What is the value of the following determinant: given that $q^2- pr<0, p>0$ then evaluate \begin{vmatrix}
p &q &px+qy\\
q &r &qx+ry\\
px+qy &qx+ry &0
\end{vmatrix}
I've tried this:
\begin{vmatrix}
p &q &px+qy\\
q &r &qx+ry\\
px+qy &qx+ry &0
\end{vmatrix}
then, i modified the rows as $R_3=R_3-(xR_1+yR_2)$ to get \begin{vmatrix}
p &q &px+qy\\
q &r &qx+ry\\
0 &0 &px^2+2qxy+ry^2
\end{vmatrix}
then I've simplified it as $(px^2+2qxy+ry^2)(pr-q^2)$, but i cannot do further. can you please give me hints?
 A: Just a typo, but it seems that you dropped a minus sign since you took $R_3 = xR_1 + yR_2 - R_3$, so you essentially replaced $R_3$ with its negative (which negates the determinant) then added multiples of other rows to it (which does not change the determinant).
Besides that, I don't see how you can simplify it any more. It seems to me that the conditions imposed on $p, q,$ and $r$ just make your answer have a positive coefficient for $x$.
A: First, $pr-q^2$ is positive.  What about $px^2+2qxy+ry^2$?  This is the quadratic form $\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}p&q\\q&r\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$.  Whether this is a positive quantity comes down to whether the eigenvalues of the matrix are positive (and whether $x,y$ are nonzero).  The eigenvalues are the roots of $\lambda^2-(p+r)\lambda-(q^2-pr)$, which are $\frac{1}{2}(p+r\pm\sqrt{p^2+4q^2-2pr+r^2})$.  The expression $p^2+4q^2-2pr+r^2$ can be written as $(p+r)^2+4(q^2-pr)<(p+r)^2$, hence the roots are both between $0$ and $p+r$ (exclusive).  The condition $p>0$ and $q^2<pr$ together imply that $r>0$ as well, so the roots are positive.
Thus, the determinant of the matrix is nonnegative, and positive if and only if $x,y$ are not both $0$.
Alternatively, instead of examining eigenvalues of the quadratic form, we can use the quadratic formula along with the intermediate value theorem.  Let's see for which $x,y$ the factor $px^2+2qxy+ry^2$ is $0$.  The quadratic formula gives $x=\frac{-2qy\pm\sqrt{4q^2y^2-4pry^2}}{2p}=y\frac{-q\pm \sqrt{q^2-pr}}{p}$.  The discriminant is negative, so $y$ must be $0$, and so $x$ must be $0$ as well.  Since the substitution $(x,y)=(1,0)$ gives $px^2+2qxy+ry^2=p>0$, the factor is always non-negative, and positive when $(x,y)\neq (0,0)$. (This is because if it were ever negative, a path between $(1,0)$ and the point where it is negative would contain a point where the factor is $0$, by the intermediate value theorem.)
