Question:easy

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 \(G(x)G(y)=\)

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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: Write the rotation matrix $G(x)$.
The standard 2D rotation matrix is $G(x) = \begin{pmatrix}\cos x & -\sin x\\ \sin x & \cos x\end{pmatrix}$, which rotates vectors by angle $x$.
Step 2: Recall the multiplication property of rotation matrices.
A fundamental property of rotation matrices is: $G(x) \cdot G(y) = G(x+y)$. This is because rotating by $x$ and then by $y$ is equivalent to rotating by $x+y$.
Step 3: Apply the condition $x + y = 0$.
Given that $x + y = 0$, we substitute directly: \[G(x) G(y) = G(x+y) = G(0).\]
Step 4: Evaluate $G(0)$.
\[G(0) = \begin{pmatrix}\cos 0 & -\sin 0\\ \sin 0 & \cos 0\end{pmatrix} = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} = I.\]
Step 5: Verify using direct matrix multiplication.
With $y = -x$: $G(y) = G(-x) = \begin{pmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{pmatrix}$. Then $G(x)G(-x) = \begin{pmatrix}\cos^2 x + \sin^2 x & \cos x \sin x - \sin x \cos x\\ \sin x \cos x - \cos x \sin x & \sin^2 x + \cos^2 x\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix} = I$.
Step 6: State the conclusion.
When $x + y = 0$, the product $G(x)G(y)$ equals the $2 \times 2$ identity matrix. \[ \boxed{G(x)G(y) = I \text{ (Identity Matrix)}} \]
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