Paper 1 · Further Pure 1 ✦

Matrices & Transformations

A matrix is a machine that moves the plane. Learn to read the geometry straight off the numbers — and to spot what the machine leaves untouched.

Syllabus 1.4At most 3 × 3MF19 gives you nothing here

What the examiner expects you to do

Add, subtract and multiply matrices (including non-square ones and products of more than two); use the terms zero matrix and identity matrix I.
Know singular and non-singular; evaluate $\det 𝐌$ and find inverses for 2 × 2 and 3 × 3 matrices.
Use $(𝐀 𝐁)^{- 1} = 𝐁^{- 1} 𝐀^{- 1}$ .
Use 2 × 2 matrices as plane transformations — rotation, reflection, enlargement, stretch, shear — including the link between $𝐀$ and $𝐀^{- 1}$ , the fact that AB means B first, the area scale factor, and finding the matrix for a given transformation or sequence.
Understand invariant points and lines, and solve simple problems about them.

§1 The algebra you must be fluent in

◇ The rules

Addition / subtraction: only for matrices of the same order; work entry by entry.
Multiplication: $(m \times n) (n \times p) = (m \times p)$ . The inner numbers must match, and they vanish; the outer numbers give the order of the answer. Entry $(i, j)$ is row $i$ of the first dotted with column $j$ of the second.
Identity $𝐈$ : ones down the leading diagonal, zeros elsewhere. $𝐀 𝐈 = 𝐈 𝐀 = 𝐀$ .
Zero matrix $𝟎$ : all entries zero. Careful — $𝐀 𝐁 = 𝟎$ does not force $𝐀 = 𝟎$ or $𝐁 = 𝟎$ .

◆ Associative, but not commutative

Matrix multiplication is associative, so $(𝐀 𝐁) 𝐂 = 𝐀 (𝐁 𝐂)$ and a product of three or more can be built up in any grouping. It is not commutative — in general

𝐀 𝐁 \neq 𝐁 𝐀 .

Order is not a technicality here; it is the whole geometry, as $§ 3$ shows.

✎ Worked example — order matters

With $𝐀 = (\begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array})$ and $𝐁 = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ :

𝐀 𝐁 = (\begin{array}{cc} 2 & 1 \\ 1 & 0 \end{array}), 𝐁 𝐀 = (\begin{array}{cc} 0 & 1 \\ 1 & 2 \end{array}) .

Same two matrices, different products. Always read the question for which comes first.

★ A* tip — check the order before you multiply

Write the orders side by side first: $(2 \times 3) (3 \times 2)$ is fine and gives $2 \times 2$ ; $(3 \times 2) (3 \times 2)$ is undefined. Examiners set at least one product a year that simply does not exist, and the mark is for saying so.

§2 Determinant, singularity and the inverse

◆ 2 × 2

𝐌 = (\begin{array}{cc} a & b \\ c & d \end{array}) \Rightarrow \det 𝐌 = a d - b c, 𝐌^{- 1} = \frac{1}{a d - b c} (\begin{array}{cc} d & - b \\ - c & a \end{array}) .

Swap the leading diagonal, negate the other diagonal, divide by the determinant. $𝐌$ is singular when $\det 𝐌 = 0$ — and then $𝐌^{- 1}$ does not exist.

◆ 3 × 3 — determinant by expansion along the first row

With the alternating sign pattern $+ - +$ , each entry is multiplied by the $2 \times 2$ determinant left when you delete its row and column (its minor):

\det (\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}) = a | \begin{array}{cc} e & f \\ h & i \end{array} | - b | \begin{array}{cc} d & f \\ g & i \end{array} | + c | \begin{array}{cc} d & e \\ g & h \end{array} | .

You may expand along any row or column — choose the one with the most zeros — provided you use the matching signs from the chequerboard

(\begin{array}{ccc} + & - & + \\ - & + & - \\ + & - & + \end{array}) .

◇ 3 × 3 inverse — the four-step recipe

Minors: replace each entry by the $2 \times 2$ determinant of what is left when you delete its row and column.
Cofactors: apply the $+ - +$ chequerboard of signs.
Adjugate: transpose the cofactor matrix (reflect it in the leading diagonal).
Divide by $\det 𝐌$ . So $𝐌^{- 1} = \frac{1}{\det 𝐌} adj 𝐌$ .

Forgetting the transpose in step 3 is the classic error — and it is invisible unless you check.

✎ Worked example — a full 3 × 3 inverse

Invert $𝐌 = (\begin{array}{ccc} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array})$ .

Determinant (first row): $2 (1 \cdot 1 - 0 \cdot 1) - 0 (1 \cdot 1 - 0 \cdot 0) + 1 (1 \cdot 1 - 1 \cdot 0) = 2 + 0 + 1 = 3 \neq 0$ , so $𝐌$ is non-singular.

Cofactors (minors with the signs already applied):

𝐂 = (\begin{array}{ccc} 1 & - 1 & 1 \\ 1 & 2 & - 2 \\ - 1 & 1 & 2 \end{array}) ⟹ adj 𝐌 = 𝐂^{𝖳} = (\begin{array}{ccc} 1 & 1 & - 1 \\ - 1 & 2 & 1 \\ 1 & - 2 & 2 \end{array}) .

Divide by the determinant:

𝐌^{- 1} = \frac{1}{3} (\begin{array}{ccc} 1 & 1 & - 1 \\ - 1 & 2 & 1 \\ 1 & - 2 & 2 \end{array}) .

Always check one row: row 1 of $𝐌$ times column 1 of the adjugate is $2 (1) + 0 (- 1) + 1 (1) = 3$ , and $3 \div 3 = 1$ . ✓ That single line catches a lost transpose.

◆ The reversal law

(𝐀 𝐁)^{- 1} = 𝐁^{- 1} 𝐀^{- 1}

Inverses come off in the opposite order.

Why the order reverses (and why it is obvious once you see it)

Multiply $𝐀 𝐁$ by the candidate $𝐁^{- 1} 𝐀^{- 1}$ and use associativity:

(𝐀 𝐁) (𝐁^{- 1} 𝐀^{- 1}) = 𝐀 ({𝐁 𝐁}^{- 1}) 𝐀^{- 1} = {𝐀 𝐈 𝐀}^{- 1} = {𝐀 𝐀}^{- 1} = 𝐈,

and in the same way $(𝐁^{- 1} 𝐀^{- 1}) (𝐀 𝐁) = 𝐈$ . Since the inverse is unique, $(𝐀 𝐁)^{- 1} = 𝐁^{- 1} 𝐀^{- 1}$ . $■$

Geometrically: $𝐀 𝐁$ does $𝐁$ then $𝐀$ . To undo it you must undo $𝐀$ first, then $𝐁$ — socks on, shoes on; shoes off, socks off.

§3 2 × 2 matrices as transformations

◆ The one idea that generates every transformation matrix

Because $(\begin{array}{cc} a & b \\ c & d \end{array}) (\begin{array}{c} 1 \\ 0 \end{array}) = (\begin{array}{c} a \\ c \end{array})$ and $(\begin{array}{cc} a & b \\ c & d \end{array}) (\begin{array}{c} 0 \\ 1 \end{array}) = (\begin{array}{c} b \\ d \end{array})$ :

𝐭𝐡𝐞 𝐜𝐨𝐥𝐮𝐦𝐧𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐦𝐚𝐭𝐫𝐢𝐱 𝐚𝐫𝐞 𝐭𝐡𝐞 𝐢𝐦𝐚𝐠𝐞𝐬 𝐨𝐟 𝐢 𝐚𝐧𝐝 𝐣 .

To build the matrix for any described transformation, just work out where $(1,0)$ and $(0,1)$ go and write them down as columns. Nothing on this topic is printed in MF19, so this is how you recover every result under exam pressure.

◇ The five named transformations

Transformation	Matrix	det
Rotation about $O$ , angle $θ$ anticlockwise	$(\begin{array}{cc} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array})$	$1$
Reflection in the line through $O$ at angle $θ$ to the $x$ -axis	$(\begin{array}{cc} \cos 2 θ & \sin 2 θ \\ \sin 2 θ & - \cos 2 θ \end{array})$	$- 1$
Enlargement, centre $O$ , scale factor $k$	$(\begin{array}{cc} k & 0 \\ 0 & k \end{array}) = k 𝐈$	$k^{2}$
Stretch factor $k$ parallel to the $x$ -axis (the $y$ -axis is fixed)	$(\begin{array}{cc} k & 0 \\ 0 & 1 \end{array})$	$k$
Shear, $x$ -axis fixed, $(0,1) \mapsto (k, 1)$	$(\begin{array}{cc} 1 & k \\ 0 & 1 \end{array})$	$1$

Special cases worth recognising instantly: reflection in the $x$ -axis $(\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$ (that is $θ = 0$ ), reflection in $y = x$ $(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ ( $θ = 45^{\circ}$ ), rotation by $90^{\circ}$ $(\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$ .

◆ Composition, inverses and area

$𝐀 𝐁$ means "do $𝐁$ first, then $𝐀$ " — because $(𝐀 𝐁) 𝐱 = 𝐀 (𝐁 𝐱)$ . Matrices compose right to left.
$𝐀^{- 1}$ is the transformation that undoes $𝐀$ : rotate by $- θ$ , reflect in the same line again, enlarge by $1 / k$ .
Area scale factor $= | \det 𝐀 |$ . A region of area $S$ maps to one of area $| \det 𝐀 | S$ .
The sign of the determinant carries the orientation: $\det > 0$ keeps it, $\det < 0$ flips it (a reflection), $\det = 0$ collapses the whole plane onto a line — which is exactly why a singular matrix has no inverse.

✎ Worked example — a sequence of transformations

Find the matrix for "reflect in the $x$ -axis, then rotate $90^{\circ}$ anticlockwise about $O$ ".

Reflection $𝐑 = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$ , rotation $𝐐 = (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$ . Reflection is first, so it goes on the right:

𝐐 𝐑 = (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}) (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}) = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}),

which is reflection in $y = x$ . Check with $𝐢$ : $(1,0) \to (1,0) \to (0,1)$ , and the answer matrix does send $(1,0)$ to $(0,1)$ . ✓

The other order, $𝐑 𝐐 = (\begin{array}{cc} 0 & - 1 \\ - 1 & 0 \end{array})$ , is reflection in $y = - x$ — a different transformation.

✎ Worked example — area

A triangle of area $6$ is transformed by $𝐀 = (\begin{array}{cc} 3 & 1 \\ 1 & 2 \end{array})$ . Since $\det 𝐀 = 6 - 1 = 5$ , the image has area $5 \times 6 = 30$ .

§4 Invariant points and invariant lines

⚠ Two different things — do not blur them

An invariant point does not move: $𝐀 𝐱 = 𝐱$ .
An invariant line maps onto itself as a whole. Its points are allowed to slide along it. A line of invariant points is the stronger condition where every point on it is fixed.

◆ Invariant points — solve

(𝐀 - 𝐈) 𝐱 = 𝟎

$𝐀 𝐱 = 𝐱 ⟺ (𝐀 - 𝐈) 𝐱 = 𝟎$ . The origin always works, so the only question is whether anything else does:

$\det (𝐀 - 𝐈) \neq 0$ ⟹ the origin only.
$\det (𝐀 - 𝐈) = 0$ (and $𝐀 \neq 𝐈$ ) ⟹ a whole line of invariant points through the origin; read it off either row.

✎ Worked example — a line of invariant points

Find the invariant points of $𝐀 = (\begin{array}{cc} 6 & 5 \\ 2 & 3 \end{array})$ .

𝐀 - 𝐈 = (\begin{array}{cc} 5 & 5 \\ 2 & 2 \end{array}), \det (𝐀 - 𝐈) = 10 - 10 = 0,

so there is more than just the origin. The first row gives $5 x + 5 y = 0$ , i.e. $y = - x$ (the second row, $2 x + 2 y = 0$ , says the same thing — a good check).

Verify: $(\begin{array}{cc} 6 & 5 \\ 2 & 3 \end{array}) (\begin{array}{c} t \\ - t \end{array}) = (\begin{array}{c} 6 t - 5 t \\ 2 t - 3 t \end{array}) = (\begin{array}{c} t \\ - t \end{array})$ . ✓ Every point of $y = - x$ is fixed.

◆ Invariant lines through the origin — one quadratic

Take a general point $(t, m t)$ on $y = m x$ . Its image is $(t (a + b m), t (c + d m))$ , and this lies on $y = m x$ for all $t$ exactly when $c + d m = m (a + b m)$ :

b m^{2} + (a - d) m - c = 0

Solve for $m$ . Two roots ⟹ two invariant lines; a repeated root ⟹ one; no real roots ⟹ none. Then check the vertical line $x = 0$ separately: it is invariant precisely when $b = 0$ .

Derive this on the spot — it takes three lines — rather than trusting a memorised formula with the signs the wrong way round.

✎ Worked example — two invariant lines, but only one invariant point

Find the invariant lines through the origin for $𝐀 = (\begin{array}{cc} 4 & - 1 \\ 2 & 1 \end{array})$ .

Here $a = 4, b = - 1, c = 2, d = 1$ , so $b m^{2} + (a - d) m - c = 0$ reads

- m^{2} + 3 m - 2 = 0 ⟺ m^{2} - 3 m + 2 = 0 ⟺ m = 1 or m = 2 .

The invariant lines are $y = x$ and $y = 2 x$ . Verify $y = 2 x$ : $(t, 2 t) \mapsto (4 t - 2 t, 2 t + 2 t) = (2 t, 4 t)$ , which is on $y = 2 x$ . ✓ Note the points moved along the line — they were not fixed.

Indeed $\det (𝐀 - 𝐈) = | \begin{array}{cc} 3 & - 1 \\ 2 & 0 \end{array} | = 2 \neq 0$ , so the only invariant point is the origin. Two invariant lines, one invariant point — no contradiction.

★ A* extension — invariant lines that miss the origin

For $y = m x + k$ , substitute $(t, m t + k)$ and demand the image satisfies $Y = m X + k$ for all $t$ . Matching the $t$ terms gives the same quadratic in $m$ ; matching the constants then fixes $k$ .

Example: for $𝐀 = (\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array})$ , the point $(t, t + k)$ maps to $(3 t + k, 3 t + 2 k)$ , and $Y - X = k$ — so every line of gradient $1$ is invariant, whatever $k$ .

§5 Interactive: see the matrix move the plane

The faint square is the unit square; the blue parallelogram is its image. Its two edges from the origin are exactly the columns of the matrix. Watch the determinant become the signed area, and watch the invariant lines stay put.

a 1

b 0

c 0

d 1

unit square image invariant line line of invariant points

the columns are where i and j land ✎

§6 Examiner traps & A* checklist

⚠ The seven most common mark-losers

Multiplying in the wrong order for a sequence of transformations. Last one described goes furthest left.
Writing $(𝐀 𝐁)^{- 1} = 𝐀^{- 1} 𝐁^{- 1}$ . It reverses.
Losing the transpose when forming the adjugate of a 3 × 3.
Sign slips in the $+ - +$ chequerboard, especially the middle entry of the top row.
Quoting the area scale factor as $\det 𝐀$ rather than $| \det 𝐀 |$ — an area is never negative.
Confusing an invariant line with a line of invariant points. Read the question wording exactly.
Forgetting to test $x = 0$ separately when hunting invariant lines — the method with $y = m x$ can never find a vertical line.

★ Before you turn the page

Did I check one row of my inverse against the original to confirm it gives $𝐈$ ?
Does the sign of my determinant match the geometry (negative for anything involving a reflection)?
Have I described the transformation in words as well as giving the matrix, when asked?

§7 Video explainers (curated)

▶3 × 3 determinants & inverses
Curated search — minors, cofactors, adjugate, worked in full.↗ ▶Matrices as transformations
Curated search — the five named types and the area scale factor.↗ ▶Invariant points & lines
Curated search — the difference, and how to find each.↗

§8 Check yourself

Score 0 / 5

Q1 · singularity

The matrix $(\begin{array}{cc} k & 3 \\ 2 & k - 1 \end{array})$ is singular. The possible values of $k$ are:

Solution

Singular means $\det = 0$ : $k (k - 1) - 6 = 0$ , so $k^{2} - k - 6 = 0$ , $(k - 3) (k + 2) = 0$ . Check $k = - 2$ : $(\begin{array}{cc} - 2 & 3 \\ 2 & - 3 \end{array})$ has $\det = 6 - 6 = 0$ . ✓

Q2 · inverse

The inverse of $(\begin{array}{cc} 4 & 3 \\ 3 & 2 \end{array})$ is:

Solution

$\det = 8 - 9 = - 1$ , so $𝐌^{- 1} = \frac{1}{- 1} (\begin{array}{cc} 2 & - 3 \\ - 3 & 4 \end{array}) = (\begin{array}{cc} - 2 & 3 \\ 3 & - 4 \end{array})$ . The near-miss option is what you get if you forget to divide by the negative determinant. Check: $4 (- 2) + 3 (3) = 1$ and $4 (3) + 3 (- 4) = 0$ . ✓

Q3 · area

A shape of area $6$ is transformed by $(\begin{array}{cc} 3 & 1 \\ 1 & 2 \end{array})$ . The image has area:

Solution

Area scale factor $= | \det | = | 6 - 1 | = 5$ , so the image has area $5 \times 6 = 30$ .

Q4 · invariant lines

The invariant lines through the origin for $(\begin{array}{cc} 3 & 2 \\ 1 & 2 \end{array})$ are:

Solution

$b m^{2} + (a - d) m - c = 0$ gives $2 m^{2} + m - 1 = 0$ , i.e. $(2 m - 1) (m + 1) = 0$ , so $m = \frac{1}{2}$ or $m = - 1$ . Check $m = \frac{1}{2}$ : $(2,1) \mapsto (3 (2) + 2 (1), 2 + 2 (1)) = (8,4)$ , which is on $y = \frac{1}{2} x$ . ✓

Q5 · full method — try it on paper first

The transformation $T$ is a rotation of $90^{\circ}$ anticlockwise about the origin, followed by a reflection in the $x$ -axis.

(i) Find the matrix $𝐌$ of $T$ . (ii) Describe $T$ as a single transformation. (iii) State the area scale factor. (iv) Find the invariant points. (v) Find the invariant lines through the origin.

Solution

(i) Rotation $𝐐 = (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$ happens first, so it goes on the right of the reflection $𝐑 = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$ :

𝐌 = 𝐑 𝐐 = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}) (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}) = (\begin{array}{cc} 0 & - 1 \\ - 1 & 0 \end{array}) .

Check with $𝐢$ : $(1,0) \overset{\overset{}{rotate}}{\to} (0,1) \overset{\overset{}{reflect}}{\to} (0, - 1)$ , and column 1 of $𝐌$ is indeed $(0, - 1)$ . ✓

(ii) Comparing with $(\begin{array}{cc} \cos 2 θ & \sin 2 θ \\ \sin 2 θ & - \cos 2 θ \end{array})$ gives $\cos 2 θ = 0, \sin 2 θ = - 1$ , so $2 θ = - 90^{\circ}$ and $θ = - 45^{\circ}$ : $T$ is a reflection in the line $y = - x$ .

(iii) $\det 𝐌 = 0 - 1 = - 1$ , so the area scale factor is $| - 1 | = 1$ — areas are unchanged, but the negative sign says orientation is reversed, as it must be for a reflection.

(iv) $𝐌 - 𝐈 = (\begin{array}{cc} - 1 & - 1 \\ - 1 & - 1 \end{array})$ has determinant $1 - 1 = 0$ , so there is a whole line of invariant points. Either row gives $- x - y = 0$ , i.e. $y = - x$ — the mirror line, exactly as expected. Verify: $(t, - t) \mapsto (0 \cdot t + (- 1) (- t), (- 1) t + 0 \cdot (- t)) = (t, - t)$ . ✓

(v) With $a = 0, b = - 1, c = - 1, d = 0$ , the quadratic $b m^{2} + (a - d) m - c = 0$ is $- m^{2} + 1 = 0$ , so $m = \pm 1$ . The invariant lines are $y = x$ and $y = - x$ : the mirror itself (fixed pointwise) and the perpendicular through $O$ (whose points swap sides but stay on the line). Since $b = - 1 \neq 0$ , $x = 0$ is not invariant. $■$

FP1 · Topic 4 · Matrices