If a 2- b 2- c 2- ab - bc - ca - 0 - a - b - c - R then value of the determinant beginvmatrix a-b-2- 2 - a- 2-b- 2 - 11 -b-c-2- 2 - b- 2-c- 2c - 2- a - 2 - 1 - c - a -2- 2lendvmatrix isA- 65B- a2-b2-c2-3C- 4a2-b2-c2D- 0

The correct option is A 65We have, a^2 +b^2 +c^2 ab bc ca ≤ 0 ⇒1/2(2a^2 +2b^2 +2c^2 2ab 2bc 2 ca) ≤ 0 ⇒1/2[(a b)^2 +(b c)^2 +(c a)^2] ≤ 0 Sum of whole squa ...

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Byju's Answer

Standard XII

Mathematics

Evaluation of a Determinant

If a 2+ b 2+ ...

Question

If $a^{2} + b^{2} + c^{2} - a b - b c - c a \leq 0 \forall a, b, c \in R$ then value of the determinant $∣ ∣ ∣ ∣ ∣ \begin{matrix} (a - b + 2)^{2} & a^{2} - b^{2} & 1 1 & (b - c + 2)^{2} & b^{2} - c^{2} c^{2} - a^{2} & 1 & (c - a + 2)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$ is

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a^{2} + b^{2} + c^{2} + 3

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4 (a^{2} + b^{2} + c^{2})

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0.0

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Solution

The correct option is A 65
We have,
$a^{2} + b^{2} + c^{2} - a b - b c - c a \leq 0$
$\Rightarrow \frac{1}{2} (2 a^{2} + 2 b^{2} + 2 c^{2} - 2 a b - 2 b c - 2 c a) \leq 0$
$\Rightarrow \frac{1}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] \leq 0$
Sum of whole squares cannot be less than zero.
$∴ \frac{1}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] = 0$
So a = b = c
Putting a = b =c in the given determinant, we get
So $∣ ∣ ∣ ∣ \begin{matrix} 4 & 0 & 1 1 & 4 & 0 0 & 1 & 4 \end{matrix} ∣ ∣ ∣ ∣ = 4 (16) + 1 (1) = 65$

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