Composite Transformation Matrix

Two Ways to Apply a Sequence of Matrix Transformations

In geometry, a transformation maps every point of a figure to an image point. Depending on the transformation, the figure's position, orientation, size, or shape may change. Applying several transformations in sequence creates 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 represented by $2 \times 2$ matrices. For example, reflection across the $x$ -axis uses this matrix:

\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}

To find the image of $(x,y)$ under this reflection, multiply the matrix by the point's position vector:

\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:

Reflection across the $x$ -axis:

$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
Reflection across the $y$ -axis:

$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$
Reflection across the line $y=x$ :

$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
Reflection across the line $y=-x$ :

$\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$
Reflection across the origin $O(0,0)$ (equivalent to a $180^\circ$ rotation):

$\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$
Rotation about the origin $(0,0)$ by an angle $\theta$ :

$\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$
Dilation about the origin $(0,0)$ with a scale factor $k$ :

$\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$

Applying Composite Transformations with 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:

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 $T_2$ :
  
  $\begin{pmatrix} x'' \\ y'' \end{pmatrix} = M_2 \begin{pmatrix} x' \\ y' \end{pmatrix}$
Substituting the first image into the second calculation gives:

$\begin{pmatrix} x'' \\ y'' \end{pmatrix} = M_2 \left( M_1 \begin{pmatrix} x \\ y \end{pmatrix} \right)$
Finding the Composite Matrix:
- 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 ( $M_2$ ) stands to the left of the matrix of the first transformation ( $M_1$ ).
- Calculate the image $P''(x'',y'')$ of $P(x,y)$ using the composite matrix $M$ :
  
  $\begin{aligned} &\begin{pmatrix} x'' \\ y'' \end{pmatrix} \\ & = M \begin{pmatrix} x \\ y \end{pmatrix} \\ & = (M_2 M_1) \begin{pmatrix} x \\ y \end{pmatrix} \end{aligned}$

Both methods give the same final image because matrix multiplication is associative: $M_2 (M_1 P) = (M_2 M_1) P$ . Here, $P$ 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 the $y$ -axis) is:

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

The matrix for $T_2$ (rotation $90^\circ$ ) is:

\begin{aligned} 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} \end{aligned}

Method 1: Sequential Transformation on the Point

Image of $P(x,y)$ under $T_1$ :

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

So $P'(-x,y)$ .
Image of $P'(-x,y)$ under $T_2$ :

$\begin{aligned} \begin{pmatrix} x'' \\ y'' \end{pmatrix} &= M_2 \begin{pmatrix} x' \\ y' \end{pmatrix} \\ &= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} -x \\ y \end{pmatrix} \\ &= \begin{pmatrix} (0)(-x)+(-1)(y) \\ (1)(-x)+(0)(y) \end{pmatrix} \\ &= \begin{pmatrix} -y \\ -x \end{pmatrix} \end{aligned}$
The final image is $P''(-y,-x)$ .

Method 2: Composite Matrix First

Composite matrix $M = M_2 M_1$ :

Write $M=(m_{ij})$ . The subscripts name the row and column of each entry. To find $m_{12}$ , for example, multiply the first row of the left matrix by the second column of the right matrix and add the products.
$\begin{aligned} &M=(m_{ij}) \\ &=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \end{aligned}$
Image of $P(x,y)$ under $M$ :

$\begin{aligned} \begin{pmatrix} x'' \\ y'' \end{pmatrix} &= M \begin{pmatrix} x \\ y \end{pmatrix} \\ &= \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ &= \begin{pmatrix} (0)(x)+(-1)(y) \\ (-1)(x)+(0)(y) \end{pmatrix} \\ &= \begin{pmatrix} -y \\ -x \end{pmatrix} \end{aligned}$
The final image is $P''(-y,-x)$ .

Both methods give the same result. If many points use the same composition, calculate $M_2 M_1$ once and apply that composite matrix to every point.

Composite Matrix Rule

Let $M_1$ and $M_2$ be the matrices for transformations $T_1$ and $T_2$ , respectively:

M_1 = \begin{pmatrix} p & q \\ r & s \end{pmatrix}

For the composition $T_2 \circ T_1$ , we apply $T_1$ first and then $T_2$ . Its matrix is:

M_2 M_1 = \begin{pmatrix} t & u \\ v & w \end{pmatrix} \begin{pmatrix} p & q \\ r & s \end{pmatrix}

The order of matrix multiplication determines the order of the transformations. The matrix for the transformation performed first ( $M_1$ ) is written on the right.

Combining Reflections and Rotations with Matrices

Combining two transformations means multiplying their matrices, and the order of that product decides the result. Reading a product from right to left applies the rightmost transformation to the point first, so swapping the two matrices usually produces a different image.

Composition of Two Reflections

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

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 $M = M_2 M_1$ .

\begin{aligned} &M=(m_{ij}) \\ &=\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{aligned}

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

\begin{aligned} \begin{pmatrix} x' \\ y' \end{pmatrix} &= M \begin{pmatrix} 2 \\ 5 \end{pmatrix} \\ &= \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 2 \\ 5 \end{pmatrix} \\ &= \begin{pmatrix} (-1)(2)+(0)(5) \\ (0)(2)+(-1)(5) \end{pmatrix} \\ &= \begin{pmatrix} -2 \\ -5 \end{pmatrix} \end{aligned}

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.

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

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

\begin{aligned} &M=(m_{ij}) \\ &=\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \end{aligned}

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

\begin{aligned} \begin{pmatrix} x' \\ y' \end{pmatrix} &= M \begin{pmatrix} -2 \\ 3 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -2 \\ 3 \end{pmatrix} \\ &= \begin{pmatrix} (1)(-2)+(0)(3) \\ (0)(-2)+(-1)(3) \end{pmatrix} \\ &= \begin{pmatrix} -2 \\ -3 \end{pmatrix} \end{aligned}

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

Composition of Three Transformations

Apply three transformations to point $P(2,5)$ . First reflect it across the $x$ -axis, then rotate it $90^\circ$ about the origin, and finally rotate it $180^\circ$ about the origin. Determine its image.

Solution:

Let:

$T_1$ : Reflection across the $x$ -axis.

The matrix is:

$M_1 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
$T_2$ : Rotation $90^\circ$ about the origin.

The matrix is:

$\begin{aligned} M_2 &= \begin{pmatrix} \cos 90^\circ & -\sin 90^\circ \\ \sin 90^\circ & \cos 90^\circ \end{pmatrix} \\ & = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \end{aligned}$
$T_3$ : Half turn ( $180^\circ$ rotation about the origin).

The matrix is:

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

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

\begin{aligned} &M_2 M_1=(n_{ij}) \\ &=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{aligned}

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

\begin{aligned} \begin{pmatrix} x' \\ y' \end{pmatrix} &= M \begin{pmatrix} 2 \\ 5 \end{pmatrix} \\ &= \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 2 \\ 5 \end{pmatrix} \\ &= \begin{pmatrix} (0)(2)+(-1)(5) \\ (-1)(2)+(0)(5) \end{pmatrix} \\ &= \begin{pmatrix} -5 \\ -2 \end{pmatrix} \end{aligned}

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

Exercise

Apply three transformations to point $P(2,5)$ . First reflect it across the $y$ -axis, then rotate it $180^\circ$ about the origin, and finally reflect it across the line $y=x$ . Determine its image.

Solution

Let:

$T_1$ : Reflection across the $y$ -axis.

The matrix is:

$M_1 = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$
$T_2$ : Rotation $180^\circ$ about the origin.

The matrix is:

$M_2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$
$T_3$ : Reflection across the line $y=x$ .

The matrix is:

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

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 $M_2 M_1$ .

\begin{aligned} &M_2 M_1=(n_{ij}) \\ &=\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \end{aligned}

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

\begin{aligned} &M=(m_{ij}) \\ &=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{aligned}

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

\begin{aligned} \begin{pmatrix} x' \\ y' \end{pmatrix} &= M \begin{pmatrix} 2 \\ 5 \end{pmatrix} \\ &= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 2 \\ 5 \end{pmatrix} \\ &= \begin{pmatrix} (0)(2)+(-1)(5) \\ (1)(2)+(0)(5) \end{pmatrix} \\ &= \begin{pmatrix} -5 \\ 2 \end{pmatrix} \end{aligned}

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