Consider the matrix f(x)= [ ccc x & - x & 0 x & x & 0 0 & 0 & 1 ]… - Vidyadip

MCQ

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

A
Statement $I$ is false but Statement $II$ is true
B
Both Statement $I$ and Statement $II$ are false
C
Statement $I$ is true but Statement $II$ is false
✓
Both Statement $I$ and Statement $II$ are true

✓

Answer

Correct option: D.

Both Statement $I$ and Statement $II$ are true

d
$f(-x)=\left[\begin{array}{ccc}\cos x & \sin x & 0 \\-\sin x & \cos x & 0 \\0 & 0 & 1\end{array}\right]$

$f(x) \cdot f(-x)=\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]=I$

Hence statement- $I$ is correct

Now, checking statement $II$

$f(y)=\left[\begin{array}{ccc}\cos y & -\sin y & 0 \\\sin y & \cos y & 0 \\0 & 0 & 1\end{array}\right]$

$f(x) \cdot f(y)=\left[\begin{array}{ccc}\cos (x+y) & -\sin (x+y) & 0 \\\sin (x+y) & \cos (x+y) & 0 \\0 & 0 & 1\end{array}\right]$

$ \Rightarrow f(x) \cdot f(y)=f(x+y)$

Hence statement -$II$ is also correct.

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