Question:
Let
\[
G(x)=
\begin{bmatrix}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{bmatrix}.
\]
If \(x+y=0\), then \(G(x)G(y)=\)
Let
\[
G(x)=
\begin{bmatrix}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{bmatrix}.
\]
If \(x+y=0\), then \(G(x)G(y)=\)
Show Hint
For rotation matrices,
\[
G(x)G(y)=G(x+y).
\]
So, if
\[
x+y=0,
\]
then
\[
G(x)G(y)=G(0)=I.
\]
Updated On: Jun 22, 2026
- Null Matrix
- Skew Symmetric Matrix
- Identity Matrix
- Symmetric Matrix
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The Correct Option is C
Solution and Explanation
Step 1: Identify the matrix type.
The given matrix is \[ G(x)= \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \] This is a rotation matrix form.
Step 2: Use the multiplication property.
For such matrices, \[ G(x)G(y)=G(x+y) \]
Step 3: Apply the given condition.
Given, \[ x+y=0 \] Therefore, \[ G(x)G(y)=G(0) \]
Step 4: Find \(G(0)\).
\[ G(0)= \begin{bmatrix} \cos 0 & -\sin 0 & 0 \\ \sin 0 & \cos 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] Since \[ \cos 0=1 \] and \[ \sin 0=0 \] we get \[ G(0)= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Step 5: Final conclusion.
Thus, \[ G(x)G(y)=I \] Therefore, \[ \boxed{\text{Identity Matrix}} \]
The given matrix is \[ G(x)= \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \] This is a rotation matrix form.
Step 2: Use the multiplication property.
For such matrices, \[ G(x)G(y)=G(x+y) \]
Step 3: Apply the given condition.
Given, \[ x+y=0 \] Therefore, \[ G(x)G(y)=G(0) \]
Step 4: Find \(G(0)\).
\[ G(0)= \begin{bmatrix} \cos 0 & -\sin 0 & 0 \\ \sin 0 & \cos 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] Since \[ \cos 0=1 \] and \[ \sin 0=0 \] we get \[ G(0)= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Step 5: Final conclusion.
Thus, \[ G(x)G(y)=I \] Therefore, \[ \boxed{\text{Identity Matrix}} \]
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