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If a, b, c are three complex numbers such that $a^{2} + b^{2} + c^{2} = 0$ and $Δ = |\begin{matrix} b^{2} + c^{2} & a b & a c \\ a b & c^{2} + a^{2} & b c \\ a c & b c & a^{2} + b^{2} \end{matrix}| = k a^{2} b^{2} c^{2},$ then the value of k is

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Correct option is D

Using a2+b2+c2=0 ,we can write ∆ as Δ=-a2abacab-b2bcacbc-c2=abc-aaab-bbcc-c [taking a, b, c common from C1, C2, C3 respectively]=a2b2c2-1111-1111-1 [taking a, b, c common from R1, R2, R3 respectively]=a2b2c202120100-1=-2a2b2c2210-1=4a2b2c2 [applying C1→C1+C3 and C3→C2+C3]=a2b2c202120100-1=-2a2b2c2210-1=4a2b2c2 [applying C1 → C1 + C3 and C3→ C2 + C3] Thus, k = 4.

Similar Questions

If

$|\begin{array}{ccc} a^{2} & b^{2} & c^{2} \\ (a + λ)^{2} & (b + λ)^{2} & (c + λ)^{2} \\ (a - λ)^{2} & (b - λ)^{2} & (c - λ)^{2} \end{array}| = k λ |\begin{array}{ccc} a^{2} & b^{2} & c^{2} \\ a & b & c \\ 1 & 1 & 1 \end{array}|$

$λ \neq 0$ then k is equal to:

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