UPSC NDA I 2022 Mathematics Solved Paper

Concept:A matrix is singular when its determinant is $0$. If two rows (or columns) of a matrix are identical, its determinant is $0$.Explanation:Start with the given matrix: $$A = \begin{bmatrix} 2\cos 2\theta & 2\cos 2\theta & 6 \\ 1 - 2\sin^2\theta & 2\cos^2\theta - 1 & 3 \\ k & 2k & 1 \end{bmatrix}$$ Use the identity $\cos 2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$. So the second row becomes $[\cos 2\theta \quad \cos 2\theta \quad 3]$. Factor $2$ from the first row: $$A = 2 \begin{bmatrix} \cos 2\theta & \cos 2\theta & 3 \\ \cos 2\theta & \cos 2\theta & 3 \\ k & 2k & 1 \end{bmatrix}$$ Now the first and second rows are identical. Therefore, the determinant $|A| = 0$ for any value of $k$.Answer:The matrix is singular for any real value of $k$. Option D: Any real value.