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I have asked the question a few time back, and due to very much edit,I deleted it and posting it again with a solution I think. Please check it.

I have a symmetric $3×3$ matrix $A$ and $Q = x^{t}Ax = x_1x_2-1, \forall $ reals, where $x= [x_1,x_2,1]^{t} $ is a column vector, Now I have to find the number of +ve eigen values and rank of $A$.

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My attempt is that as $Q$ can be negative and positive for certain non zero values of $x_1,x_2$, so $A$ is indefinite. Hence $A$ will have a negative principal minor of odd order and a positive principal minor of odd order.(Hence rank of $A$ is 3) .

Now I think the matrix $$ A = \begin{pmatrix} 0 & 1/2 & 0 \\ 1/2 & 0 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix} $$ Here I have considered $x_3 = 1$

Now $\text{det}A = 1/4 $, so it has 2 negative and one positive eigen value.

Is my process of solving right ? I am just eager to know it . If this is not the right way , then please let me know the way of solving it correctly. Thank you.

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