Search any question & find its solution
Question:
Answered & Verified by Expert
Let $G(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$. If $x+y=0$, then $G(x) G(y)=$
Options:
Solution:
1377 Upvotes
Verified Answer
The correct answer is:
identity Matrix
Here, $G(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$
$G(y)=\left[\begin{array}{ccc}\cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1\end{array}\right]$
So, $x+y=0 \Rightarrow y=(-x)$
Now, $G(-x)=\left[\begin{array}{ccc}\cos (-x) & -\sin (-x) & 0 \\ \sin (-x) & \cos (-x) & 0 \\ 0 & 0 & 1\end{array}\right]$
$G(x) \cdot G(y)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}\cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I$
$G(y)=\left[\begin{array}{ccc}\cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1\end{array}\right]$
So, $x+y=0 \Rightarrow y=(-x)$
Now, $G(-x)=\left[\begin{array}{ccc}\cos (-x) & -\sin (-x) & 0 \\ \sin (-x) & \cos (-x) & 0 \\ 0 & 0 & 1\end{array}\right]$
$G(x) \cdot G(y)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}\cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.