If a 2 + b 2 + c 2 + ab + b c + c a ≤ 0 , evaluate the determinant: det ( a + b + 2 ) 2 1 c 2 + a 2 a 2 + b 2 ( b + c + 2 ) 2 1 1 b 2 + c 2 ( c + a + 2 ) 2

Twice the condition is ( a + b ) 2 + ( b + c ) 2 + ( c + a ) 2 ≤ 0 , so a = b = c = 0 . The determinant is 65 .

Mathematics - Determinant Question with Solution | TestHub

MathematicsDeterminantSpecial det.Medium2 minQB

MathematicsMediumsingle choice

If $a^{2} + b^{2} + c^{2} + ab + b c + c a \leq 0$ , evaluate the determinant:

$det (a + b + 2)^{2} 1 c^{2} + a^{2} a^{2} + b^{2} (b + c + 2)^{2} 1 1 b^{2} + c^{2} (c + a + 2)^{2}$

Answer:

Solution:

Twice the condition is $(a + b)^{2} + (b + c)^{2} + (c + a)^{2} \leq 0$ , so $a = b = c = 0$ . The determinant is 65 .

Stream:JEESubject:MathematicsTopic:DeterminantSubtopic:Special det.

⏱ 2mℹ️ Source: QB

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