Matrix Transformations

A matrix transformation moves points in the plane by multiplying each point, written as a column, by a 2 × 2 matrix. Different matrices rotate, reflect, stretch or shrink a figure. To transform a whole shape, put its vertices in the columns of one matrix and multiply once.

Before this lesson: Matrix Multiplication, Rotations, Reflections and Dilations

How a matrix moves a point

Write the point (x, y) as a column, \begin{bmatrix} x \\ y \end{bmatrix}. Multiply it by a 2 × 2 matrix. The product is a new column, and that column is the image point.

\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} ax + by \\ cx + dy \end{bmatrix}

The columns of the matrix tell you everything. The first column is where the point (1, 0) lands, and the second column is where (0, 1) lands. Every other point follows along, because the matrix keeps lines straight and keeps the origin fixed.

Transformation matrices

For a rotation by any angle θ counterclockwise about the origin, use this matrix:

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

Two more facts are useful. To do one transformation and then another, multiply their matrices with the second one on the left: BA means “do A first, then B.” Also, the determinant tells you how areas change. The image’s area is |\det A| times the original area.

How to transform a figure with a matrix

1. Choose the transformation matrix from the table, or build it from where (1, 0) and (0, 1) go.
2. Write the figure’s vertices as the columns of a point matrix, in order.
3. Multiply, with the transformation matrix on the left and the point matrix on the right.
4. Read each column of the product as an image point.

Matrix transformation examples

Example 1: Rotate a triangle 90°

Triangle ABC has vertices A(1, 1), B(4, 1) and C(1, 3). Rotate it 90° counterclockwise about the origin.

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

P = \begin{bmatrix} 1 & 4 & 1 \\ 1 & 1 & 3 \end{bmatrix}

RP = \begin{bmatrix} 0(1) + (-1)(1) & 0(4) + (-1)(1) & 0(1) + (-1)(3) \\ 1(1) + 0(1) & 1(4) + 0(1) & 1(1) + 0(3) \end{bmatrix}

RP = \begin{bmatrix} -1 & -1 & -3 \\ 1 & 4 & 1 \end{bmatrix}

Answer: A′(−1, 1), B′(−1, 4) and C′(−3, 1). Each image point follows the rule (x, y) \to (-y, x).

Example 2: Reflect a point across y = x

Reflect the point (2, −5) across the line y = x.

F = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}

F\begin{bmatrix} 2 \\ -5 \end{bmatrix} = \begin{bmatrix} 0(2) + 1(-5) \\ 1(2) + 0(-5) \end{bmatrix}

F\begin{bmatrix} 2 \\ -5 \end{bmatrix} = \begin{bmatrix} -5 \\ 2 \end{bmatrix}

Answer: (−5, 2). A reflection across y = x swaps the two coordinates.

Example 3: Build the matrix from two points

A transformation T sends (1, 0) to (2, 1) and (0, 1) to (−1, 3). Find its matrix, the image of (4, −2), and how T changes areas.

T = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}

T\begin{bmatrix} 4 \\ -2 \end{bmatrix} = \begin{bmatrix} 2(4) + (-1)(-2) \\ 1(4) + 3(-2) \end{bmatrix}

T\begin{bmatrix} 4 \\ -2 \end{bmatrix} = \begin{bmatrix} 8 + 2 \\ 4 - 6 \end{bmatrix}

T\begin{bmatrix} 4 \\ -2 \end{bmatrix} = \begin{bmatrix} 10 \\ -2 \end{bmatrix}

\det T = 2(3) - (-1)(1)

\det T = 6 + 1

\det T = 7

Answer: The matrix is \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}, the point (4, −2) maps to (10, −2), and T multiplies every area by 7.

Example 4: Rotate a point by 60°

Rotate the point (2, 2) by 60° counterclockwise about the origin. Give exact coordinates.

R_{60} = \begin{bmatrix} \cos 60^\circ & -\sin 60^\circ \\ \sin 60^\circ & \cos 60^\circ \end{bmatrix}

R_{60} = \begin{bmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix}

R_{60}\begin{bmatrix} 2 \\ 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}(2) - \frac{\sqrt{3}}{2}(2) \\ \frac{\sqrt{3}}{2}(2) + \frac{1}{2}(2) \end{bmatrix}

R_{60}\begin{bmatrix} 2 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 - \sqrt{3} \\ \sqrt{3} + 1 \end{bmatrix}

Answer: (1 - \sqrt{3}, 1 + \sqrt{3}) \approx (-0.73, 2.73). As a check, a rotation keeps the distance from the origin. The original distance squared is 2^2 + 2^2 = 8, and the new one is (1 - \sqrt{3})^2 + (1 + \sqrt{3})^2 = 8 too.

Example 5: Word problem with two moves

In a video game, a ship sits at (5, 2) on the screen’s grid. The game turns the ship’s position 90° counterclockwise about the origin, then reflects it across the x-axis. Find one matrix that does both moves, and the ship’s new position.

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

F = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}

FR = \begin{bmatrix} 1(0) + 0(1) & 1(-1) + 0(0) \\ 0(0) + (-1)(1) & 0(-1) + (-1)(0) \end{bmatrix}

FR = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}

FR\begin{bmatrix} 5 \\ 2 \end{bmatrix} = \begin{bmatrix} 0(5) + (-1)(2) \\ (-1)(5) + 0(2) \end{bmatrix}

FR\begin{bmatrix} 5 \\ 2 \end{bmatrix} = \begin{bmatrix} -2 \\ -5 \end{bmatrix}

Answer: The combined matrix is \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}, and the ship moves to (−2, −5). Check it one move at a time: the rotation sends (5, 2) to (−2, 5), and the reflection sends (−2, 5) to (−2, −5).

Common mistakes

Matrices and the rules from geometry

Every coordinate rule from Geometry has a matrix. The matrix version makes it easy to combine moves: multiply the matrices once, as in compositions of transformations, instead of moving every point twice.

The rotation matrix comes from the unit circle. Rotating (1, 0) by θ lands on (\cos\theta, \sin\theta), which is the first column. Rotating (0, 1) by θ lands on (-\sin\theta, \cos\theta), the second column.

Practice problems

1. Rotate the point (2, 5) by 90° counterclockwise about the origin.
2. Reflect the point (−3, 4) across the y-axis.
3. Rotate the point (6, −1) by 180° about the origin.
4. Dilate the triangle with vertices (1, 2), (3, 0) and (0, −2) by a factor of 3.
5. Find the image of (3, −4) under the matrix \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}.
6. A transformation sends (1, 0) to (0, −1) and (0, 1) to (1, 0). Write its matrix and name the transformation.
7. The matrix \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix} is applied to a region with an area of 5 square units. Find the area of the image.
8. Rotate the point (4, 0) by 30° counterclockwise about the origin. Give exact coordinates.
9. Find one matrix that rotates 180° and then reflects across the y-axis. Use it to transform (2, 7).
10. A designer stretches a picture with the matrix \begin{bmatrix} 1.5 & 0 \\ 0 & 2 \end{bmatrix}. The picture is a rectangle with corners (0, 0), (8, 0), (8, 5) and (0, 5). Find the image of the corner (8, 5) and the new area.

R\begin{bmatrix} 2 \\ 5 \end{bmatrix} = \begin{bmatrix} 0(2) + (-1)(5) \\ 1(2) + 0(5) \end{bmatrix}

R\begin{bmatrix} 2 \\ 5 \end{bmatrix} = \begin{bmatrix} -5 \\ 2 \end{bmatrix}

F\begin{bmatrix} -3 \\ 4 \end{bmatrix} = \begin{bmatrix} (-1)(-3) + 0(4) \\ 0(-3) + 1(4) \end{bmatrix}

F\begin{bmatrix} -3 \\ 4 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \end{bmatrix}

R\begin{bmatrix} 6 \\ -1 \end{bmatrix} = \begin{bmatrix} (-1)(6) + 0(-1) \\ 0(6) + (-1)(-1) \end{bmatrix}

R\begin{bmatrix} 6 \\ -1 \end{bmatrix} = \begin{bmatrix} -6 \\ 1 \end{bmatrix}

DP = \begin{bmatrix} 3(1) + 0(2) & 3(3) + 0(0) & 3(0) + 0(-2) \\ 0(1) + 3(2) & 0(3) + 3(0) & 0(0) + 3(-2) \end{bmatrix}

DP = \begin{bmatrix} 3 & 9 & 0 \\ 6 & 0 & -6 \end{bmatrix}

A\begin{bmatrix} 3 \\ -4 \end{bmatrix} = \begin{bmatrix} 2(3) + 1(-4) \\ 0(3) + 1(-4) \end{bmatrix}

A\begin{bmatrix} 3 \\ -4 \end{bmatrix} = \begin{bmatrix} 6 - 4 \\ 0 - 4 \end{bmatrix}

A\begin{bmatrix} 3 \\ -4 \end{bmatrix} = \begin{bmatrix} 2 \\ -4 \end{bmatrix}

T = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}

T\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} y \\ -x \end{bmatrix}

\det A = 3(4) - 1(2)

\det A = 12 - 2

\det A = 10

\text{new area} = 10(5) = 50

R_{30} = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}

R_{30}\begin{bmatrix} 4 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2}(4) - \frac{1}{2}(0) \\ \frac{1}{2}(4) + \frac{\sqrt{3}}{2}(0) \end{bmatrix}

R_{30}\begin{bmatrix} 4 \\ 0 \end{bmatrix} = \begin{bmatrix} 2\sqrt{3} \\ 2 \end{bmatrix}

F_yR = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}

F_yR = \begin{bmatrix} (-1)(-1) + 0(0) & (-1)(0) + 0(-1) \\ 0(-1) + 1(0) & 0(0) + 1(-1) \end{bmatrix}

F_yR = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}

F_yR\begin{bmatrix} 2 \\ 7 \end{bmatrix} = \begin{bmatrix} 1(2) + 0(7) \\ 0(2) + (-1)(7) \end{bmatrix}

F_yR\begin{bmatrix} 2 \\ 7 \end{bmatrix} = \begin{bmatrix} 2 \\ -7 \end{bmatrix}

S\begin{bmatrix} 8 \\ 5 \end{bmatrix} = \begin{bmatrix} 1.5(8) + 0(5) \\ 0(8) + 2(5) \end{bmatrix}

S\begin{bmatrix} 8 \\ 5 \end{bmatrix} = \begin{bmatrix} 12 \\ 10 \end{bmatrix}

\det S = 1.5(2) - 0(0)

\det S = 3

\text{new area} = 3(8 \cdot 5)

\text{new area} = 3(40) = 120

Matrix transformations FAQ

What is a transformation matrix?

It is a matrix that moves points when you multiply it by them. A 2 × 2 matrix times a point (x, y), written as a column, gives the image point.

What is the matrix for a 90 degree rotation?

For a 90° counterclockwise rotation about the origin, the matrix is \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}. It sends (x, y) to (−y, x).

How do you combine two transformations with matrices?

Multiply their matrices, with the second transformation on the left. The product is one matrix that does both moves in order.

What does the determinant tell you about a transformation?

Its absolute value is the area scale factor. A negative determinant also means the transformation flips the figure over, as a reflection does.

Summary

• To transform a point, write it as a column and multiply, with the matrix on the left.
• The first column of the matrix is the image of (1, 0), and the second is the image of (0, 1).
• A rotation by θ uses \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}.
• For “A first, then B,” use the product BA.
• Areas are multiplied by |\det A|.

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