Consider the matrix f(x)=[[|gathered x - x 0 x x 0 0 0 1 gathered]]|.

Question

Consider the matrix $f(x)=[[|\begin{gathered} \cos x \quad  -\sin x \quad  0 \ \sin x \quad  \cos x \quad  0 \ 0 \quad  0 \quad  1 \end{gathered}]]|$.

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 :

Accepted Answer

$\begin{gathered} f(-x)=[[|\begin{array}{ccc}\cos x \quad  \sin x \quad  0 \ -\sin x \quad  \cos x \quad  0 \ 0 \quad  0 \quad  1\end{array}]]| \ f(x)\cdot f(-x)=[[|\begin{array}{ccc}1 \quad  0 \quad  0 \ 0 \quad  1 \quad  0 \ 0 \quad  0 \quad  1\end{array}]]|=I \end{gathered}$

Hence statement- I is correct

Now, checking statement II

$\begin{gathered} f(y)=[[|\begin{array}{ccc}\cos y \quad  -\sin y \quad  0 \ \sin y \quad  \cos y \quad  0 \ 0 \quad  0 \quad  1\end{array}]]| \ f(x)\cdot f(y)=[[|\begin{array}{ccc}\cos (x+y) \quad  -\sin (x+y) \quad  0 \ \sin (x+y) \quad  \cos (x+y) \quad  0 \ 0 \quad  0 \quad  1\end{array}]]| \ \Rightarrow f(x)\cdot f(y)=f(x+y) \end{gathered}$

Hence statement-II is also correct.

Solutions · 1