If a, b and c are distinct positive real numbers such that Δ 1=| a b c b c a c a b | and Δ 2=| bc-a2 ac-b2 ab-c2 ac-b2 ab-c2 bc-a2 ab-c2 bc-a2 ac-b2 | , then

Question

If a, b and c are distinct positive real numbers such that Δ 1=| a b c b c a c a b | and Δ 2=| bc-a2 ac-b2 ab-c2 ac-b2 ab-c2 bc-a2 ab-c2 bc-a2 ac-b2 | , then (A) Δ 1=Δ 2 (B) Δ 12+Δ 2=0 (C) Δ 12=Δ 2 (D) Δ 12=Δ 22

Tardigrade Admin · Accepted Answer

Elements of Δ 2 are co-factors of the elements of Δ 1 Hence Δ 12=Δ 2

Q. If $a, b$ and $c$ are distinct positive real numbers such that $Δ_{1} = ∣ ∣ a b c b c a c a b ∣ ∣$ and $Δ_{2} = ∣ ∣ b c - a^{2} a c - b^{2} ab - c^{2} a c - b^{2} ab - c^{2} b c - a^{2} ab - c^{2} b c - a^{2} a c - b^{2} ∣ ∣$ , then

$Δ_{1} = Δ_{2}$

$Δ_{1}^{2} + Δ_{2} = 0$

$Δ_{1}^{2} = Δ_{2}$

$Δ_{1}^{2} = Δ_{2}^{2}$

Elements of $Δ_{2}$ are co-factors of the elements of $Δ_{1}$
Hence $Δ_{1}^{2} = Δ_{2}$

Q. If a,b and c are distinct positive real numbers such that Δ1​=∣∣​abc​bca​cab​∣∣​ and Δ2​=∣∣​bc−a2ac−b2ab−c2​ac−b2ab−c2bc−a2​ab−c2bc−a2ac−b2​∣∣​ , then

Δ1​=Δ2​

Δ12​+Δ2​=0

Δ12​=Δ2​

Δ12​=Δ22​