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JEE Main 201212 MayOnline
Exam
JEE Main
Year
2012
Date
12 May
Mode
Online
Subject
Mathematics
Chapter
Determinants

JEE Main 2012 · 12 May · Online · Mathematics · Determinants

If | array ccc -2 a & a+b & a+c b+a & -2 b & b+c c+a & b+c & -2 c...

If \left|−2aa+ba+cb+a−2bb+cc+ab+c−2c\begin{array}{ccc}-2 a & a+b & a+c \\ b+a & -2 b & b+c \\ c+a & b+c & -2 c\end{array}$\right|=\alpha(a+b() b+c() c+a) ≠ 0 then α\alpha is equal to$

Options

  1. Aa+b+ca+b+c
  2. Babca b c
  3. C4
  4. D1

Correct answer

C. 4

Step-by-step solution

Let \Delta=\left|−2aa+ba+cb+a−2bb+cc+ab+c−2c\begin{array}{ccc}-2 a & a+b & a+c \\ b+a & -2 b & b+c \\ c+a & b+c & -2 c\end{array}\right| Applying C1+C3andC_1+C_3 andC_2+C_3\Delta=\left|−a+c2a+b+ca+c2b+a+c−b+cb+ca−cb−c−2c\begin{array}{ccc} -a+c & 2 a+b+c & a+c \\ 2 b+a+c & -b+c & b+c \\ a-c & b-c & -2 c \end{array}\right| Now, applying R1+R3andR_1+R_3 andR_2+R_3\Delta=\left|02(a+b)a−c2(a+b0b−ca−cb−c−2c\begin{array}{ccc}0 & 2(a+b & ) a-c \\ 2(a+b & 0 & b-c \\ a-c & b-c & -2 c \end{array}\right| On expanding, we get Δ=−2(a+b){−2c[2(a+b)]−(a−c)(b−c)}+(a−c)[2(a+b)(b−c)]Δ=8c(a+b)(a+b)+4(a+b)(a−c)(b−c)\begin{aligned}\Delta=& -2(a+b)\{-2 c[2(a+b)] \\ & -(a-c)(b-c)\} \\ & +(a-c)[2(a+b)(b-c)] \\ \Delta=& 8 \mathrm{c}(a+b)(a+b) \\ & +4(a+b)(a-c)(b-c) \end{aligned}=4(a+b)[2ac+2bc+ab−bc−ac+c2]=4(a+b)[ac+bc+ab+c2]=4(a+b)[c(a+c)+b(a+c)]=4(a+b)(b+c)(c+a)=α(a+b)(b+c)(c+a)\begin{aligned}& =4(a+b)\left[2 a c+2 b c+a b-b c-a c+c²\right] \\ & =4(a+b)\left[a c+b c+a b+c²\right] \\ & =4(a+b)[c(a+c)+b(a+c)] \\ & =4(a+b)(b+c)(c+a) \\ & =\alpha(a+b)(b+c)(c+a) \end{aligned}Hence, α=4\alpha=4
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