Notes · Further MathematicsUK · A-Levels

Matrices

Matrices provide a compact algebra for linear transformations and for systems of linear equations. Further Mathematics develops matrix arithmetic up to $3\times 3$ , determinants and inverses, the representation of geometric transformations by matrices, invariant points and lines, and the solution and geometric interpretation of simultaneous equations. The determinant emerges as the area or volume scale factor of the transformation, tying the algebra to the geometry.

5 sections~15 min reading time3 competenciesLevel Standard 1 · Advanced 4

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Exam profile

AO1 · Carry out matrix algebra, evaluate determinants and find inverses accuratelyAO2 · Connect matrices to geometric transformations and justify the determinant as an area or volume scale factorAO3 · Solve and interpret systems of linear equations, including the geometry of intersecting planes

Operators:findshow thatsolveprovededucehencedeterminedescribe

basic level

AS Further Mathematics covers matrix arithmetic, $2\times 2$ determinants and inverses, and transformations of the plane.

higher level

The full A-Level adds $3\times 3$ determinants and inverses, transformations of three-dimensional space, invariant lines, and the geometric interpretation of systems of three equations in three unknowns.

Depth

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Contents · 5 sections

Matrices

Matrix algebra#

StandardAQA 7367 §3.3DfE Further Mathematics — Matrices

A matrix is a rectangular array of numbers with a fixed number of rows and columns. Addition and subtraction are defined only for matrices of the same order and act entry by entry, while scalar multiplication multiplies every entry by the scalar. These operations obey the familiar commutative and associative laws, so at this level matrices behave like ordinary numbers.

Matrix multiplication is where the subject acquires its character. The product

\mathbf{AB}

is defined only when the number of columns of

\mathbf{A}

equals the number of rows of

\mathbf{B}

, and the

(i,j)

entry of

\mathbf{AB}

is the dot product of row

i

\mathbf{A}

with column

j

\mathbf{B}

. Multiplication is associative,

\mathbf{(AB)C} = \mathbf{A(BC)}

, and distributes over addition, but it is emphatically not commutative: in general

\mathbf{AB} \neq \mathbf{BA}

, and for many pairs only one of the two products is even defined.

The identity matrix

\mathbf{I}

, with

1

s on the leading diagonal and

0

s elsewhere, satisfies

\mathbf{AI} = \mathbf{IA} = \mathbf{A}

and plays the role of the number

1

. The zero matrix plays the role of

0

. A warning that flows from non-commutativity: the usual algebraic shortcuts fail —

(\mathbf{A}+\mathbf{B})^2 = \mathbf{A}^2 + \mathbf{AB} + \mathbf{BA} + \mathbf{B}^2

, and this cannot be shortened to

\mathbf{A}^2 + 2\mathbf{AB} + \mathbf{B}^2

unless

\mathbf{A}

and

\mathbf{B}

happen to commute.

Because multiplication represents the composition of transformations (developed later in this topic), the order in which matrices are multiplied corresponds to the order in which the transformations are applied — and, following function-composition convention,

\mathbf{AB}

means 'do

\mathbf{B}

first, then

\mathbf{A}

'. Keeping this reading in mind makes the non-commutativity intuitive: rotating then reflecting is generally not the same as reflecting then rotating.

(\mathbf{AB})_{ij} = \sum_{k} \mathbf{A}_{ik}\,\mathbf{B}_{kj}

Matrix multiplication

The $(i,j)$ entry is the dot product of row $i$ of $\mathbf{A}$ with column $j$ of $\mathbf{B}$ .

\mathbf{AI} = \mathbf{IA} = \mathbf{A}, \qquad \mathbf{AB} \neq \mathbf{BA} \ \text{in general}

Identity and non-commutativity

$\mathbf{I}$ behaves like $1$ ; multiplication is associative but not commutative.

Worked example

A product depends on order

With $\mathbf{A} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$ , find $\mathbf{AB}$ and $\mathbf{BA}$ .

Compute AB
$\mathbf{AB} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} 1{+}6 & 0{+}2 \\ 0{+}3 & 0{+}1 \end{pmatrix} = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}$ .
Compute BA
$\mathbf{BA} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 3 & 7 \end{pmatrix}$ .
$\mathbf{AB} = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix} \neq \begin{pmatrix} 1 & 2 \\ 3 & 7 \end{pmatrix} = \mathbf{BA}$
Conclude
The two products differ, confirming that matrix multiplication is not commutative.

Result: $\mathbf{AB} = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}$ , $\mathbf{BA} = \begin{pmatrix} 1 & 2 \\ 3 & 7 \end{pmatrix}$ ; they are unequal.

Typical mistakes

Multiplying matrices entry by entry as if it were addition, instead of using row-times-column.
Assuming multiplication commutes, so that identities such as $(\mathbf{A}+\mathbf{B})^2 = \mathbf{A}^2 + 2\mathbf{AB} + \mathbf{B}^2$ are applied incorrectly.

Active revision

Given $\mathbf{A} = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix} 1 & -1 \\ 4 & 2 \end{pmatrix}$ , compute $\mathbf{AB}$ and $\mathbf{BA}$ and verify that they are different.

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Determinants and inverses#

AdvancedAQA 7367 §3.3DfE Further Mathematics — Matrices

The determinant as an area scale factor

Fig. 1The columns of

\mathbf{M}=\begin{pmatrix}2&1\\0&1\end{pmatrix}

span a parallelogram of area

|\det\mathbf{M}| = 2

For a

2\times 2

matrix

\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

the determinant is

\det\mathbf{M} = ad - bc

. It measures how the transformation represented by

\mathbf{M}

scales area: the unit square of area

1

maps to a parallelogram of area

|\det\mathbf{M}|

. A negative determinant signals that the transformation reverses orientation (includes a reflection), and a zero determinant means the whole plane is squashed onto a line, so area collapses to zero.

The inverse of a non-singular

2\times 2

matrix is

\mathbf{M}^{-1} = \dfrac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

: swap the leading-diagonal entries, negate the other two, and divide by the determinant. The inverse satisfies

\mathbf{M}\mathbf{M}^{-1} = \mathbf{M}^{-1}\mathbf{M} = \mathbf{I}

and exists precisely when

\det\mathbf{M} \neq 0

. A matrix with zero determinant is singular and has no inverse — geometrically, the collapse onto a line cannot be undone.

For a

3\times 3

matrix the determinant is evaluated by expansion along a row or column:

\det\mathbf{M} = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13}

, where each

M_{1j}

is the

2\times 2

minor obtained by deleting the row and column of

a_{1j}

, and the alternating signs follow the checkerboard pattern. The determinant now measures the volume scale factor of the transformation of three-dimensional space.

The inverse of a non-singular

3\times 3

matrix is found from the matrix of cofactors: transpose the cofactor matrix to form the adjugate, then divide by the determinant,

\mathbf{M}^{-1} = \dfrac{1}{\det\mathbf{M}}\operatorname{adj}\mathbf{M}

. A useful algebraic identity is

(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}

— the inverse of a product reverses the order — which mirrors the fact that to undo 'do

\mathbf{B}

then

\mathbf{A}

' you undo

\mathbf{A}

first.

\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc, \qquad \begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

2x2 determinant and inverse

The inverse exists iff $ad-bc\neq 0$ .

(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}, \qquad \det(\mathbf{AB}) = \det\mathbf{A}\,\det\mathbf{B}

Products

The inverse of a product reverses order; determinants multiply.

Worked example

Inverse of a 3×3 matrix (determinant first)

Find $\det\mathbf{M}$ for $\mathbf{M} = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$ and state whether $\mathbf{M}^{-1}$ exists.

Expand along row 1
$\det\mathbf{M} = 1\begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} - 2\begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} + 3\begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix}$ .
Evaluate the minors
$= 1(0-24) - 2(0-20) + 3(0-5) = -24 + 40 - 15$ .
$\det\mathbf{M} = -24 + 40 - 15 = 1$
Conclude
Since $\det\mathbf{M} = 1 \neq 0$ , the matrix is non-singular and $\mathbf{M}^{-1}$ exists (and, because $\det\mathbf{M}=1$ , the inverse equals the adjugate).

Result: $\det\mathbf{M} = 1$ , so $\mathbf{M}^{-1}$ exists.

Typical mistakes

Forgetting the alternating signs (the cofactor pattern) when expanding a $3\times 3$ determinant or forming cofactors.
Failing to divide by the determinant, or attempting to invert a singular matrix (whose determinant is zero).

Active revision

Find the determinant and inverse of $\mathbf{M} = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$ , and verify that $\mathbf{M}\mathbf{M}^{-1} = \mathbf{I}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Matrices as linear transformations#

AdvancedAQA 7367 §3.3DfE Further Mathematics — Matrices

A shear transformation of the unit square

Fig. 2Under

\mathbf{M}=\begin{pmatrix}2&1\\0&1\end{pmatrix}

the unit square (solid) maps to the parallelogram (dashed), doubling in area.

2\times 2

matrix

\mathbf{M}

acts on the plane by sending the position vector

\begin{pmatrix} x \\ y \end{pmatrix}

\mathbf{M}\begin{pmatrix} x \\ y \end{pmatrix}

. Such a map is linear: it fixes the origin and sends straight lines to straight lines, preserving ratios along them. The columns of

\mathbf{M}

are precisely the images of the basis vectors

\mathbf{i}=\begin{pmatrix}1\\0\end{pmatrix}

and

\mathbf{j}=\begin{pmatrix}0\\1\end{pmatrix}

, which gives a quick way both to build a transformation matrix and to read one off.

The standard plane transformations have standard matrices. A rotation anticlockwise through

\theta

about the origin is

\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

; a reflection in the line

y = x

\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

; an enlargement by factor

k

k\mathbf{I}

; and stretches and shears have their own characteristic forms. Recognising a matrix as one of these — by finding the images of

\mathbf{i}

and

\mathbf{j}

— is a routine exam demand.

Composing transformations corresponds to multiplying their matrices, in the reverse order of application: applying

S

then

T

is represented by

\mathbf{T}\mathbf{S}

. This is the geometric meaning of non-commutativity — a rotation followed by a reflection generally differs from the reflection followed by the rotation. The determinant of a composite is the product of the determinants, so scale factors multiply, and a transformation is a pure rotation exactly when its matrix is orthogonal with determinant

+1

The same framework extends to three dimensions with

3\times 3

matrices: rotations about the coordinate axes, reflections in the coordinate planes, and enlargements all have standard

3\times 3

forms, and the determinant is now the volume scale factor. A determinant of

-1

indicates an orientation-reversing map (a reflection), and a determinant of

0

collapses space onto a plane or line. Throughout, the geometric picture and the matrix algebra reinforce one another.

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

Anticlockwise rotation about the origin

The columns are the images of $\mathbf{i}$ and $\mathbf{j}$ under the rotation.

\text{'do } S \text{ then } T\text{'} \ \longleftrightarrow\ \mathbf{TS}, \qquad \det(\mathbf{TS}) = \det\mathbf{T}\det\mathbf{S}

Composition

Compose by multiplying, in the reverse order of application; scale factors multiply.

Worked example

Identify a transformation from its matrix

Describe fully the transformation represented by $\mathbf{M} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ .

Image of i
$\mathbf{M}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}0\\1\end{pmatrix}$ , so $\mathbf{i}$ maps to $\mathbf{j}$ .
Image of j
$\mathbf{M}\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}-1\\0\end{pmatrix}$ , so $\mathbf{j}$ maps to $-\mathbf{i}$ .
Recognise the pattern
Comparing with $R(\theta)$ , $\cos\theta = 0$ and $\sin\theta = 1$ , so $\theta = 90^{\circ}$ ; the determinant is $0\cdot 0 - (-1)(1) = 1$ , confirming a pure rotation.
$\mathbf{M} = \begin{pmatrix} \cos 90^{\circ} & -\sin 90^{\circ} \\ \sin 90^{\circ} & \cos 90^{\circ} \end{pmatrix}$
Describe
It is an anticlockwise rotation of $90^{\circ}$ about the origin.

Result: A $90^{\circ}$ anticlockwise rotation about the origin ( $\det = 1$ , area preserved).

Typical mistakes

Multiplying the transformation matrices in the wrong order when composing, giving the reverse transformation.
Writing the rotation matrix with the sines' signs the wrong way round, which turns an anticlockwise rotation into a clockwise one.

Active revision

The transformation $T$ is a rotation of $90^{\circ}$ anticlockwise about the origin and $S$ is a reflection in the $x$ -axis. Find the single matrix representing 'apply $S$ then $T$ ', and describe the resulting transformation geometrically.

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Invariant points and lines#

AdvancedAQA 7367 §3.3DfE Further Mathematics — Matrices

An invariant line through the origin

Fig. 3The line

y = x

is invariant under

\mathbf{M}=\begin{pmatrix}2&1\\1&2\end{pmatrix}

: points on it map to points on it (here scaled by 3).

An invariant point of a transformation

\mathbf{M}

is a point that maps to itself:

\mathbf{M}\mathbf{x} = \mathbf{x}

. The origin is always invariant for a linear transformation, but there may be others. Finding them means solving

(\mathbf{M} - \mathbf{I})\mathbf{x} = \mathbf{0}

; if

\det(\mathbf{M}-\mathbf{I}) \neq 0

the only solution is the origin, while if the determinant is zero there is a whole line of invariant points.

An invariant line is subtler: it is a line that maps onto itself, though the individual points on it need not be fixed. Every point of the line maps to some point of the same line. Reflections have their mirror line as a line of invariant points and the perpendicular direction as an invariant line; rotations (other than

0^{\circ}

and

180^{\circ}

) have no real invariant lines at all, since they turn every direction.

To find invariant lines through the origin of the form

y = mx

, require that the image of a general point

(x, mx)

also satisfies

y = mx

. Writing

\mathbf{M}\begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} X \\ Y \end{pmatrix}

and imposing

Y = mX

gives an equation for the gradient

m

— usually a quadratic — whose (real) roots are the gradients of the invariant lines. This links directly to eigenvectors: an invariant line through the origin is spanned by an eigenvector of

\mathbf{M}

The number of invariant lines therefore depends on how many real gradients solve that equation: two distinct lines, one repeated line, or none. Interpreting the outcome geometrically — a shear fixes one line and shears everything parallel to it; an enlargement fixes every line through the origin — is part of the understanding examined. Distinguishing a line of invariant points (every point fixed) from an invariant line (the line maps to itself) is a frequent source of confusion and of marks.

\text{Invariant point:}\ \mathbf{M}\mathbf{x} = \mathbf{x} \iff (\mathbf{M}-\mathbf{I})\mathbf{x} = \mathbf{0}

Invariant points

A non-trivial solution requires $\det(\mathbf{M}-\mathbf{I}) = 0$ , giving a line of invariant points.

\text{Invariant line } y=mx:\ \ \mathbf{M}\begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} X \\ Y \end{pmatrix} \ \text{with}\ Y = mX

Invariant lines through the origin

Imposing the image on the same line gives an equation for the gradient $m$ .

Worked example

Finding invariant lines

Find the invariant lines through the origin of $\mathbf{M} = \begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix}$ .

Set up the image
A point $(x, mx)$ maps to $\begin{pmatrix} 3x + 2mx \\ mx \end{pmatrix} = \begin{pmatrix} X \\ Y \end{pmatrix}$ , so $X = (3+2m)x$ and $Y = mx$ .
Impose the line condition
For the image to lie on $y = mx$ we need $Y = mX$ , i.e. $mx = m(3+2m)x$ . For $x\neq 0$ : $m = m(3+2m)$ .
Solve for m
$m = 3m + 2m^2 \Rightarrow 2m^2 + 2m = 0 \Rightarrow 2m(m+1) = 0$ , giving $m = 0$ or $m = -1$ .
$2m^2 + 2m = 0 \ \Rightarrow\ m = 0 \ \text{or}\ m = -1$
State the lines
The invariant lines through the origin are $y = 0$ (the $x$ -axis) and $y = -x$ .

Result: The invariant lines are $y = 0$ and $y = -x$ .

Typical mistakes

Confusing an invariant line (maps to itself) with a line of invariant points (every point fixed).
Only looking for lines through the origin and forgetting that a translation-free linear map may still have invariant directions not obvious by inspection.

Active revision

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

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Systems of linear equations and their geometry#

AdvancedAQA 7367 §3.3DfE Further Mathematics — Matrices

A system of linear equations can be written as a single matrix equation

\mathbf{M}\mathbf{x} = \mathbf{b}

, where

\mathbf{M}

holds the coefficients,

\mathbf{x}

the unknowns and

\mathbf{b}

the constants. When

\mathbf{M}

is a non-singular square matrix, the system has a unique solution

\mathbf{x} = \mathbf{M}^{-1}\mathbf{b}

. This is both a practical solution method and a conceptual one: solving equations is inverting a transformation.

The behaviour when

\det\mathbf{M} = 0

is where the geometry becomes interesting. A singular coefficient matrix means the system either has no solution or infinitely many, never exactly one. For three equations in three unknowns, each equation represents a plane in space, and the solution set is the intersection of the three planes: a single point (unique solution, non-singular case), a common line or common plane (infinitely many solutions), or no common point at all (inconsistent).

The classification of the singular cases has a vivid geometric reading. Three planes may meet in a line (a 'sheaf'), forming a consistent system with a one-parameter family of solutions; they may form a triangular prism, with each pair meeting in a line but no point common to all three (inconsistent); or two of them may be parallel. Deciding which case occurs is done by examining the equations for consistency once the determinant is found to be zero.

In practice, one solves a non-singular

3\times 3

system either by computing

\mathbf{M}^{-1}\mathbf{b}

or by Gaussian elimination, and the two methods agree. Being able to move between the algebra (row-reduce, or invert) and the geometry (how three planes sit in space) is exactly the synthesis of AO1, AO2 and AO3 that this topic is designed to develop, and it recurs when planes are studied again in Further Vectors.

\mathbf{M}\mathbf{x} = \mathbf{b}, \qquad \det\mathbf{M} \neq 0 \ \Rightarrow\ \mathbf{x} = \mathbf{M}^{-1}\mathbf{b} \ \text{(unique)}

Matrix form of a linear system

A non-singular coefficient matrix gives a unique solution; the three planes meet in a point.

\det\mathbf{M} = 0 \ \Rightarrow\ \text{no solution (inconsistent) or infinitely many (a line or plane of solutions)}

The singular case

Never exactly one solution; classify by testing consistency.

Worked example

Solving a 3×3 system

Solve $\begin{cases} x + y + z = 6 \\ 2x - y + z = 3 \\ x + 2y - z = 2 \end{cases}$ and describe the configuration of the planes.

Eliminate to reduce
Subtract twice the first from the second: $(2x-y+z) - 2(x+y+z) = 3 - 12$ gives $-3y - z = -9$ , so $3y + z = 9$ . Subtract the first from the third: $(x+2y-z)-(x+y+z) = 2-6$ gives $y - 2z = -4$ .
Solve the 2×2 system
From $3y + z = 9$ and $y - 2z = -4$ : from the first $z = 9 - 3y$ ; substitute: $y - 2(9-3y) = -4 \Rightarrow 7y - 18 = -4 \Rightarrow y = 2$ , then $z = 9 - 6 = 3$ .
$y = 2, \quad z = 3$
Back-substitute
From $x + y + z = 6$ : $x = 6 - 2 - 3 = 1$ .
Interpret
The determinant of the coefficient matrix is non-zero, so the three planes meet in the single point $(1, 2, 3)$ .

Result: $x = 1,\ y = 2,\ z = 3$ ; the three planes meet in one point.

Typical mistakes

Concluding 'no solution' as soon as the determinant is zero, without checking whether the system is consistent (it may have infinitely many).
Arithmetic slips in the $3\times 3$ inverse; solving by elimination is often safer and should be checked by substitution.

Active revision

Solve the system $x + y + z = 6$ , $2x - y + z = 3$ , $x + 2y - z = 2$ , and describe the geometric configuration of the three planes.

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

References & sources

Sources

Department for Education

Further mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Further Mathematics 7367 specification

Matrices

Key points

A product depends on order

Key points

Inverse of a 3×3 matrix (determinant first)

Key points

Identify a transformation from its matrix

Key points

Finding invariant lines

Key points

Solving a 3×3 system

Sources