Q. For a quadratic form in $3$ variables over $\Bbb R$, let $r$ be the rank and $s$ be the signature. The number of possible pairs $(r,s)$ is$-$ $\quad$ (a) $13$, (b) $9$, (c) $10$, (d) $16.$
According to the question, $r\le 3.$ Since $s$ equals the number of positive eigenvalues minus that of negative ones, $s$ could be $3$ at most when $r=3$ and all the eigenvalues are positive. Furthermore, we have $s=p-(r-p)\Rightarrow s+r=2p$ (an even number) where $p$ denotes the index of the quadratic form. Thus, according to me, all the possible pairs are $(1,1), (1,3), (2,2), (3,3)$, and $(3,1).$ Among these, the pair $(1,3)$ is redundant. So, there are $4$ pairs in total. What am I missing?