,0)$: you left out the linear and constant terms (which all happen to be zero in this case). The $3\times3$ matrix in your question is that of a quadratic form $q(x,y,z)$ that contains only second-degree terms. It does correspond to a quadric surface, but one that’s defined by the equation $q(x,y,z)=1$.