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NDA 2014
Exam
NDA
Year
2014
Subject
Mathematics
Chapter
Determinants

NDA 2014 · Mathematics · Determinants

One of the roots of | array ccc x+a & b & c a & x+b & c a & b & x+c...

One of the roots of \left|x+abcax+bcabx+c\begin{array}{ccc}x+a & b & c \\ a & x+b & c \\ a & b & x+c\end{array}\right|=0 is

Options

  1. Aabca b c
  2. Ba+b+ca+b+c
  3. C−(a+b+c)-(a+b+c)
  4. D−abc-a b c

Correct answer

C. −(a+b+c)-(a+b+c)

Step-by-step solution

∣x+abcax+bcabx+c∣=0(x+a)[(x+b)(x+c)−bc]−b[a(x+c)−ac]+c[ab−a(x+b)]=(x+a)[x2+xc+bx+bc−bc]−b[ax+ac−ac]+c[ab−ax−ab]=(x+a)[x2+xc+bx]−abx−acx=x3+x2c+bx2+ax2+acx+abx−abx−acx=x3+x2c+bx2+ax2=x2(x+a+b+c)\begin{aligned} & \left|\begin{array}{ccc} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{array}\right|=0 \\ & (x+a)[(x+b)(x+c)-b c]-b[a(x+c)-a c] \\ & +c[a b-a(x+b)] \\ & =(x+a)\left[x^2+x c+b x+b c-b c\right]-b[a x+a c-a c] \\ & +c[a b-a x-a b] \\ & =(x+a)\left[x^2+x c+b x\right]-a b x-a c x \\ & =x^3+x^2 c+b x^2+a x^2+a c x+a b x-a b x-a c x \\ & =x^3+x^2 c+b x^2+a x^2 \\ & =x^2(x+a+b+c) \end{aligned} Now, x2(x+a+b+c)=0x^2(x+a+b+c)=0⇒x+a+b+c=0⇒x=−a−b−c\begin{aligned} & \Rightarrow x+a+b+c=0 \\ & \Rightarrow x=-a-b-c \end{aligned}$Hence,oneoftherootisHence, one of the root is-(a+b+c)$.
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