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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Determinants

NDA 2016 · Mathematics · Determinants

The system of linear equations kx+y+z=1, x+ky+z=1 and x+y+k z=1 has...

The system of linear equations kx+y+z=1,x+ky+z=1\mathrm{kx}+\mathrm{y}+\mathrm{z}=1, \mathrm{x}+\mathrm{ky}+\mathrm{z}=1 and x+y+kz=1x+y+k z=1 has a unique solution under which one of the following conditions?

Options

  1. Ak≠1\mathrm{k} \neq 1 and k≠−2\mathrm{k} \neq-2
  2. Bk≠1\mathrm{k} \neq 1 and k≠2\mathrm{k} \neq 2
  3. Ck≠−1\mathrm{k} \neq-1 and k≠−2\mathrm{k} \neq-2
  4. Dk≠−1k \neq-1 and k≠2k \neq 2

Correct answer

A. k≠1\mathrm{k} \neq 1 and k≠−2\mathrm{k} \neq-2

Step-by-step solution

Linear equations Kx+y+z=1x+Ky+z=1x+y+Kz=1[K111K111K][xyz]=[111]AX=BX=A−1B\begin{aligned} & K x+y+z=1 \\ & x+K y+z=1 \\ & x+y+K z=1 \\ & {\left[\begin{array}{ccc} K & 1 & 1 \\ 1 & K & 1 \\ 1 & 1 & K \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]} \\ & \quad A X=B \\ & X=A^{-1} B \end{aligned} Linear equation will have unique solution when A^{-1}exist:exist :∣A∣≠0[K111K111K]≠0K(K2−1)−1(K−1)+1(1−K)≠0⇒K3−K−K+1+1−K≠0⇒K3−3K+2≠0\begin{aligned} & |A| \neq 0 \\ & {\left[\begin{array}{ccc} K & 1 & 1 \\ 1 & K & 1 \\ 1 & 1 & K \end{array}\right] \neq 0} \\ & K\left(K^2-1\right)-1(K-1)+1(1-K) \neq 0 \\ & \Rightarrow K^3-K-K+1+1-K \neq 0 \\ & \Rightarrow K^3-3 K+2 \neq 0 \end{aligned}$\begin{aligned} & (K-1)\left(K^2+K-2\right) \neq 0 \\ & (K-1)(K-1)(K+2) \neq 0 \\ & K \neq 1, K \neq 1 \text{ and } K \neq-2 \\ & \Rightarrow K \neq 1 \text{ and } K \neq-2\end{aligned}$
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