Composite Transformation Matrix

Composition of Transformations Using Matrices

In geometry, a transformation is an operation that moves or changes the shape of an object. When multiple transformations are applied sequentially to an object, this is called a composition of transformations.

We can use matrices to represent many geometric transformations and also to find the result of the composition of these transformations.

We will focus on transformations that can be represented by $2 \times 2$ matrices. For example, reflection across the X-axis can be represented by the matrix $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$. If the point $(x,y)$ is reflected across the X-axis, its image can be found by multiplying this matrix by the position vector of the point: $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$.

Here are some basic transformations along with their matrices that are often used in the composition of transformations:

1. Reflection across the X-axis: $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
2. Reflection across the Y-axis: $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$
3. Reflection across the line $y=x$: $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
4. Reflection across the line $y=-x$: $\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$
5. Reflection across the origin $O(0,0)$ (equivalent to a $180^\circ$ rotation): $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$
6. Rotation about the origin $(0,0)$ by an angle $\theta$: $\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$
7. Dilation about the origin $(0,0)$ with a scale factor $k$: $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$

Operating Composition of Transformations Using Matrices

Composition of transformations means performing several transformations in sequence. If transformation $T_1$ is followed by transformation $T_2$, we denote it as $T_2 \circ T_1$. This means $T_1$ is applied first, then its result is transformed by $T_2$.

Suppose the matrix corresponding to $T_1$ is $M_1$, and the matrix corresponding to $T_2$ is $M_2$. To find the image of point $P(x,y)$ under the composition $T_2 \circ T_1$, there are two equivalent methods:

1. Applying Transformations Sequentially to the Point:

   • Calculate the image $P'(x',y')$ of $P(x,y)$ under $T_1$: $\begin{pmatrix} x' \\ y' \end{pmatrix} = M_1 \begin{pmatrix} x \\ y \end{pmatrix}$.
   • Then, calculate the image $P''(x'',y'')$ of $P'(x',y')$ under $T_2$: $\begin{pmatrix} x'' \\ y'' \end{pmatrix} = M_2 \begin{pmatrix} x' \\ y' \end{pmatrix}$.

   If we substitute step (a) into (b), we get: $\begin{pmatrix} x'' \\ y'' \end{pmatrix} = M_2 \left( M_1 \begin{pmatrix} x \\ y \end{pmatrix} \right)$.

2. Finding the Composite Matrix First:

   • Determine the matrix $M$ that represents the composite transformation $T_2 \circ T_1$. This matrix is the product $M_2 M_1$.

     Note the order: the matrix of the second transformation ($M_2$) is multiplied from the left by the matrix of the first transformation ($M_1$).

   • Calculate the image $P''(x'',y'')$ of $P(x,y)$ using the composite matrix $M$: $\begin{pmatrix} x'' \\ y'' \end{pmatrix} = M \begin{pmatrix} x \\ y \end{pmatrix} = (M_2 M_1) \begin{pmatrix} x \\ y \end{pmatrix}$.

Both methods yield the same final image due to the associative property of matrix multiplication, i.e., $M_2 (M_1 P) = (M_2 M_1) P$, where $P$ is the column vector $\begin{pmatrix} x \\ y \end{pmatrix}$.

Illustrative Example:

Suppose $T_1$ is a reflection across the Y-axis, and $T_2$ is a rotation about the origin $O$ by $\frac{1}{2}\pi$ radians ($90^\circ$). We want to find the image of point $P(x,y)$ under $T_2 \circ T_1$.

The matrix for $T_1$ (reflection across Y-axis) is $M_1 = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$.

The matrix for $T_2$ (rotation $90^\circ$) is $M_2 = \begin{pmatrix} \cos 90^\circ & -\sin 90^\circ \\ \sin 90^\circ & \cos 90^\circ \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.

Method 1: Sequential Transformation on the Point

• Image of $P(x,y)$ under $T_1$:

  $$\begin{pmatrix} x' \\ y' \end{pmatrix} = M_1 \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -x \\ y \end{pmatrix}$$

  So $P'(-x,y)$.

• Image of $P'(-x,y)$ under $T_2$:

  $$\begin{pmatrix} x'' \\ y'' \end{pmatrix} = M_2 \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} -x \\ y \end{pmatrix} = \begin{pmatrix} (0)(-x)+(-1)(y) \\ (1)(-x)+(0)(y) \end{pmatrix} = \begin{pmatrix} -y \\ -x \end{pmatrix}$$

• The final image is $P''(-y,-x)$.

Method 2: Composite Matrix First

• Composite matrix $M = M_2 M_1$:

  $$M = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} (0)(-1)+(-1)(0) & (0)(0)+(-1)(1) \\ (1)(-1)+(0)(0) & (1)(0)+(0)(1) \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$$

• Image of $P(x,y)$ under $M$:

  $$\begin{pmatrix} x'' \\ y'' \end{pmatrix} = M \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} (0)(x)+(-1)(y) \\ (-1)(x)+(0)(y) \end{pmatrix} = \begin{pmatrix} -y \\ -x \end{pmatrix}$$

• The final image is $P''(-y,-x)$.

Both methods give the same result. Using the composite matrix ($M_2 M_1$) is often more efficient if we need to transform many points with the same composition.

Composite Matrix Rule

Suppose the matrices related to transformations $T_1$ and $T_2$ are $M_1 = \begin{pmatrix} p & q \\ r & s \end{pmatrix}$ and $M_2 = \begin{pmatrix} t & u \\ v & w \end{pmatrix}$ respectively.

Then, the matrix related to the composition of transformations $T_2 \circ T_1$ (Transformation $T_1$ followed by $T_2$) is $M_2 M_1 = \begin{pmatrix} t & u \\ v & w \end{pmatrix} \begin{pmatrix} p & q \\ r & s \end{pmatrix}$.

Remember that the order of matrix multiplication is important. The matrix for the transformation performed first ($M_1$) is written on the right.

Application Examples

Composition of Two Reflections

Determine the image of the point $(2,5)$ reflected across the X-axis and then reflected across the Y-axis.

Alternative Solution:

Let $T_1$ be the reflection across the X-axis, and $T_2$ be the reflection across the Y-axis.

The matrix for $T_1$ ($M_1$) is $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.

The matrix for $T_2$ ($M_2$) is $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$.

The composition of transformations $T_2 \circ T_1$ has the matrix $M = M_2 M_1$.

The image of the point $(2,5)$ is:

So, the image of the point is $(-2,-5)$.

Composition of Reflection and Rotation

Determine the image of the point $(-2,3)$ transformed by the composition of a reflection across the Y-axis followed by a $180^\circ$ rotation about the origin.

Alternative Solution:

Let $T_1$ be the reflection across the Y-axis, and $T_2$ be the $180^\circ$ rotation about the origin.

The matrix for $T_1$ ($M_1$) is $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$.

The matrix for $T_2$ ($M_2$) is $\begin{pmatrix} \cos 180^\circ & -\sin 180^\circ \\ \sin 180^\circ & \cos 180^\circ \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$.

The composition of transformations $T_2 \circ T_1$ has the matrix $M = M_2 M_1$.

The image of the point $(-2,3)$ is:

So, the image of the point is $(-2,-3)$.

Composition of Three Transformations

Suppose you want to perform three transformations on a point $P(2,5)$, namely reflection across the X-axis, rotation $90^\circ$ about the origin, and a half turn ($180^\circ$ rotation about the origin). Determine its image!

Alternative Solution:

Let:

• $T_1$: Reflection across the X-axis.

  Matrix $M_1 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

• $T_2$: Rotation $90^\circ$ about the origin.

  Matrix $M_2 = \begin{pmatrix} \cos 90^\circ & -\sin 90^\circ \\ \sin 90^\circ & \cos 90^\circ \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$

• $T_3$: Half turn ($180^\circ$ rotation about the origin).

  Matrix $M_3 = \begin{pmatrix} \cos 180^\circ & -\sin 180^\circ \\ \sin 180^\circ & \cos 180^\circ \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$

The composition of transformations is $T_3 \circ T_2 \circ T_1$. Its matrix is $M = M_3 M_2 M_1$.

The image of $P(2,5)$ is:

So, the image of the point is $(-5,-2)$.

Exercise

Suppose we want to perform three transformations on a point $P(2,5)$, namely reflection across the Y-axis, rotation $180^\circ$ about the origin, and reflection across the line $y=x$. Determine its image!

Answer Key

Let:

• $T_1$: Reflection across the Y-axis.

  Matrix $M_1 = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$.

• $T_2$: Rotation $180^\circ$ about the origin.

  Matrix $M_2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$.

• $T_3$: Reflection across the line $y=x$ .

  Matrix $M_3 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

The composition of transformations is $T_3 \circ T_2 \circ T_1$. Its matrix is $M = M_3 M_2 M_1$.

Step 1: Calculate $M_2 M_1$.

Step 2: Calculate $M = M_3 (M_2 M_1)$.

The image of $P(2,5)$ is:

So, the image of the point is $(-5,2)$.