Linear Transformations in 2D

A 2×2 matrix is not merely a grid of numbers — it is a geometric machine. Multiplying a position vector by a matrix moves, rotates, reflects, stretches or shears the whole plane, and every such linear transformation of the plane corresponds to exactly one matrix. This is the bridge between the algebra of matrices (Lessons 1–5) and the geometry of the plane, and it is one of the most visual, intuition-rich topics in Further Pure. The single idea that unlocks everything is this: the columns of the matrix are the images of the basis vectors. If you know where the matrix sends $\binom{1}{0}$ and $\binom{0}{1}$ , you know the matrix — and conversely, reading the columns tells you instantly what the matrix does. Master that and you can write down any transformation matrix, recognise any given one, compose transformations by multiplying, and reverse them by inverting. This lesson builds all of that rigorously, with the standard library of matrices derived (not just quoted), the all-important determinant–area link, and a full treatment of invariant points and invariant lines that examiners love to test.

1. Where this sits in AQA 7367

Linear transformations in 2D are compulsory pure content for Papers 1 and 2. The topic is unusually balanced across the assessment objectives: deriving or quoting a standard matrix and applying it is AO1; recognising an unknown matrix as a particular geometric transformation, or describing the effect of $\det M$ , is AO2 (reason and interpret); and the invariant-line work — solving for which lines map to themselves — is genuinely AO3 problem-solving. It draws directly on the matrix-multiplication lesson (composition is multiplication), the determinant lesson ( $\det$ = area scale factor), and the inverse lesson (undoing a transformation). It also reaches forward to eigenvectors (Lesson 8): an invariant line through the origin is exactly the direction of a real eigenvector.

2. What "linear" means, and why a matrix captures it

A transformation $T:\mathbb{R}^2\to\mathbb{R}^2$ is linear if it respects vector addition and scalar multiplication:

T(\mathbf{u}+\mathbf{v}) = T(\mathbf{u})+T(\mathbf{v}), \qquad T(k\mathbf{u}) = k\,T(\mathbf{u}).

Geometrically these two conditions force straight lines to stay straight, the origin to stay fixed, and parallel lines to stay parallel — the transformation can rotate, reflect, stretch and shear, but it cannot bend or translate. The pay-off is a theorem: every linear map of the plane is multiplication by a unique $2\times2$ matrix. Here is why. Write any vector in terms of the standard basis $\mathbf{e}_1=\binom{1}{0}$ , $\mathbf{e}_2=\binom{0}{1}$ :

\binom{x}{y} = x\,\mathbf{e}_1 + y\,\mathbf{e}_2.

Apply $T$ and use linearity twice:

T\binom{x}{y} = T(x\mathbf{e}_1 + y\mathbf{e}_2) = x\,T(\mathbf{e}_1) + y\,T(\mathbf{e}_2) = \bigl(\,T(\mathbf{e}_1)\ \ T(\mathbf{e}_2)\,\bigr)\binom{x}{y}.

So if we build the matrix $M$ whose columns are $T(\mathbf{e}_1)$ and $T(\mathbf{e}_2)$ , then $T(\mathbf{x}) = M\mathbf{x}$ for every $\mathbf{x}$ . This is the master key:

\boxed{\;M = \bigl(\,T(\mathbf{e}_1)\ \mid\ T(\mathbf{e}_2)\,\bigr)\;}

To find a transformation's matrix: track where $\binom{1}{0}$ and $\binom{0}{1}$ go, and put those images in as columns. To read a matrix: its first column tells you where $\binom{1}{0}$ lands, its second where $\binom{0}{1}$ lands.

3. The standard library — derived, not memorised

Every standard matrix below comes straight from the master key. Picture the unit vectors and ask where they go.

Rotation by $\theta$ anticlockwise about $O$ . The vector $\binom{1}{0}$ sweeps to $\binom{\cos\theta}{\sin\theta}$ ; the vector $\binom{0}{1}$ sweeps to $\binom{-\sin\theta}{\cos\theta}$ . Hence

R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}.

Reflections. Reflecting in the $x$ -axis fixes $\binom{1}{0}$ and flips $\binom{0}{1}\to\binom{0}{-1}$ ; reflecting in $y=x$ swaps the two basis vectors. The general reflection in the line through $O$ at angle $\theta$ to the $x$ -axis is $\begin{pmatrix}\cos2\theta & \sin2\theta \\ \sin2\theta & -\cos2\theta\end{pmatrix}$ , a result worth knowing for harder questions.

Transformation	Matrix	$\det$
Reflection in $x$ -axis	$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$	$-1$
Reflection in $y$ -axis	$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$	$-1$
Reflection in $y=x$	$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$	$-1$
Reflection in $y=-x$	$\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$	$-1$
Rotation $\theta$ anticlockwise	$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$	$+1$
Enlargement, factor $k$	$\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$	$k^2$
Stretch in $x$ , factor $a$	$\begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}$	$a$
Shear parallel to $x$ -axis, factor $k$	$\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$	$1$

Notice the determinant column already tells a story: reflections have $\det=-1$ (orientation reversed), rotations $\det=+1$ (orientation and area preserved), and shears $\det=1$ (area preserved even though shape is sheared). That is the next section.

The image below shows the unit square (black) and its image (blue) under the shear $\binom{1\ \,2}{0\ \,1}$ : the base stays put, the top slides right, the area is unchanged.

4. The determinant as the area scale factor

For a transformation with matrix $M$ , the unit square (area 1) maps to a parallelogram whose area is exactly $\lvert\det M\rvert$ . More generally, every region's area is multiplied by $\lvert\det M\rvert$ . The sign of $\det M$ records orientation:

$\det M > 0$ : orientation preserved (a "right-handed" shape stays right-handed);
$\det M < 0$ : orientation reversed (the shape is flipped — a reflection is involved);
$\det M = 0$ : the plane is squashed onto a line (or the origin), area $\to 0$ , and the transformation has no inverse.

This is why an enlargement of factor $k$ multiplies area by $k^2$ (its determinant is $k^2$ ), while a rotation leaves area untouched ( $\det = \cos^2\theta+\sin^2\theta = 1$ ).

5. Composition is multiplication — and order matters

If transformation $P$ has matrix $A$ and transformation $Q$ has matrix $B$ , then doing $P$ first and then $Q$ has matrix $BA$ — the second transformation is written on the left. The reason is the order in which the matrices hit the vector: $Q\bigl(P(\mathbf{x})\bigr) = B(A\mathbf{x}) = (BA)\mathbf{x}$ . Because matrix multiplication is not commutative, $BA\neq AB$ in general, so the order of transformations genuinely changes the result. The mantra:

\boxed{\text{"}P \text{ then } Q\text{" has matrix } BA \;-\; \textbf{first applied is rightmost.}}

Worked Example 1 — find a rotation matrix (with mark scheme)

Find the matrix representing rotation by $90^\circ$ anticlockwise about the origin, and use it to find the image of $(4,1)$ .

With $\theta = 90^\circ$ , $\cos\theta = 0$ and $\sin\theta = 1$ , so

R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \qquad R\binom{4}{1} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\binom{4}{1} = \binom{-1}{4}.

So $(4,1)\mapsto(-1,4)$ . (Sanity check: a $90^\circ$ anticlockwise turn should send a point in the first quadrant into the second — it does.)

Mark scheme: M1 quote/derive $R(\theta)$ and substitute $\theta=90^\circ$ ; A1 correct matrix $\binom{0\ -1}{1\ \ 0}$ ; M1 multiply by $\binom{4}{1}$ ; A1 image $(-1,4)$ .

Worked Example 2 — composition (with mark scheme)

Find the single matrix representing a reflection in the $x$ -axis followed by a rotation of $90^\circ$ anticlockwise, and describe the resulting transformation.

Reflection in the $x$ -axis: $A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ . Rotation $90^\circ$ anticlockwise: $B = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ . "Reflection then rotation" means apply $A$ first, so the single matrix is $BA$ :

BA = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.

This is the matrix for reflection in $y=x$ . (Check: $\det(BA) = -1$ , consistent with a reflection; and indeed reflect-then-rotate composes two orientation-reversing/preserving maps to give an orientation-reversing one.)

Mark scheme: M1 write down both matrices correctly; M1 form the product in the correct order $BA$ (not $AB$ ); A1 $\binom{0\ 1}{1\ 0}$ ; A1 identify it as reflection in $y=x$ . A candidate who computes $AB$ instead loses the order mark and reaches $\binom{0\,-1}{-1\ \ 0}$ , reflection in $y=-x$ — the wrong answer.

6. Inverse transformations

The inverse transformation — the one that undoes $M$ — has matrix $M^{-1}$ , which exists precisely when $\det M \neq 0$ (i.e. when the transformation does not collapse the plane). Two facts are worth carrying into the exam:

If $M = R(\theta)$ is a rotation by $\theta$ , then $M^{-1} = R(-\theta)$ — rotate back. Indeed $R(\theta)^{-1} = \begin{pmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{pmatrix} = R(-\theta)$ .
If $M$ is a reflection, then $M^{-1} = M$ : reflecting twice in the same line returns every point to where it started, so $M^2 = I$ and $M$ is self-inverse (an involution).

A transformation with $\det M = 0$ (such as a projection onto a line) has no inverse — information is lost when the plane is flattened, so it cannot be undone.

7. Invariant points and invariant lines

This is the AO3 heart of the topic, and the place students most often go wrong.

An invariant point is a point $\mathbf{p}$ fixed by the transformation: $M\mathbf{p}=\mathbf{p}$ . The origin is always invariant for a linear map ( $M\mathbf{0}=\mathbf{0}$ ). Finding other invariant points means solving $(M-I)\mathbf{p}=\mathbf{0}$ : if $\det(M-I)\neq0$ the origin is the only one; if $\det(M-I)=0$ there is a whole line of invariant points.

An invariant line is a line that maps onto itself as a set — points may slide along it, but no point leaves it. Crucially this is weaker than "a line of invariant points": on an invariant line the individual points can move; on a line of invariant points each point is fixed. Examiners deliberately exploit this distinction.

To find invariant lines through the origin, $y=mx$ : a point $\binom{x}{mx}$ must map to another point on the same line. Compute $M\binom{x}{mx}=\binom{X}{Y}$ and impose $Y=mX$ ; solving for $m$ gives the gradients of the invariant lines. (Invariant lines need not pass through the origin in general, but at this level the origin-lines are the standard case, and they coincide with real eigenvector directions — the link to Lesson 8.)

It is worth pausing on why these two ideas are so often confused, because the distinction is where marks are won. A reflection in $y=x$ , for example, fixes every single point of the mirror line — so $y=x$ is a line of invariant points. But the perpendicular line $y=-x$ is also mapped to itself by that same reflection: a point on $y=-x$ is sent to the diametrically opposite point on $y=-x$ , still on the line. So $y=-x$ is an invariant line but emphatically not a line of invariant points — its points all move (except the origin). The test in each case is different: for a line of invariant points you solve $M\mathbf{p}=\mathbf{p}$ (each point fixed); for an invariant line you only require that the image of each point stays somewhere on the line. Whenever a question says "invariant", read carefully which of the two it means, and state explicitly which you have found.

Worked Example 3 — invariant lines of a shear (with mark scheme)

Find all invariant lines through the origin of the shear $M = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ .

Linear Transformations in 2D

Linear Transformations in 2D

1. Where this sits in AQA 7367

2. What "linear" means, and why a matrix captures it

3. The standard library — derived, not memorised

4. The determinant as the area scale factor

5. Composition is multiplication — and order matters

Worked Example 1 — find a rotation matrix (with mark scheme)

Worked Example 2 — composition (with mark scheme)

6. Inverse transformations

7. Invariant points and invariant lines

Worked Example 3 — invariant lines of a shear (with mark scheme)

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