Consider the matrix $$f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$$. Given below are two statements : Statement I: $$f(-x)$$ is the inverse of the matrix $$f(x)$$. Statement II: $$f(x) f(y) = f(x + y)$$. In the light of the above statements, choose the correct answer from the options given below

We need to verify two statements about the matrix $$f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$$.

We observe that this is a rotation matrix representing a rotation by angle $$x$$ about the z-axis.

Let us first analyze the claim that $$f(-x)$$ is the inverse of $$f(x)$$.

First, we compute $$f(-x)$$:

$$ f(-x) = \begin{bmatrix} \cos(-x) & -\sin(-x) & 0 \\ \sin(-x) & \cos(-x) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

Next, we compute the product $$f(x) \cdot f(-x)$$:

$$ f(x) \cdot f(-x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

Entry (1,1): $$\cos^2 x + \sin^2 x = 1$$

Entry (1,2): $$\cos x \sin x - \sin x \cos x = 0$$

Entry (2,1): $$\sin x \cos x - \cos x \sin x = 0$$

Entry (2,2): $$\sin^2 x + \cos^2 x = 1$$

All other entries work out to give the identity matrix.

$$ f(x) \cdot f(-x) = I_3 $$

Hence, $$f(-x) = [f(x)]^{-1}$$, proving the first statement.

Now consider the second statement: $$f(x)f(y) = f(x+y)$$.

We compute the product $$f(x) \cdot f(y)$$:

$$ f(x) \cdot f(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

Entry (1,1): $$\cos x \cos y - \sin x \sin y = \cos(x+y)$$

Entry (1,2): $$-\cos x \sin y - \sin x \cos y = -\sin(x+y)$$

Entry (2,1): $$\sin x \cos y + \cos x \sin y = \sin(x+y)$$

Entry (2,2): $$-\sin x \sin y + \cos x \cos y = \cos(x+y)$$

$$ f(x) \cdot f(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} = f(x+y) $$

Thus, the second statement also holds.

The correct answer is Option (4): Both Statement I and Statement II are true.