# Nakafa Framework: LLM

URL: https://nakafa.com/en/subject/high-school/11/mathematics/geometric-transformation/composite-transformation-matrix
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---

export const metadata = {
  title: "Composite Transformation Matrix",
  description: "Master composite transformation matrices with step-by-step examples. Learn to combine reflections, rotations & dilations using matrix multiplication.",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "05/10/2025",
  subject: "Geometric Transformation",
};

## 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 <InlineMath math="2 \times 2" /> matrices. For example, reflection across the X-axis can be represented by the matrix <InlineMath math="\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}" />. If the point <InlineMath math="(x,y)" /> is reflected across the X-axis, its image can be found by multiplying this matrix by the position vector of the point: <InlineMath math="\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: <InlineMath math="\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}" />
2.  Reflection across the Y-axis: <InlineMath math="\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}" />
3.  Reflection across the line <InlineMath math="y=x" />: <InlineMath math="\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}" />
4.  Reflection across the line <InlineMath math="y=-x" />: <InlineMath math="\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}" />
5.  Reflection across the origin <InlineMath math="O(0,0)" /> (equivalent to a <InlineMath math="180^\circ" /> rotation): <InlineMath math="\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}" />
6.  Rotation about the origin <InlineMath math="(0,0)" /> by an angle <InlineMath math="\theta" />: <InlineMath math="\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}" />
7.  Dilation about the origin <InlineMath math="(0,0)" /> with a scale factor <InlineMath math="k" />: <InlineMath math="\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 <InlineMath math="T_1" /> is followed by transformation <InlineMath math="T_2" />, we denote it as <InlineMath math="T_2 \circ T_1" />. This means <InlineMath math="T_1" /> is applied first, then its result is transformed by <InlineMath math="T_2" />.

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

1.  **Applying Transformations Sequentially to the Point:**

    - Calculate the image <InlineMath math="P'(x',y')" /> of <InlineMath math="P(x,y)" /> under <InlineMath math="T_1" />: <InlineMath math="\begin{pmatrix} x' \\ y' \end{pmatrix} = M_1 \begin{pmatrix} x \\ y \end{pmatrix}" />.
    - Then, calculate the image <InlineMath math="P''(x'',y'')" /> of <InlineMath math="P'(x',y')" /> under <InlineMath math="T_2" />: <InlineMath math="\begin{pmatrix} x'' \\ y'' \end{pmatrix} = M_2 \begin{pmatrix} x' \\ y' \end{pmatrix}" />.

    If we substitute step (a) into (b), we get: <InlineMath math="\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 <InlineMath math="M" /> that represents the composite transformation <InlineMath math="T_2 \circ T_1" />. This matrix is the product <InlineMath math="M_2 M_1" />.

      **Note the order:** the matrix of the second transformation (<InlineMath math="M_2" />) is multiplied from the left by the matrix of the first transformation (<InlineMath math="M_1" />).

    - Calculate the image <InlineMath math="P''(x'',y'')" /> of <InlineMath math="P(x,y)" /> using the composite matrix <InlineMath math="M" />: <InlineMath math="\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., <InlineMath math="M_2 (M_1 P) = (M_2 M_1) P" />, where <InlineMath math="P" /> is the column vector <InlineMath math="\begin{pmatrix} x \\ y \end{pmatrix}" />.

**Illustrative Example:**

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

The matrix for <InlineMath math="T_1" /> (reflection across Y-axis) is <InlineMath math="M_1 = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}" />.

The matrix for <InlineMath math="T_2" /> (rotation <InlineMath math="90^\circ" />) is <InlineMath math="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 <InlineMath math="P(x,y)" /> under <InlineMath math="T_1" />:

  <BlockMath math="\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 <InlineMath math="P'(-x,y)" />.

- Image of <InlineMath math="P'(-x,y)" /> under <InlineMath math="T_2" />:

  <BlockMath math="\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 <InlineMath math="P''(-y,-x)" />.

**Method 2: Composite Matrix First**

- Composite matrix <InlineMath math="M = M_2 M_1" />:

  <BlockMath math="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 <InlineMath math="P(x,y)" /> under <InlineMath math="M" />:

  <BlockMath math="\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 <InlineMath math="P''(-y,-x)" />.

Both methods give the same result. Using the composite matrix (<InlineMath math="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 <InlineMath math="T_1" /> and <InlineMath math="T_2" /> are <InlineMath math="M_1 = \begin{pmatrix} p & q \\ r & s \end{pmatrix}" /> and <InlineMath math="M_2 = \begin{pmatrix} t & u \\ v & w \end{pmatrix}" /> respectively.

Then, the matrix related to the composition of transformations <InlineMath math="T_2 \circ T_1" /> (Transformation <InlineMath math="T_1" /> followed by <InlineMath math="T_2" />) is <InlineMath math="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 (<InlineMath math="M_1" />) is written on the right.

## Application Examples

### Composition of Two Reflections

Determine the image of the point <InlineMath math="(2,5)" /> reflected across the X-axis and then reflected across the Y-axis.

**Alternative Solution:**

Let <InlineMath math="T_1" /> be the reflection across the X-axis, and <InlineMath math="T_2" /> be the reflection across the Y-axis.

The matrix for <InlineMath math="T_1" /> (<InlineMath math="M_1" />) is <InlineMath math="\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}" />.

The matrix for <InlineMath math="T_2" /> (<InlineMath math="M_2" />) is <InlineMath math="\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}" />.

The composition of transformations <InlineMath math="T_2 \circ T_1" /> has the matrix <InlineMath math="M = M_2 M_1" />.

<BlockMath math="M = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} (-1)(1)+(0)(0) & (-1)(0)+(0)(-1) \\ (0)(1)+(1)(0) & (0)(0)+(1)(-1) \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}" />

The image of the point <InlineMath math="(2,5)" /> is:

<BlockMath math="\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}" />

So, the image of the point is <InlineMath math="(-2,-5)" />.

### Composition of Reflection and Rotation

Determine the image of the point <InlineMath math="(-2,3)" /> transformed by the composition of a reflection across the Y-axis followed by a <InlineMath math="180^\circ" /> rotation about the origin.

**Alternative Solution:**

Let <InlineMath math="T_1" /> be the reflection across the Y-axis, and <InlineMath math="T_2" /> be the <InlineMath math="180^\circ" /> rotation about the origin.

The matrix for <InlineMath math="T_1" /> (<InlineMath math="M_1" />) is <InlineMath math="\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}" />.

The matrix for <InlineMath math="T_2" /> (<InlineMath math="M_2" />) is <InlineMath math="\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 <InlineMath math="T_2 \circ T_1" /> has the matrix <InlineMath math="M = M_2 M_1" />.

<BlockMath math="M = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} (-1)(-1)+(0)(0) & (-1)(0)+(0)(1) \\ (0)(-1)+(-1)(0) & (0)(0)+(-1)(1) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}" />

The image of the point <InlineMath math="(-2,3)" /> is:

<BlockMath math="\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}" />

So, the image of the point is <InlineMath math="(-2,-3)" />.

### Composition of Three Transformations

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

**Alternative Solution:**

Let:

- <InlineMath math="T_1" />: Reflection across the X-axis.

  Matrix <InlineMath math="M_1 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}" />

- <InlineMath math="T_2" />: Rotation <InlineMath math="90^\circ" /> about the
  origin.

  Matrix <InlineMath math="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}" />

- <InlineMath math="T_3" />: Half turn (<InlineMath math="180^\circ" /> rotation
  about the origin).

  Matrix <InlineMath math="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 <InlineMath math="T_3 \circ T_2 \circ T_1" />. Its matrix is <InlineMath math="M = M_3 M_2 M_1" />.

<div className="flex flex-col gap-4">
  <BlockMath math="M_2 M_1 = \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}" />
  <BlockMath math="M = M_3 (M_2 M_1) = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} (-1)(0)+(0)(1) & (-1)(1)+(0)(0) \\ (0)(0)+(-1)(1) & (0)(1)+(-1)(0) \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}" />
</div>

The image of <InlineMath math="P(2,5)" /> is:

<BlockMath math="\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}" />

So, the image of the point is <InlineMath math="(-5,-2)" />.

## Exercise

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

### Answer Key

Let:

- <InlineMath math="T_1" />: Reflection across the Y-axis.

  Matrix <InlineMath math="M_1 = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}" />.

- <InlineMath math="T_2" />: Rotation <InlineMath math="180^\circ" /> about the
  origin.

  Matrix <InlineMath math="M_2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}" />.

- <InlineMath math="T_3" />: Reflection across the line <InlineMath math="y=x" />
  .

  Matrix <InlineMath math="M_3 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}" />

The composition of transformations is <InlineMath math="T_3 \circ T_2 \circ T_1" />. Its matrix is <InlineMath math="M = M_3 M_2 M_1" />.

**Step 1:** Calculate <InlineMath math="M_2 M_1" />.

<BlockMath math="M_2 M_1 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} (-1)(-1)+(0)(0) & (-1)(0)+(0)(1) \\ (0)(-1)+(-1)(0) & (0)(0)+(-1)(1) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}" />

**Step 2:** Calculate <InlineMath math="M = M_3 (M_2 M_1)" />.

<BlockMath math="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}" />

The image of <InlineMath math="P(2,5)" /> is:

<BlockMath math="\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}" />

So, the image of the point is <InlineMath math="(-5,2)" />.