Showing that a certain symmetric 3-by-3 determinant is negative

Question

Given that $q^2$$ - pr <0$ and $p>0$, we have to prove that $$\left| \begin{array}{ccc} p & q & px+qy \\ q & r & qx+ry \\ px+qy & qx+ry & 0 \\ \end{array}\right|<0.$$ I tried to solve the determinant, but didn't get a result and wasn't able to use the inequalities.

Answer 1

Hint 1: The value of a determinant does not change under a row-multiply operation, i.e. a multiple of one row is subtracted from another. Two such operations can be used to simplify the last row of your matrix and thereby get a result which is easier to compute.

Hint 2: Ultimately, you'll need to show the that the quadratic $p x^2 + 2 q x y + r y^2$ is nonnegative. This is most easily established by multiplying by $p$ (which is positive!) and then completing the square in $x$; the bound should then be obvious.

Answer 2

$$D=\begin{vmatrix} p & q & px+qy \\ q & r & qx+ry \\ px+qy & qx+ry & 0 \\ \end{vmatrix}= 2q(px+qy)(qx+ry)-r(px+qy)^{2}-p(qx+ry)^2$$

once $q^2<pr$ and $p>0$ then $r>0$ $\Rightarrow$ $-\sqrt{pr}<q<\sqrt{pr}$. Then

If $(px+qy)(qx+ry)>0$ we have:

$$D>-2\sqrt{pr}(px+qy)(qx+ry)-r(px+qy)^{2}-p(qx+ry)^2$$

$$D<2\sqrt{pr}(px+qy)(qx+ry)-r(px+qy)^{2}-p(qx+ry)^2$$

what give us

$$−[\sqrt{r}(px+qy)+\sqrt{p}(qx+ry)]^2<D<−[\sqrt{r}(px+qy)-\sqrt{p}(qx+ry)]^2$$

If $(px+qy)(qx+ry)<0$ we get:

$$−[\sqrt{r}(px+qy)-\sqrt{p}(qx+ry)]^2<D<−[\sqrt{r}(px+qy)+\sqrt{p}(qx+ry)]^2$$