Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

Given that $b^2-ac <0, a > 0$. The value of $Δ=\begin{vmatrix}a&b&ax+by\\b&c&bx+cy\\ax+by&bx+cy&0\end{vmatrix}$, is

Options:

zero

positive

negative

$b^2 + ac$

Correct Answer:

negative

Explanation:

We have,

$Δ=\begin{vmatrix}a&b&ax+by\\b&c&bx+cy\\ax+by&bx+cy&0\end{vmatrix}$

$⇒Δ=\begin{vmatrix}a&b&ax+by\\b&c&bx+cy\\0&0&-(ax^2+2bxy+cy^2)\end{vmatrix}$  [Applying $R_3 → R_3-xR_1-yR_2$]

$⇒Δ=(b^2-ac) (ax^2 + 2bxy + cy^2)$

Now,

$b^2-ac <0$ and $a > 0$

⇒ Discriminant of $ax^2 + 2bxy + cy^2$ is negative and $a > 0$.

$⇒ax^2+2bxy+cy^2 > 0$ for all $x, y ∈ R$.

$⇒ Δ=(b^2-ac) (ax^2 + 2bxy + cy^2) <0$