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JEE Main 201219 MayOnline
Exam
JEE Main
Year
2012
Date
19 May
Mode
Online
Subject
Mathematics
Chapter
Determinants

JEE Main 2012 · 19 May · Online · Mathematics · Determinants

If a, b, c , are non zero complex numbers satisfying a^2+b^2+c^2=0...

If a,b,ca, b, c, are non zero complex numbers satisfying a^2+b^2+c^2=0 and \left|b2+c2abacabc2+a2bcacbca2+b2\begin{array}{ccc}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{array}\right|=k a^2 b^2 c^2, then kk is equal to

Options

  1. A1
  2. B3
  3. C4
  4. D2

Correct answer

C. 4

Step-by-step solution

Let \Delta=\left|b2+c2abacabc2+a2bcacbca2+b2\begin{array}{ccc}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{array}\right| Multiply C1C_1 by a,C2a, C_2 by b and C3C_3 by candhencedivideby\mathrm{c} and hence divide bya b c.=. =\frac{1}{a b c}\left|a(b2+c2ab2)ac2a2bb(c2+a2bc2)a2cb2cc(a2+b2\begin{array}{ccc}a\left(b^2+c²\right. & \left.a b²\right) & a c² \\ a² b & b\left(c^2+a²\right. & \left.b c²\right) \\ a² c & b² c & c\left(a^2+b²\right. \end{array}\right| Take out a,b,ca, b, c common from R1,R2R_1, R_2 and R3R_3 respectively. ∴Δ=abcabc∣b2+c2b2c2a2c2+a2c2a2b2a2+b2∣ Apply C1→C1−C2−C3Δ=∣0b2c2−2c2c2+a2c2−2b2b2a2+b2∣=−2∣0b2c2c2c2+a2c2b2b2a2+b2∣\begin{aligned}& \therefore \quad \Delta=\frac{a b c}{a b c}\left|\begin{array}{ccc}b^2+c² & b² & c² \\ a² & c²+a² & c² \\ a² & b² & a²+b² \end{array}\right| \\ & \text{ Apply }C_1 → C_1-C_2-C_3 \\ & \Delta=\left|\begin{array}{ccc}0 & b² & c² \\ -2 c² & c²+a² & c² \\ -2 b² & b² & a²+b² \end{array}\right| \\ & =-2\left|\begin{array}{ccc}0 & b² & c² \\ c² & c²+a² & c² \\ b² & b² & a²+b² \end{array}\right| \\ & \end{aligned}Apply C2−C1C_2-C_1 and C3−C1C_3-C_1=−2∣0b2c2c2a20b20a2∣=−2[−b2(c2a2)+c2(−a2b2)]=2a2b2c2+2a2b2c2=4a2b2c2 But Δ=ka2b2c2∴k=4\begin{aligned}& =-2\left|\begin{array}{ccc}0 & b² & c² \\ c² & a² & 0 \\ b² & 0 & a² \end{array}\right| \\ & =-2\left[-b^2\left(c^2 a²\right)+c²\left(-a^2 b²\right)\right] \\ & =2 a² b² c²+2 a² b² c²=4 a² b² c² \\ & \text{ But } \Delta=k a² b² c² \therefore k=4 \end{aligned}$
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