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NDA 2013
Exam
NDA
Year
2013
Subject
Mathematics
Chapter
Determinants

NDA 2013 · Mathematics · Determinants

Consider the following statements: 1. The matrix ( array ccc 1 & 2...

Consider the following statements: 1. The matrix \left(121a2a1b2b1\begin{array}{ccc}1 & 2 & 1 \\ a & 2 a & 1 \\ b & 2 b & 1\end{array}\right) is singular. 2. The matrix \left(c2c1a2a1b2b1\begin{array}{ccc}c & 2 c & 1 \\ a & 2 a & 1 \\ b & 2 b & 1\end{array}\right) is non-singular. Which of the above statements is/are correct?

Options

  1. A1 only
  2. B2 only
  3. CBoth 1 and 2
  4. DNeither 1 nor 2

Correct answer

A. 1 only

Step-by-step solution

A square matrix is said to be singular if its determinant is zero \& non-singular if its determinant is not equal to zero. ∣121a2a1b2b1∣=∣121a−b2(a−b)01−b2(1−b)0∣{ Applying R2→R2−R3;R3→R1−R3}=(a−b)(1−b)∣121120120∣=(a−b)(1−b)(2−2)=0 It is singular. \begin{aligned} & \left|\begin{array}{ccc} 1 & 2 & 1 \\ a & 2 a & 1 \\ b & 2 b & 1 \end{array}\right|=\left|\begin{array}{ccc} 1 & 2 & 1 \\ a-b & 2(a-b) & 0 \\ 1-b & 2(1-b) & 0 \end{array}\right| \\ & \left\{\text{ Applying } \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3 ; \mathrm{R}_3 \rightarrow \mathrm{R}_1-\mathrm{R}_3\right\} \\ & =(a-b)(1-b)\left|\begin{array}{ccc} 1 & 2 & 1 \\ 1 & 2 & 0 \\ 1 & 2 & 0 \end{array}\right| \\ & =(a-b)(1-b)(2-2)=0 \text{ It is singular. } \end{aligned} Again, ∣c2c1a2a1b2b1∣=∣c2c1a−b2(a−b)0b−c2(b−c)0∣{ Applying R2→R2−R3;R3→R3−R2}=(a−b)(b−c)∣c2c1120120∣=(a−b)(b−c)(2−2)=0\begin{aligned} & \text{ Again, }\left|\begin{array}{lll} c & 2 c & 1 \\ a & 2 a & 1 \\ b & 2 b & 1 \end{array}\right|=\left|\begin{array}{ccc} c & 2 c & 1 \\ a-b & 2(a-b) & 0 \\ b-c & 2(b-c) & 0 \end{array}\right| \\ & \quad\left\{\text{ Applying } \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3 ; \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_2\right\} \\ & =(a-b)(b-c)\left|\begin{array}{ccc} c & 2 c & 1 \\ 1 & 2 & 0 \\ 1 & 2 & 0 \end{array}\right| \\ & =(a-b)(b-c)(2-2)=0 \end{aligned} It is singular.
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