Matrices and Determinants of Order 2

Lessons

Senior 4 · Mathematics

Matrices and Determinants of Order 2

5 lessons

Lesson 1 of 5

Linear transformations and their matrices

25 min

Moving every point at once

When a phone rotates a photo, or a design program makes upright letters lean to the right, it does not move the points one at a time by hand. It applies a single rule to every point $(x, y)$ of the picture. Many fonts, for example, make their slanted ("oblique") style with a rule of the form $(x, y) \mapsto (x + ky,\ y)$ for a small number $k$ : points high up slide further to the right than points near the base line.

Rules of this kind send vectors of $\mathbb{R}^2$ to vectors of $\mathbb{R}^2$ . The most important of them respect the two operations of the vector space, and every one of those can be stored in a small table of four numbers: a matrix.

Definition

Linear transformation: a function $f : \mathbb{R}^2 \to \mathbb{R}^2$ is linear when, for all vectors $\vec{u}, \vec{v}$ and every real $k$ ,

f(\vec{u} + \vec{v}) = f(\vec{u}) + f(\vec{v}) \quad \text{and} \quad f(k\vec{u}) = k f(\vec{u}).

The vector $f(\vec{u})$ is the image of $\vec{u}$ . Taking $k = 0$ shows that every linear transformation satisfies $f(\vec{0}) = \vec{0}$ .

Example: $f(x, y) = (2x + y,\ 3y)$ is linear; $g(x, y) = (x + 1,\ y)$ is not, because $g(0, 0) = (1, 0) \ne \vec{0}$ .

Worked example 1: testing for linearity

Decide whether (a) $f(x, y) = (2x + y,\ 3y)$ and (b) $h(x, y) = (xy,\ x)$ are linear.

Solution. (a) Let $\vec{u} = (x_1, y_1)$ , $\vec{v} = (x_2, y_2)$ . Then

f(\vec{u} + \vec{v}) = \big(2(x_1 + x_2) + (y_1 + y_2),\ 3(y_1 + y_2)\big) = (2x_1 + y_1,\ 3y_1) + (2x_2 + y_2,\ 3y_2) = f(\vec{u}) + f(\vec{v}),

and $f(k\vec{u}) = (2kx_1 + ky_1,\ 3ky_1) = k f(\vec{u})$ . So $f$ is linear.

(b) $h (2 (1, 1)) = h (2, 2) = (4, 2)$ , but $2 h (1, 1) = 2 (1, 1) = (2, 2)$ . The second condition fails, so $h$ is not linear. A product $x y$ of the coordinates always spoils linearity.

Two images decide everything

Every vector is $(x, y) = x\vec{e}_1 + y\vec{e}_2$ . If $f$ is linear, the two conditions give

f(x, y) = x\,f(\vec{e}_1) + y\,f(\vec{e}_2).

So once you know where the two standard basis vectors go, you know where every vector goes. Write $f(\vec{e}_1) = (a, c)$ and $f(\vec{e}_2) = (b, d)$ . Then $f(x, y) = (ax + by,\ cx + dy)$ , and we store the four numbers with the two images as columns:

M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \qquad M\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}.

The diagram shows $M = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}$ : its first column $(2, 0)$ is where $\vec{e}_1$ lands and its second column $(1, 1)$ is where $\vec{e}_2$ lands, so the unit square becomes a parallelogram.

Definition

Matrix of order 2: a square array $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ of four real numbers, its entries, in 2 rows and 2 columns. The entry in row $i$ and column $j$ is written $m_{ij}$ : here $m_{11} = a$ , $m_{12} = b$ , $m_{21} = c$ , $m_{22} = d$ .

Matrix of a linear transformation: the matrix whose first column is $f(\vec{e}_1)$ and whose second column is $f(\vec{e}_2)$ . Then $f(\vec{v}) = M\vec{v}$ for every $\vec{v}$ , with $\vec{v}$ written as a column.

Example: $f(x, y) = (3x - y,\ x + 2y)$ has matrix $\begin{pmatrix} 3 & -1 \\ 1 & 2 \end{pmatrix}$ : read the coefficients of $x$ and $y$ row by row.

Worked example 2: a matrix from two images

(a) A linear transformation $f$ has $f (1, 0) = (2, 5)$ and $f (0, 1) = (- 1, 3)$ . Write its matrix and find $f (4, 2)$ .

(b) A linear transformation $g$ has $g (1, 1) = (3, 1)$ and $g (1, - 1) = (1, 5)$ . Find its matrix.

Solution.

(a) The images are the columns: $M = \begin{pmatrix} 2 & -1 \\ 5 & 3 \end{pmatrix}$ . Then $M\begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 - 2 \\ 20 + 6 \end{pmatrix} = \begin{pmatrix} 6 \\ 26 \end{pmatrix}$ .

(b) Since $\vec{e}_1 = \frac{1}{2}\big[(1, 1) + (1, -1)\big]$ and $\vec{e}_2 = \frac{1}{2}\big[(1, 1) - (1, -1)\big]$ , linearity gives

g(\vec{e}_1) = \tfrac{1}{2}\big[(3, 1) + (1, 5)\big] = (2, 3), \qquad g(\vec{e}_2) = \tfrac{1}{2}\big[(3, 1) - (1, 5)\big] = (1, -2).

So $g$ has matrix $\begin{pmatrix} 2 & 1 \\ 3 & -2 \end{pmatrix}$ . Check: $\begin{pmatrix} 2 & 1 \\ 3 & -2 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ . ✓

Matrices of geometric transformations

Many transformations from S2 and S3 geometry are linear because they fix the origin and keep straight lines and parallel lines. To find the matrix, ask where $\vec{e}_1 = (1, 0)$ and $\vec{e}_2 = (0, 1)$ go, and write the answers as columns.

Transformation (centre or axis through $O$ )	$\vec{e}_1 \mapsto$	$\vec{e}_2 \mapsto$	Matrix
Reflection in the $x$ -axis	$(1, 0)$	$(0, - 1)$	$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
Reflection in the $y$ -axis	$(- 1, 0)$	$(0, 1)$	$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$
Reflection in the line $y = x$	$(0, 1)$	$(1, 0)$	$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
Rotation of $90^\circ$ anticlockwise about $O$	$(0, 1)$	$(- 1, 0)$	$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$
Rotation of $180^\circ$ about $O$	$(- 1, 0)$	$(0, - 1)$	$\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$
Enlargement, centre $O$ , scale factor $k$	$(k, 0)$	$(0, k)$	$\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$
Shear parallel to the $x$ -axis, factor $k$	$(1, 0)$	$(k, 1)$	$\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$
Projection onto the $x$ -axis	$(1, 0)$	$(0, 0)$	$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$

A translation $(x, y) \mapsto (x + p,\ y + q)$ with $(p, q) \ne \vec{0}$ moves the origin, so it is not linear and has no $2 \times 2$ matrix.

Worked example 3: the image of a shape

Find the image of the triangle $A (1, 1)$ , $B (4, 1)$ , $C (1, 3)$ under the rotation of $90^\circ$ anticlockwise about $O$ .

Solution. The matrix is $R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ , so $R\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -y \\ x \end{pmatrix}$ . Hence $A^{'} = (- 1, 1)$ , $B^{'} = (- 1, 4)$ , $C^{'} = (- 3, 1)$ . The side $A B$ was horizontal with length 3; its image $A^{'} B^{'}$ is vertical with length 3, as a rotation of $90^\circ$ should give.

Warning

Columns, not rows. The image of $\vec{e}_1$ is the first column of the matrix. Writing the images as rows gives the matrix of a different transformation (the transpose, which you will meet later in this unit). For the rotation of $90^\circ$ anticlockwise, $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ is right; $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ rotates the other way.

Tip

To find the matrix of any linear transformation, ask two questions: "Where does $(1, 0)$ go? Where does $(0, 1)$ go?" Write the two answers side by side as columns. Check with one more point.

3Blue1Brown: Linear transformations and matrices, Chapter 3 of Essence of Linear Algebra (about 11 min). Watch for how the columns of a matrix record where the two basis vectors land.

✦

Practice

Work each one, then open the answers.

Which are linear? (a) $f(x, y) = (4x,\ x - y)$ (b) $g(x, y) = (x + 2,\ y)$ (c) $h(x, y) = (x^2,\ y)$ (d) $k(x, y) = (0,\ 5x)$
Write the matrix of $f(x, y) = (x - 4y,\ 2x + 3y)$ and find $f (3, - 2)$ .
A linear transformation sends $(1, 0)$ to $(3, - 1)$ and $(0, 1)$ to $(2, 4)$ . Write its matrix and find the image of $(- 2, 5)$ .
A linear transformation $f$ has $f (1, 1) = (5, 2)$ and $f (1, - 1) = (1, 0)$ . Find its matrix.
Write the matrix of (a) the reflection in the $y$ -axis (b) the enlargement with centre $O$ and scale factor $- 2$ (c) the projection onto the $y$ -axis.
Find the image of the triangle $A (1, 0)$ , $B (3, 1)$ , $C (2, 3)$ under $M = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ , and name the transformation.
The shear $(x, y) \mapsto (x + ky,\ y)$ sends $(2, 3)$ to $(8, 3)$ . Find $k$ and the matrix of the shear.
Prove that a linear transformation sends $\vec{0}$ to $\vec{0}$ , and deduce that the translation by $(2, 1)$ is not linear.

Show answersHide answers

(a) linear (b) not: $g (0, 0) = (2, 0)$ (c) not: $h(2, 0) = (4, 0) \ne 2h(1, 0) = (2, 0)$ (d) linear.
$\begin{pmatrix} 1 & -4 \\ 2 & 3 \end{pmatrix}$ ; $f(3, -2) = (3 + 8,\ 6 - 6) = (11, 0)$ .
$\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}$ ; image $- 2 (3, - 1) + 5 (2, 4) = (4, 22)$ .
$f(\vec{e}_1) = \frac{1}{2}\big[(5, 2) + (1, 0)\big] = (3, 1)$ , $f(\vec{e}_2) = \frac{1}{2}\big[(5, 2) - (1, 0)\big] = (2, 1)$ : matrix $\begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix}$ .
(a) $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ (b) $\begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix}$ (c) $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ .
$M\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} y \\ -x \end{pmatrix}$ : $A^{'} = (0, - 1)$ , $B^{'} = (1, - 3)$ , $C^{'} = (3, - 2)$ . It is the rotation of $90^\circ$ clockwise about $O$ .
$2 + 3 k = 8$ , so $k = 2$ ; matrix $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ .
$f(\vec{0}) = f(0 \cdot \vec{0}) = 0 \cdot f(\vec{0}) = \vec{0}$ . The translation sends $(0, 0)$ to $(2, 1) \ne \vec{0}$ , so it is not linear.

Quick check

Matrices are written row by row. Which matrix represents the reflection in the line y = x?

Key points

$f$ is linear when $f(\vec{u} + \vec{v}) = f(\vec{u}) + f(\vec{v})$ and $f(k\vec{u}) = kf(\vec{u})$ ; then $f(\vec{0}) = \vec{0}$ .
The matrix of $f$ has columns $f(\vec{e}_1)$ and $f(\vec{e}_2)$ , and $f(\vec{v}) = M\vec{v}$ .
$\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}$ .
Reflections in lines through $O$ , rotations and enlargements about $O$ , shears and projections are linear; translations are not.

Sources and credits

Written by Ganzaa: the explanations, the diagrams, the worked examples, the practice items and the questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
Video: Linear transformations and matrices, Chapter 3, Essence of linear algebra, 3Blue1Brown, embedded from YouTube.

1Moving every point at once

2Two images decide everything

3Matrices of geometric transformations

Moving every point at once

Two images decide everything

Matrices of geometric transformations