Q46 / 160 · AP EAMCET 2022 (05 Jul Shift 1) · Mathematics

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 $G(x) G(y)=$

  • A
    null Matrix
  • B
    skew-symmetric Matrix
  • Correct answer
    C
    identity Matrix
  • D
    symmetric Matrix

Solution

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

Hence, the correct answer is option: 'C'.

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