UPSC NDA 2 2016 Mathematics Solved Paper

Concept:Matrix multiplication and determinant properties using trigonometric identities and even function definition.Explanation:Given $f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$.Statement 1: $f(\theta) \times f(\phi) = f(\theta + \phi)$.Multiply the two matrices. The $(1,1)$ entry is $\cos\theta\cos\phi - \sin\theta\sin\phi = \cos(\theta+\phi)$. The $(1,2)$ entry is $-\cos\theta\sin\phi - \sin\theta\cos\phi = -\sin(\theta+\phi)$. The $(2,1)$ entry is $\sin\theta\cos\phi + \cos\theta\sin\phi = \sin(\theta+\phi)$. The $(2,2)$ entry is $-\sin\theta\sin\phi + \cos\theta\cos\phi = \cos(\theta+\phi)$. The bottom row and column give $0$ and $1$ as required. Thus the product equals $f(\theta+\phi)$, so statement 1 is correct.Statement 2: Determinant of $f(\theta) \times f(\phi)$ is $1$.Since $f(\theta) \times f(\phi) = f(\theta+\phi)$, its determinant is $\det f(\theta+\phi) = \cos^2(\theta+\phi) + \sin^2(\theta+\phi) = 1$. Alternatively, $\det f(\theta) = \det f(\phi) = 1$, so product determinant is $1 \times 1 = 1$. Hence statement 2 is correct.Statement 3: The determinant of $f(x)$ is an even function.$\det f(x) = \cos^2 x + \sin^2 x = 1$. Replacing $x$ with $-x$ gives $\det f(-x) = \cos^2(-x) + \sin^2(-x) = \cos^2 x + \sin^2 x = 1 = \det f(x)$. Since $\det f(-x) = \det f(x)$ for all $x$, it is an even function. Thus statement 3 is correct.All three statements are correct.Answer: D. 1, 2 and 3