Krdytkyu

For $alphainmathbb{R}$, let $q(x_1, x_2) = x_1^2
+ 2alpha x_1x_2 + dfrac{1}{2}2x_2^2$, for $(x_1, x_2) in mathbb{R^2}$.Find all values of $alpha$ for which the signature of $q$ is 1.

The matrix associated with the quadratic form is $begin{bmatrix} 1& alpha\alpha&frac12end{bmatrix}$. To find the signature we reduce it to a diagonal form by simultaneous row and column transformations as $$begin{bmatrix} 1& alpha\alpha&frac12end{bmatrix}xrightarrow[C_2rightarrow C_2-alpha C_1]{R_2rightarrow R_2-alpha R_1}begin{bmatrix} 1& 0\0&frac12-alpha^2end{bmatrix}.$$

Signature of a matrix is defined as number of positive entries - total no of negative entries.
Signature of matrix of $q$ is $1$ if ${displaystylealphageqfrac{1}{2}}cup{alphaleqfrac{-1}{2}} $. But there is also another formula so to say viz $2p-r$ where $p$ is the number of positive entries and $r$ is the rank of the diagonal form. For $2p-r=1$ we require $p=1$ and $r=1$. Thus the values of $alpha=pmdfrac{1}{2}$. Is it weird that we are getting two ranges for $alpha$ for the same quadratic form? I am sure I have made a mistake. Can anyone please clarify?

$r$ isn't the rank of the matrix. It's the rank of a quadratic form. If the matrix is non-degenerate and $frac 1 2 - alpha^2 < 0$ then its signature is $1$.

For definitions see here.