Question

Let $T: \mathbb{R}^2(\mathbb{R}) \to \mathbb{R}^2(\mathbb{R})$ be a linear transformation such that $T$ is a rotation at angle of $90^\circ$ in clockwise direction in $\mathbb{R}^2$, then}

• A. $T(1,0) = (0,1)$
• B. $T(-1,0) = (1,-1)$
• C. $T(1,0) = (0,-1)$
• D. $T(-1,0) = (1,1)$

Hint:

Visualizing basic rotations on a coordinate plane is often faster than using matrices:
- Start at the positive x-axis: (1, 0).
- Rotate \( 90^\circ \) clockwise $\rightarrow$ points straight down to (0, -1).
- Rotate \( 90^\circ \) counter-clockwise $\rightarrow$ points straight up to (0, 1).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept

A linear transformation \(T:\mathbb{R}^2 \rightarrow \mathbb{R}^2\) can be represented by multiplication with a \(2 \times 2\) matrix. A rotation in the Cartesian plane is a special linear transformation that rotates every vector by a fixed angle about the origin.

Key Formula:

The standard matrix for a counter-clockwise rotation through an angle \(\theta\) is

\[ R_{\theta}= \begin{bmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix}. \]

A clockwise rotation by an angle \(\theta\) is equivalent to a counter-clockwise rotation by \(-\theta\). Using the identities

\[ \cos(-\theta)=\cos\theta, \qquad \sin(-\theta)=-\sin\theta, \]

the clockwise rotation matrix becomes

\[ R_{\text{clockwise}}= \begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix}. \]

Step 2: Detailed Explanation

For a clockwise rotation of \(\theta=90^\circ\),

\[ \cos90^\circ=0, \qquad \sin90^\circ=1. \]

Substituting these values into the rotation matrix gives

\[ R= \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix}. \]

Now apply this transformation to the vector \((1,0)\):

\[ \begin{aligned} T(1,0) &= \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \begin{bmatrix} 1\\ 0 \end{bmatrix} \\[4pt] &= \begin{bmatrix} 0\cdot1+1\cdot0\\ -1\cdot1+0\cdot0 \end{bmatrix} \\[4pt] &= \begin{bmatrix} 0\\ -1 \end{bmatrix}. \end{aligned} \]

Hence,

\[ T(1,0)=(0,-1). \]

Geometrically, the vector \((1,0)\) lies on the positive x-axis. After a \(90^\circ\) clockwise rotation, it points along the negative y-axis, whose coordinates are \((0,-1)\).

Step 3: Final Answer

Therefore, the image of the vector \((1,0)\) under the given transformation is

\[ \boxed{T(1,0)=(0,-1)}. \]

Hence, the correct answer is Option (C).