Consider the matrix $f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1…
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.
Statement I is false but Statement II is true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Both Statement I and Statement II are true
Solution
Given: $f(x) = \begin{aligned} \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned}$
$\Rightarrow f(-x) = \begin{aligned} \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned}$
$\Rightarrow f(x) \times f(-x) = \begin{aligned} \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned}$
$\Rightarrow f(x) \times f(-x) = \begin{aligned} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned}$
$\Rightarrow f(x) \times f(-x) = I$
$\Rightarrow f(-x) = f^{-1}(x)$
Hence statement-I is correct
Now, checking statement II
$\Rightarrow f(y) = \begin{aligned} \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned}$
$\Rightarrow f(x)f(y) = \begin{aligned} \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned}$
$\Rightarrow f(x)f(y) = \begin{aligned} \begin{bmatrix} \cos x \cos y - \sin x \sin y & -\cos x \sin y - \sin x \cos y & 0 \\ \sin x \cos y + \cos x \sin y & -\sin x \sin y + \cos x \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned}$
$\Rightarrow f(x)f(y) = \begin{aligned} \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned}$
$\Rightarrow f(x) \cdot f(y) = f(x+y)$
Hence statement-II is also correct.