Question

f $b^2 - ac < 0, a < 0$ then the value of
$\begin{vmatrix}a&b&ax+by\\ b&c&bx+cy\\ ax+by&bx+cy&0\end{vmatrix}$ is

• A. Zero
• B. Positive
• C. Negative
• D. $b^2 + ac$

Answer 1

Answer: (B)

Let $\Delta = \begin{vmatrix}a&b&ax+by\\ b&c&bx+cy\\ ax+by&bx+cy&0\end{vmatrix}$
$= \begin{vmatrix}a&b&ax+by\\ b&c&bx+cy\\ 0&0&-\left(ax^{2}+2bxy+cy^{2}\right)\end{vmatrix}$
[Applying $R_{3} \to R_{3} - xR_{1} -yR_{2}$]
$= \left(b^{2}-ac\right)\left(ax^{2}+2bxy +cy^{2}\right)$
Now, $ b^{2} - ac < 0$ and $a < 0$
$\Rightarrow $ Discriminant of $ax^{2} + 2bxy + cy^{2}$ is negative and $a < 0$.
$\Rightarrow ax^{2} + 2bxy + cy^{2} < 0$ for all $x, \,y \in R$ [See Quadratics]
$\Rightarrow \Delta = \left(b^{2}-ac\right) \left(ax^{2}+2bxy +cy^{2}\right) > 0.$