Consider the matrix $f(x)=\left[\begin{array}{ccc}\cos <!--\; \; -->x& -<!--\; -\; -->\sin <!--\; \; -->x& 0\\ \sin <!--\; \; -->x& \cos <!--\; \; -->x& 0\\ 0& 0& 1\end{array}\right]$. 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

The correct option is: Both Statement I and Statement II are true.

Let us analyze both statements step-by-step by working with the given matrix:
$f(x)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$

Step 1: Verify Statement I
First, we find $f(-x)$ by substituting $-x$ in place of $x$.
Using the trigonometric identities $\cos (-x)=\cos x$ and $\sin (-x)=-\sin x$, we have:
$f(-x)=\left[\begin{array}{ccc}\cos (-x)& -\sin (-x)& 0\\ \sin (-x)& \cos (-x)& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}\cos x& \sin x& 0\\ -\sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$

To check if $f(-x)$ is the inverse of $f(x)$, we multiply the two matrices:
$f(x)\cdot f(-x)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\cos x& \sin x& 0\\ -\sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$
Multiplying row by column:
- Row 1, Column 1: ${\cos }^{2}x+{\sin }^{2}x+0=1$
- Row 1, Column 2: $\cos x\sin x-\sin x\cos x+0=0$
- Row 1, Column 3: $0+0+0=0$
- Row 2, Column 1: $\sin x\cos x-\cos x\sin x+0=0$
- Row 2, Column 2: ${\sin }^{2}x+{\cos }^{2}x+0=1$
- Row 2, Column 3: $0+0+0=0$
- Row 3: Since the third row of the first matrix is $[0,0,1]$ and the third column of the second matrix is ${[0,0,1]}^T$, the third row of the product yields $[0,0,1]$.
Thus:
$f(x)\cdot f(-x)=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I$
Since the product is the identity matrix $I$, we have ${(f(x))}^{-1}=f(-x)$. Therefore, Statement I is true.

Step 2: Verify Statement II
Now let us compute the product $f(x)f(y)$:
$f(x)f(y)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\cos y& -\sin y& 0\\ \sin y& \cos y& 0\\ 0& 0& 1\end{array}\right]$
Let's find the entries of the product matrix:
- Entry (1,1): $\cos x\cos y-\sin x\sin y+0=\cos (x+y)$
- Entry (1,2): $-\cos x\sin y-\sin x\cos y+0=-(\sin x\cos y+\cos x\sin y)=-\sin (x+y)$
- Entry (1,3): $0+0+0=0$
- Entry (2,1): $\sin x\cos y+\cos x\sin y+0=\sin (x+y)$
- Entry (2,2): $-\sin x\sin y+\cos x\cos y+0=\cos (x+y)$
- Entry (2,3): $0+0+0=0$
- Entry (3,1): $0+0+0=0$
- Entry (3,2): $0+0+0=0$
- Entry (3,3): $0+0+1=1$
Putting these elements together in a matrix gives:
$f(x)f(y)=\left[\begin{array}{ccc}\cos (x+y)& -\sin (x+y)& 0\\ \sin (x+y)& \cos (x+y)& 0\\ 0& 0& 1\end{array}\right]=f(x+y)$
Therefore, Statement II is also true.

Conclusion
Both Statement I and Statement II are true.