A matrix moves every point
A 2 × 2 matrix moves the points of the plane. To find where it sends the point (x, y), write the point as a column and multiply it by the matrix. The matrix M = (2 1; 0 3) sends (1, 1) to (2 × 1 + 1 × 1; 0 × 1 + 3 × 1) = (3; 3), which is the point (3, 3).
Two arrows matter most: i = (1, 0), one step right, and j = (0, 1), one step up. Every point is made of them. The point (2, 1) is 2i + j: two steps right and one step up.
The columns are where i and j land
Multiply M by the column for i: (2 × 1 + 1 × 0; 0 × 1 + 3 × 0) = (2; 0). That is the first column of M. Multiply M by the column for j: (2 × 0 + 1 × 1; 0 × 0 + 3 × 1) = (1; 3), the second column. This is true of every 2 × 2 matrix: its first column is where i lands, and its second column is where j lands.
So a matrix can be read straight off its columns. The matrix R = (0 −1; 1 0) has first column (0, 1), so it sends i to (0, 1), one step up. Its second column is (−1, 0), so it sends j to (−1, 0), one step left.
The unit square, solid, and its image under (2 1; 0 3), dashed. The corner (1, 0) lands on (2, 0), the first column; (0, 1) lands on (1, 3), the second column; and (1, 1) lands on (3, 3), their sum.
Why the columns are enough
The point (x, y) is x i + y j, and multiplying by a matrix respects sums and multiples: M(x i + y j) = x(M i) + y(M j). So M sends (x, y) to x times column 1 plus y times column 2.
For R, the point (2, 1) is 2i + j, so it lands on 2 × (0, 1) + 1 × (−1, 0) = (−1, 2). Multiplying gives the same: (0 × 2 + (−1) × 1; 1 × 2 + 0 × 1) = (−1; 2). Once i has turned up to (0, 1) and j has turned left to (−1, 0), every point turns with them, so R turns the whole plane a quarter turn counterclockwise about the origin.
R turns the triangle a quarter turn counterclockwise about the origin: the corner (2, 0) lands on (0, 2) and the corner (2, 1) lands on (−1, 2).
Reflections
Reflecting in the x-axis leaves i where it is and sends j to (0, −1), on the opposite side of the axis. So its matrix is (1 0; 0 −1), and it sends (x, y) to (x, −y).
Reflecting in the y-axis sends i to (−1, 0) and leaves j alone, so its matrix is (−1 0; 0 1). Reflecting in the line y = x swaps i and j: i goes to (0, 1) and j goes to (1, 0), so the matrix is (0 1; 1 0), and it swaps the two coordinates of every point.
The matrix (1 0; 0 −1) reflects the triangle in the x-axis: (3, 1) lands on (3, −1), and every point keeps its x-coordinate.
Enlargements and shears
The matrix (3 0; 0 3) sends i to (3, 0) and j to (0, 3), three times as far, so it sends every point to the point three times as far from the origin in the same direction: an enlargement with scale factor 3, centered on the origin.
The matrix (1 2; 0 1) leaves i where it is and sends j to (2, 1), sliding it 2 sideways. It sends (x, y) to (x + 2y, y): every point slides sideways by twice its height, so the x-axis stays put and the unit square leans over into a parallelogram. That is a shear.
The shear (1 2; 0 1) keeps the bottom edge of the unit square on the x-axis and slides the top edge 2 to the right, so the corner (0, 1) lands on (2, 1).
What a matrix cannot do
Any 2 × 2 matrix times the column (0; 0) is (0; 0), so a matrix never moves the origin. A translation moves every point, the origin included, so no 2 × 2 matrix is a translation. Every rotation a 2 × 2 matrix makes is about the origin, and every reflection is in a line through the origin.
The usual mistakes
Reading the rows instead of the columns. The rows of R are 0, −1 and 1, 0; taken as the images of i and j they give (0 1; −1 0), which turns the plane a quarter turn clockwise.
Reading the second column as the image of i. The second column of R, (−1, 0), is where j lands.
Writing the point as a row on the left, (2 1) times R. The point goes in as a column on the right of the matrix.
Taking a reflection for a turn. (1 0; 0 −1) leaves i where it is, so nothing turns.
A logo and a flower bed
In the first application below, a logo on a phone screen is turned a quarter turn. The matrix is read from where i and j land, and each corner of the logo is multiplied by it. In the second, a square flower bed is turned about a fountain through an angle whose cosine is , and its matrix comes from and in the same way.
Worked example: A Logo on a Phone Screen Turned Through a Quarter Turn
Question A phone shows a triangular logo with corners A(1, 1), B(4, 1) and C(1, 3), in centimeters from the center of the screen. When the phone is turned, the logo is rotated through 90° counterclockwise about the center. (a) Find where the rotation sends i = 10 and j = 01, and write down the matrix R of the rotation. (b) Find the corners A', B' and C' of the rotated logo, and check that the side AB keeps its length.
1.A quarter turn counterclockwise sends i, which points along the x-axis, to 01, up the y-axis. It sends j, which points up the y-axis, to −10, along the negative x-axis.
A quarter turn counterclockwise sends i to 01 and j to −10. 2.(a) The columns of the matrix are the images of i and j, in that order, so R = 0−110.
(a) The images of i and j are the columns: R = 0−110. 3.Write the corner A as a column and multiply it by R: R11 = 0 × 1 − 1 × 11 × 1 + 0 × 1 = −11, so A' is (−1, 1).
R11 = −11, so A' is (−1, 1): A has turned through 90° about the center. 4.In the same way, R41 = 0 − 14 + 0 = −14 and R13 = 0 − 31 + 0 = −31.
R41 = −14 and R13 = −31. 5.(b) The rotated logo has corners A'(−1, 1), B'(−1, 4) and C'(−3, 1). Check: AB runs 4 − 1 = 3 cm across, and A'B' runs 4 − 1 = 3 cm up, so the side keeps its length of 3 cm, as it must under a rotation.
(b) The rotated logo has corners A'(−1, 1), B'(−1, 4) and C'(−3, 1), and A'B' = AB = 3 cm.
Answer: (a) i goes to 01 and j to −10, so R = 0−110; (b) A'(−1, 1), B'(−1, 4) and C'(−3, 1), and A'B' = AB = 3 cm
Common mistakes
- Writing the images of i and j as the rows instead of the columns, which gives 01−10. That is the quarter turn clockwise, and it sends A(1, 1) to (1, −1), below the x-axis.
- Multiplying the corner as a row on the left, 11R. The point must be a column on the right of the matrix; the row product also gives the clockwise image, (1, −1).
More matrices as transformations problems, worked step by step →
Worked example: A Square Flower Bed Turned About a Fountain to Line Up with a New Path
Question On a park plan, a square flower bed has corners A(5, 5), B(10, 5), C(10, 10) and D(5, 10), in meters from a fountain at the origin O. To line it up with a new path, the bed is turned about the fountain through an angle θ counterclockwise, where cos θ = 35 and sin θ = 45. (a) Write down the rotation matrix R, and find the corners of the turned bed. (b) Show that the bed keeps its size: find the length of A'B' and find det R.
1.The columns are the images of i and j: R = cos θ−sin θsin θcos θ = 35−454535.
The columns are cos θsin θ and −sin θcos θ: R = 35−454535. 2.Multiply the first two corners by R: R55 = 3 − 44 + 3 = −17 and R105 = 6 − 48 + 3 = 211.
R55 = −17 and R105 = 211: the bed turns through θ about the fountain. 3.(a) In the same way C(10, 10) goes to (6 − 8, 8 + 6) = (−2, 14) and D(5, 10) goes to (3 − 8, 4 + 6) = (−5, 10). The turned bed has corners A'(−1, 7), B'(2, 11), C'(−2, 14) and D'(−5, 10).
(a) C'(−2, 14) and D'(−5, 10) complete the turned bed A'B'C'D'. 4.From A' to B' is 2 − (−1) = 3 m across and 11 − 7 = 4 m up, so by Pythagoras A'B' = √32 + 42 = √25 = 5 m, the same as AB = 10 − 5 = 5 m.
A'B' runs 3 m across and 4 m up, so A'B' = √32 + 42 = 5 m, the same as AB. 5.(b) A'B' = 5 m, and det R = 35 × 35 − (−45) × 45 = 925 + 1625 = 1. So the bed keeps its side of 5 m and its area of 5 × 5 = 25 square meters: the turn moves the bed without changing its size or its shape.
(b) det R = 925 + 1625 = 1: the bed keeps its side of 5 m and its area of 25 square meters.
Answer: (a) R = 35−454535, and the corners are A'(−1, 7), B'(2, 11), C'(−2, 14) and D'(−5, 10); (b) A'B' = 5 m, the same as AB, and det R = 1, so the area stays 25 square meters
Common mistakes
- Putting the minus sign on the wrong sin θ, which gives 3545−4535. That turns the bed clockwise and sends A(5, 5) to (7, −1), away from the path.
- Adding the coordinates' changes and taking A'B' as 3 + 4 = 7 m. The side runs 3 m across and 4 m up at the same time, so its length is the hypotenuse, √32 + 42 = 5 m.
More matrices as transformations problems, worked step by step →