Rotations and general reflections
Master rotations and general reflections for Year 11 Specialist Mathematics in Queensland (QCAA). These transformations about the origin are represented by 2×2 matrices: the rotation matrix \(R(\theta)\) and the general reflection matrix \(M(\theta)\).
You will learn to build each matrix from exact trigonometric values, find the image of a point by matrix multiplication, and use the determinant to describe the effect on area and orientation — key skills for composing transformations and applying matrix inverses later in the course.
Theory
A rotation or a general reflection about the origin is a linear transformation represented by a 2×2 matrix. In Year 11 Specialist Mathematics (QCAA, Queensland) you build the rotation matrix \(R(\theta)\) and the reflection matrix \(M(\theta)\), apply them to points by matrix multiplication, and read off what the determinant says about area and orientation.
A transformation of the plane that fixes the origin can be written as a 2×2 matrix. To find the image \((x',y')\) of a point \((x,y)\), write the point as a column vector and multiply: \(\begin{pmatrix}x'\\y'\end{pmatrix}=T\begin{pmatrix}x\\y\end{pmatrix}\). The columns of the matrix are the images of the basis vectors \(\mathbf{i}=(1,0)\) and \(\mathbf{j}=(0,1)\).
A rotation anticlockwise about the origin through an angle \(\theta\) sends \(\mathbf{i}\) to \((\cos\theta,\sin\theta)\) and \(\mathbf{j}\) to \((-\sin\theta,\cos\theta)\). These become the columns of the rotation matrix \(R(\theta)\).
A reflection in a line through the origin that makes an angle \(\theta\) with the positive \(x\)-axis has the general reflection matrix \(M(\theta)\). Notice the entries use \(2\theta\), not \(\theta\): the mirror line at angle \(\theta\) turns a ray at angle \(\alpha\) into a ray at angle \(2\theta-\alpha\).
The determinant measures the effect on area and orientation. A rotation has \(\det R=1\) (area preserved, orientation kept); a reflection has \(\det M=-1\) (area preserved, orientation reversed). Because a mirror image reverses when reflected again, a reflection is its own inverse: \(M^2=I\).
Rotation anticlockwise about the origin through \(\theta\):
Reflection in a line through the origin at angle \(\theta\) to the \(x\)-axis:
Apply a transformation to a point by multiplying its column vector:
Determinants fix the area and orientation effect:
How to build and use a rotation or reflection matrix
- Choose the matrix: a rotation about the origin uses \(R(\theta)\); a reflection in a line through the origin uses \(M(\theta)\) with its \(2\theta\) entries.
- Substitute the angle: put in the exact values of \(\cos\theta,\sin\theta\) (or \(\cos 2\theta,\sin 2\theta\)), keeping surds and fractions exact.
- Apply to the point: write the point as a column vector and multiply, row-by-column, to get the image \((x',y')\).
- Check with the determinant: \(\det R=1\) or \(\det M=-1\) confirms the type; the image area equals \(|\det|\times\) the original area (here, unchanged).
Write \(R(\theta)\) and substitute \(\theta=150^\circ\):
| \(R(150^\circ)\) | \(=\) | \(\begin{pmatrix}\cos 150^\circ & -\sin 150^\circ\\ \sin 150^\circ & \cos 150^\circ\end{pmatrix}\) |
Use the exact values \(\cos 150^\circ=-\dfrac{\sqrt{3}}{2}\) and \(\sin 150^\circ=\dfrac{1}{2}\):
| \(=\) | \(\begin{pmatrix}-\dfrac{\sqrt{3}}{2} & -\dfrac{1}{2}\\[4pt] \dfrac{1}{2} & -\dfrac{\sqrt{3}}{2}\end{pmatrix}\) |
\(R(150^\circ)=\begin{pmatrix}-\dfrac{\sqrt{3}}{2} & -\dfrac{1}{2}\\[4pt] \dfrac{1}{2} & -\dfrac{\sqrt{3}}{2}\end{pmatrix}\).
Build \(R(30^\circ)\) with \(\cos 30^\circ=\dfrac{\sqrt{3}}{2}\), \(\sin 30^\circ=\dfrac{1}{2}\):
| \(R(30^\circ)\) | \(=\) | \(\begin{pmatrix}\dfrac{\sqrt{3}}{2} & -\dfrac{1}{2}\\[4pt] \dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\end{pmatrix}\) |
Multiply the matrix by the point as a column vector:
| \(\begin{pmatrix}x'\\y'\end{pmatrix}\) | \(=\) | \(\begin{pmatrix}\dfrac{\sqrt{3}}{2} & -\dfrac{1}{2}\\[4pt] \dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\end{pmatrix}\begin{pmatrix}4\\0\end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix}\dfrac{\sqrt{3}}{2}(4)-\dfrac{1}{2}(0)\\[4pt] \dfrac{1}{2}(4)+\dfrac{\sqrt{3}}{2}(0)\end{pmatrix}\) | |
| \(=\) | \(\begin{pmatrix}2\sqrt{3}\\ 2\end{pmatrix}\) |
The image is \((2\sqrt{3},\,2)\).
Use \(M(\theta)\) with \(\theta=60^\circ\), so \(2\theta=120^\circ\):
| \(M(60^\circ)\) | \(=\) | \(\begin{pmatrix}\cos 120^\circ & \sin 120^\circ\\ \sin 120^\circ & -\cos 120^\circ\end{pmatrix}\) |
Substitute \(\cos 120^\circ=-\dfrac{1}{2}\), \(\sin 120^\circ=\dfrac{\sqrt{3}}{2}\):
| \(=\) | \(\begin{pmatrix}-\dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\\[4pt] \dfrac{\sqrt{3}}{2} & \dfrac{1}{2}\end{pmatrix}\) |
Now multiply by \((2,0)\) as a column vector:
| \(\begin{pmatrix}x'\\y'\end{pmatrix}\) | \(=\) | \(\begin{pmatrix}-\dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\\[4pt] \dfrac{\sqrt{3}}{2} & \dfrac{1}{2}\end{pmatrix}\begin{pmatrix}2\\0\end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix}-\dfrac{1}{2}(2)+\dfrac{\sqrt{3}}{2}(0)\\[4pt] \dfrac{\sqrt{3}}{2}(2)+\dfrac{1}{2}(0)\end{pmatrix}\) | |
| \(=\) | \(\begin{pmatrix}-1\\ \sqrt{3}\end{pmatrix}\) |
\(M(60^\circ)=\begin{pmatrix}-\dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\\[4pt] \dfrac{\sqrt{3}}{2} & \dfrac{1}{2}\end{pmatrix}\), and the image of \((2,0)\) is \((-1,\sqrt{3})\).
Find the determinant \(ad-bc\):
| \(\det T\) | \(=\) | \(\dfrac{1}{\sqrt{2}}\cdot\dfrac{1}{\sqrt{2}}-\left(-\dfrac{1}{\sqrt{2}}\right)\dfrac{1}{\sqrt{2}}\) |
| \(=\) | \(\dfrac{1}{2}+\dfrac{1}{2}\) | |
| \(=\) | \(1\) |
A determinant of \(1\) points to a rotation, so compare with \(R(\theta)\):
| \(\cos\theta\) | \(=\) | \(\dfrac{1}{\sqrt{2}}\) |
| \(\sin\theta\) | \(=\) | \(\dfrac{1}{\sqrt{2}}\) |
| \(\theta\) | \(=\) | \(45^\circ\) |
\(\det T=1\); \(T\) is a rotation of \(45^\circ\) anticlockwise about the origin.
Common pitfalls
Frequently asked questions
What is the matrix for a rotation about the origin?
An anticlockwise rotation through \(\theta\) about the origin has matrix \(R(\theta)=\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}\); its columns are the images of \((1,0)\) and \((0,1)\).
How do you find the image of a point under a rotation or reflection?
Write the point as a column vector and multiply it by the transformation matrix: \(\begin{pmatrix}x'\\y'\end{pmatrix}=T\begin{pmatrix}x\\y\end{pmatrix}\), using the row-by-column rule.
What is the general reflection matrix for a line through the origin?
A reflection in a line at angle \(\theta\) to the \(x\)-axis has matrix \(M(\theta)=\begin{pmatrix}\cos 2\theta & \sin 2\theta\\ \sin 2\theta & -\cos 2\theta\end{pmatrix}\).
Why does the reflection matrix use \(2\theta\)?
Reflecting a ray at angle \(\alpha\) in a mirror at angle \(\theta\) sends it to angle \(2\theta-\alpha\), so the matrix entries depend on \(2\theta\) rather than \(\theta\).
What does the determinant tell you about the transformation?
The image area equals \(|\det|\) times the original area. A rotation has \(\det R=1\) (area and orientation preserved); a reflection has \(\det M=-1\) (area preserved but orientation reversed).
How do you rotate a point clockwise instead of anticlockwise?
Use a negative angle: a clockwise rotation through \(\theta\) is \(R(-\theta)\), which changes the sign of both \(\sin\theta\) entries.