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NDA 2013
Exam
NDA
Year
2013
Subject
Mathematics
Chapter
Determinants

NDA 2013 · Mathematics · Determinants

What is the value of the determinant | array lll 1 & b c & a(b+c) 1...

What is the value of the determinant \left|1bca(b+c)1cab(c+a)1abc(a+b)\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right| ?

Options

  1. A0
  2. Babc
  3. Cab +bc +ca
  4. Dabc (a +b +c)

Correct answer

A. 0

Step-by-step solution

∣1bca(b+c)1cab(c+a)1abc(a+b)∣&=∣1bcab+bc+ca1cabc+ab+ca1abac+bc+ab∣(ApplyingC3=C3+C2)=(ab+bc+ca)∣1bc11ca11ab1∣=(ab+bc+ca)∣1bc01ca01ab0∣(ApplyingC3=C3−C1)=(ab+bc+ca)=0\begin{aligned} & \left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right| \& =\left|\begin{array}{lll}1 & b c & a b+b c+c a \\ 1 & c a & b c+a b+c a \\ 1 & a b & a c+b c+a b{} \end{array}\right|\quad{(Applying \quad C_3=C_3+C_2)} \\ & =(a b+b c+c a)\left|\begin{array}{lll}1 & b c & 1 \\ 1 & c a & 1 \\ 1 & a b & 1\end{array}\right| \\ & =(a b+b c+c a)\left|\begin{array}{lll}1 & b c & 0 \\ 1 & c a & 0 \\ 1 & a b & 0\end{array}\right|\quad{(Applying \quad C_3=C_3-C_1)} \\ & =(a b+b c+c a)=0\end{aligned}
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