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NDA 2011
Exam
NDA
Year
2011
Subject
Mathematics
Chapter
Determinants

NDA 2011 · Mathematics · Determinants

. If the matrix A=( array ccc 2-x & 1 & 1 1 & 3-x & 0 -1 & -3 & -x...

. If the matrix A=\left(2−x1113−x0−1−3−x\begin{array}{ccc}2-x & 1 & 1 \\ 1 & 3-x & 0 \\ -1 & -3 & -x\end{array}\right) is singular, then what is the solution set SS ?

Options

  1. AS={0,2,3}S=\{0,2,3\}
  2. BS={−1,2,3}S=\{-1,2,3\}
  3. CS={1,2,3}S=\{1,2,3\}
  4. DS={2,3}S=\{2,3\}

Correct answer

A. S={0,2,3}S=\{0,2,3\}

Step-by-step solution

Since the matrix is singular therefore, \left|2−x1113−x0−1−3−x\begin{array}{ccc} 2-x & 1 & 1 \\ 1 & 3-x & 0 \\ -1 & -3 & -x \end{array}\right|=0⇒R2→(R3+R2)∣2−x1113−x00−x−x∣=0\begin{aligned} & \Rightarrow \mathrm{R}_2 \rightarrow\left(\mathrm{R}_3+\mathrm{R}_2\right) \\ & \left|\begin{array}{ccc}2-x & 1 & 1 \\ 1 & 3-x & 0 \\ 0 & -x & -x\end{array}\right|=0\end{aligned}$\begin{aligned} & \left.\Rightarrow(2-x)\left\{-3 x+x^2\right)\right\}-1(-x+x)=0 \\ & \Rightarrow(2-x)\left(x^2-3 x\right)=0 \\ & \Rightarrow x(2-x),(x-3)=0 \\ & \Rightarrow x=0, x=2, x=3\end{aligned}$
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