If $α, β, γ$ are the roots of $p x^{3} + q x^{2} + r = 0$ , then the value of the determinant $∣ ∣ ∣ ∣ ∣ \begin{matrix} α β & β γ & γ α β γ & γ α & α β γ α & α β & β γ \end{matrix} ∣ ∣ ∣ ∣ ∣$ is

p

No worries! We‘ve got your back. Try BYJU‘S free classes today!

q

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $0$
Given $α, β, γ$ are the roots of $p x^{3} + q x^{2} + r = 0$
$\Rightarrow (α β + β γ + γ α) = 0$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} α β & β γ & γ α β γ & γ α & α β γ α & α β & β γ \end{matrix} ∣ ∣ ∣ ∣ ∣$
$C_{1} \to C_{1} + C_{2} + C_{3}$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} α β + β γ + γ α & β γ & γ α α β + β γ + γ α & γ α & α β α β + β γ + γ α & α β & β γ \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= (α β + β γ + γ α) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & β γ & γ α 1 & γ α & α β 1 & α β & β γ \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= 0$

Suggest Corrections

Similar questions

If α, β and γ are the roots of {px}^{3} + {qx}^{2} + r = 0, then the value of ⎛ ⎜ ⎝ \begin{matrix} αβ & βγ & γα βγ & γα & αβ γα & αβ & βγ \end{matrix} ∣ ∣ ∣ ∣ ∣

is

If α, β and γ are the roots of {px}^{3} + {qx}^{2} + r = 0, then the value of ⎛ ⎜ ⎝ \begin{matrix} αβ & βγ & γα βγ & γα & αβ γα & αβ & βγ \end{matrix} ∣ ∣ ∣ ∣ ∣

is

(α, β), (β, γ), (γ, α)

x^{2} - 2 p x + 2 = 0; x^{2} - 2 q x + 3 = 0; x^{2} - 2 r x + 6 = 0

α, β, γ

p + q + r

α, β, γ

x^{3} + p x + q = 0

∣ ∣ ∣ ∣ ∣ \begin{matrix} α & β & γ β & γ & α γ & α & β \end{matrix} ∣ ∣ ∣ ∣ ∣

α, β, γ

x^{3} + p x + q = 0

p, q \neq= 0

∣ ∣ ∣ ∣ ∣ \begin{matrix} α & β & γ β & γ & α γ & α & β \end{matrix} ∣ ∣ ∣ ∣ ∣

Join BYJU'S Learning Program