Matrix Transformations

The single fact that unlocks this whole topic. A linear transformation $T$ of the plane is completely determined by what it does to the two basis vectors $\mathbf{i} = (1,0)$ and $\mathbf{j} = (0,1)$. The matrix that represents $T$ has those two images as its columns:

$\mathbf{M} = \Big(\, T(\mathbf{i}) \;\big|\; T(\mathbf{j}) \,\Big)$

Why? Apply $\mathbf{M}$ to $(1,0)$ and you pick out the first column; apply it to $(0,1)$ and you pick out the second column. So if you are told (or can work out) where $(1,0)$ and $(0,1)$ go, you can write the matrix straight down.

A point becomes a column vector. To transform a point $P(x,y)$ you write it as a column and multiply the matrix on the left:

$\mathbf{r}' = \mathbf{M}\begin{pmatrix} x \\ y \end{pmatrix}$

The diagram shows the unit square (blue) and its image (red) under a shear $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$: the bottom edge along the $x$-axis is fixed, while the top edge slides right.

Cambridge tip. Whenever you are unsure what a matrix does, multiply it by $(1,0)$ and $(0,1)$ — the two answers are exactly the first and second columns, and they tell you the geometry instantly.

These are the matrices you must recognise and reproduce. Each one is verified by checking the images of $(1,0)$ and $(0,1)$.

Rotation about $O$, anticlockwise by $\theta$: $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ Check: $(1,0)\mapsto(\cos\theta, \sin\theta)$ — a unit vector turned anticlockwise by $\theta$. (For clockwise, use $-\theta$; the off-diagonal signs swap.)

Reflections:

• in the $x$-axis: $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$    (in the $y$-axis: $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$)
• in $y = x$: $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$    (in $y = -x$: $\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$)

Every reflection has $\det = -1$.

Enlargement, centre $O$, scale factor $k$: $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$. A one-way stretch scales just one axis: $\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}$ (parallel to $x$-axis) or $\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}$ (parallel to $y$-axis). A two-way stretch has both diagonal entries different from $1$.

Shears (area-preserving, $\det = 1$): $\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ shears parallel to the $x$-axis ($x$-axis invariant); $\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}$ shears parallel to the $y$-axis.

The diagram contrasts the unit square (blue) with its image under a $90^\circ$ anticlockwise rotation (red).

Cambridge tip. Don't memorise reflections in $y=x$ and $y=-x$ as separate facts — derive them in two seconds from the columns: in $y=x$, $(1,0)\mapsto(0,1)$ so the first column is $\binom{0}{1}$.

To apply transformation $P$ and then transformation $Q$ to a point $\mathbf{r}$, you compute:

$Q\big(P(\mathbf{r})\big) = \mathbf{Q}\big(\mathbf{P}\mathbf{r}\big) = (\mathbf{Q}\mathbf{P})\,\mathbf{r}$

So the single matrix for "$P$ first, then $Q$" is the product $\mathbf{Q}\mathbf{P}$ — the matrix of the first transformation sits on the right, nearest the vector, because it acts first.

This is the single biggest trap in the topic. Because we read left-to-right, "$P$ then $Q$" feels like it should be $\mathbf{P}\mathbf{Q}$, but matrix multiplication is not commutative: in general $\mathbf{P}\mathbf{Q} \neq \mathbf{Q}\mathbf{P}$, and the two products are usually different transformations.

A concrete warning. Let $P$ = reflection in the $x$-axis $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ and $Q$ = rotation $90^\circ$ $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. Then: $\mathbf{Q}\mathbf{P} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \;\;(\text{reflection in } y=x), \qquad \mathbf{P}\mathbf{Q} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \;\;(\text{reflection in } y=-x)$ Same two transformations, opposite order, genuinely different result.

The diagram tracks a point through "$P$ then $Q$": the object (blue) is first reflected, then rotated to its final image (red).

Cambridge tip. Write the order out before you multiply: "$P$ then $Q$ $\Rightarrow$ $\mathbf{Q}\mathbf{P}$". If still unsure, push a test point through both transformations by hand and confirm it matches your combined matrix.

For a transformation matrix $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant $\det\mathbf{M} = ad - bc$ has a beautiful geometric meaning.

Area scale factor. Any region of area $A$ is transformed into a region of area $|\det\mathbf{M}|\times A$: $\text{area of image} = \left|\det\mathbf{M}\right| \times \text{original area}$ The unit square (area $1$) maps to a parallelogram spanned by the columns of $\mathbf{M}$, whose area is exactly $|\det\mathbf{M}|$.

Why the modulus? Area can never be negative, so you always take the absolute value for the area itself.

Orientation from the sign. The sign of the determinant carries extra information:

• $\det\mathbf{M} > 0$: orientation preserved (anticlockwise stays anticlockwise) — rotations, enlargements, stretches and shears all have positive determinant.
• $\det\mathbf{M} < 0$: orientation reversed (the shape is "flipped over") — every reflection has $\det = -1$.
• $\det\mathbf{M} = 0$: the transformation is singular — it collapses the whole plane onto a line (or a point), and is not invertible.

Quick checks worth remembering: rotations and shears have $\det = 1$ (areas unchanged); an enlargement scale factor $k$ has $\det = k^2$ (since both directions scale by $k$).

Cambridge tip. When a question asks for the area of a transformed shape, compute $|\det\mathbf{M}|$ once and multiply — you do not need to find the image coordinates at all. Keep the sign in mind separately to answer any "is the orientation reversed?" follow-up.