Question

Give that $q^2-pr<0,p>0,$ the value of $\begin{vmatrix} p & q & px+qy \\[0.3em] q & r & qx+ry \\[0.3em] px+qy & qx+ry& 0 \end{vmatrix}$is

• A. 0
• B. > 0
• C. < 0
• D. $q^2+pr$

Answer 1

Answer: (C)

Let $\Delta = \begin{vmatrix}p & q&px+qy\\ q&r&qx+ry\\ px+qy&qx+ry&0\end{vmatrix}$
Operate $R_3-x \, R_1 - y \, R_2$, we get
$\therefore $ $\Delta = \begin{vmatrix}p & q&px+qy\\ q&r&qx+ry\\ 0&0&-(px^2 + 2\,qxy +ry^2)\end{vmatrix}$
= $(q^2 -pr)(px^2 + 2 \, qxy + ry^2)$
Since $q^2 - pr < 0 , p > 0$
$\therefore $ $px^2 + 2 \, qxy + ry^2 > 0$ and hence $\Delta$ is -ve