Q156 / 270 · NDA 2022 (Phase 1) · Mathematics

For what value of $k$ is the matrix $\left[\begin{array}{ccc}2 \cos 2 \theta & 2 \cos 2 \theta & 6 \\ 1-2 \sin ^2 \theta & 2 \cos ^2 \theta-1 & 3 \\ k & 2 k & 1\end{array}\right]$ singular?

  • A
    0 only
  • B
    1 only
  • C
    2 only
  • Correct answer
    D
    Any real value

Solution

$\left|\begin{array}{ccc}2 \cos 2 \theta & 2 \cos 2 \theta & 6 \\ 1-2 \sin ^2 \theta & 2 \cos ^2 \theta-1 & 3 \\ k & 2 k & 1\end{array}\right|$ $=\left|\begin{array}{ccc}2 \cos ^2 \theta & 2 \cos ^2 \theta & 6 \\ \cos ^2 \theta & \cos ^2 \theta & 3 \\ k & 2 k & 1\end{array}\right|=0$ $\left(\because R_1\right.$ and $R_2$ are identical $)$ Hence, for any real value of $k$ the given matrix is singular.

Hence, the correct answer is option: 'D'.

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