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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

LearnYFM01WFM01 · Further Pure Mathematics 1Matrix algebraMatrix algebra and inverses

Matrix algebra · Subtopic 1 of 1

Matrix Algebra and Inverses

Matrices multiply row-by-column. The determinant measures how a matrix scales area (or volume); when it is zero the matrix has no inverse. For $2\times2$ , $\mathbf A^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ , and inverses solve systems of equations in one step.

Cambridge 9231 Paper 1Edexcel YFM01 WFM01Edexcel 9FM0 Paper 1Edexcel 9FM0 Paper 2AQA 7367 Paper 1AQA 7367 Paper 2≈ 45 minRevision 1 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Add, subtract and multiply matrices, and know when a product is defined.
Find determinants of $2\times2$ and $3\times3$ matrices.
Find inverses and use $(\mathbf{AB})^{-1}=\mathbf B^{-1}\mathbf A^{-1}$ .
Solve systems of linear equations with inverses and interpret the result.

Pause and think

Is $\mathbf{AB}$ always equal to $\mathbf{BA}$ for matrices?

Section 2 of 14

Core idea

Start hereThree equations, one step

A shop sells three bundles of pens, pencils and rubbers at different prices. Three receipts give three equations in three unknowns. Solving by elimination is long and fiddly. Packing the coefficients into a grid, a matrix, turns the whole system into one equation $\mathbf A\mathbf x=\mathbf b$ , solved in one step by "dividing" by $\mathbf A$ .

The big idea

Matrices multiply row by column; the product depends on order. The determinant is the area (volume) scale factor; if it is $0$ the matrix is singular and has no inverse. Otherwise $\mathbf x=\mathbf A^{-1}\mathbf b$ .

1. Multiplying matrices

(\mathbf{AB})_{ij}=\sum_k a_{ik}b_{kj}

$(\mathbf{AB})_{ij}$: entry in row $i$ , column $j$ of the product.
$a_{ik}b_{kj}$: row $i$ of $\mathbf A$ times column $j$ of $\mathbf B$ , entry by entry, then added.

When is a product defined?

$\mathbf A:\ m\times n,\quad\mathbf B:\ n\times p$
Each row of $\mathbf A$ has $n$ entries; each column of $\mathbf B$ must also have $n$ .
$\mathbf{AB}:\ m\times p$
One entry for each (row of $\mathbf A$ , column of $\mathbf B$ ) pair.

\mathbf{BA}

may not even be defined, and when it is, it is usually different.

Worked exampleA 2×2 product both ways

$\mathbf A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$ , $\mathbf B=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ . Find $\mathbf{AB}$ and $\mathbf{BA}$ .

Goal: Both products.
Strategy: Row by column, carefully.

$\mathbf{AB}=\begin{pmatrix}0+2&1+0\\0+4&3+0\end{pmatrix}=\begin{pmatrix}2&1\\4&3\end{pmatrix}$
$\mathbf B$ on the right swaps the columns of $\mathbf A$ .
$\mathbf{BA}=\begin{pmatrix}3&4\\1&2\end{pmatrix}$
$\mathbf B$ on the left swaps the rows of $\mathbf A$ .

Answer

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

✓ Check

\det(\mathbf{AB})=6-4=2

and

\det\mathbf A\det\mathbf B=(-2)(-1)=2

✓.

Common misconceptionCommutativity Never swap the order of a matrix product.

\mathbf{AB}=\mathbf{BA}

only in special cases (e.g. with

\mathbf I

or powers of the same matrix).

2. The determinant

Why

ad-bc

is an area

$\mathbf i\mapsto\begin{pmatrix}a\\c\end{pmatrix},\ \mathbf j\mapsto\begin{pmatrix}b\\d\end{pmatrix}$
The columns are the images of the unit vectors.
$\text{unit square}\mapsto\text{parallelogram on these vectors}$
Linear maps send straight lines to straight lines.
$\text{area}=(a+b)(c+d)-2bc-ac-bd$
Bounding rectangle minus the corner triangles and rectangles (for the standard picture).
$=ad-bc$
Expand and simplify.

|\det\mathbf A|

is the area scale factor; a negative sign means the orientation is flipped.

\det\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)

$+,-,+$: alternating signs along the top row.
$ei-fh$ etc.: the 2×2 minors left after deleting that entry's row and column.

Worked exampleA 3×3 with a parameter

Find $k$ for which $\begin{pmatrix}1&2&0\\0&k&1\\3&1&2\end{pmatrix}$ is singular.

Goal: $k$ .
Strategy: Expand along the top row; set to zero.

$1(2k-1)$
First minor: delete row 1, column 1.
$-2(0\cdot2-1\cdot3)=+6$
Second minor, with the minus sign.
$+0(\ldots)=0$
Third term vanishes.
$2k+5=0\Rightarrow k=-\tfrac52$
Singular means determinant zero.

Answer

k=-\frac52

✓ CheckExpanding down column 1 instead:

1(2k-1)-0+3(2-0)=2k+5

✓.

3. Inverses

The

2\times2

inverse

$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}=\begin{pmatrix}ad-bc&0\\0&ad-bc\end{pmatrix}$
Multiply out: off-diagonal terms cancel.
$=(ad-bc)\mathbf I$
A multiple of the identity.
$\mathbf A^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$
Divide by the determinant (only possible if it is non-zero).

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

$(\mathbf{AB})(\mathbf B^{-1}\mathbf A^{-1})=\mathbf A(\mathbf{BB}^{-1})\mathbf A^{-1}$
Matrix multiplication is associative.
$=\mathbf{AIA}^{-1}=\mathbf{AA}^{-1}=\mathbf I$
So $\mathbf B^{-1}\mathbf A^{-1}$ undoes $\mathbf{AB}$ .

Worked exampleSolving a system

Solve $3x+y=7$ , $5x+2y=12$ .

Goal: $x,y$ .
Strategy: $\mathbf x=\mathbf A^{-1}\mathbf b$ .

$\mathbf A=\begin{pmatrix}3&1\\5&2\end{pmatrix},\ \det\mathbf A=1$
Coefficient matrix.
$\mathbf A^{-1}=\begin{pmatrix}2&-1\\-5&3\end{pmatrix}$
Swap, negate, divide by 1.
$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2&-1\\-5&3\end{pmatrix}\begin{pmatrix}7\\12\end{pmatrix}=\begin{pmatrix}2\\1\end{pmatrix}$
Multiply.

Answer

x=2

y=1

✓ Check

6+1=7

✓,

10+2=12

✓.

Three equations in three unknowns

Drawing diagram…

The determinant decides first; consistency decides the rest.

Edge case3×3 inverses For

3\times3

\mathbf A^{-1}=\frac1{\det\mathbf A}\operatorname{adj}\mathbf A

, where the adjugate is the transposed matrix of cofactors. Calculators are allowed to find it on most papers, but show the determinant.

Pause and think

If $\det\mathbf A=0$ , can $\mathbf A\mathbf x=\mathbf b$ still have solutions?

Summary

Row by column; order matters.
$\det$ = area/volume scale factor.
$2\times2$ inverse: swap, negate, divide.
$(\mathbf{AB})^{-1}=\mathbf B^{-1}\mathbf A^{-1}$ ; $\det(\mathbf{AB})=\det\mathbf A\det\mathbf B$ .

Singular: Determinant zero; no inverse.
Identity: $\mathbf I$ : leaves every vector unchanged.
Cofactor: Signed minor used in determinants and inverses.

HistoryCayley and Sylvester. James Sylvester coined the word "matrix" (Latin for "womb") in 1850; Arthur Cayley defined matrix multiplication and inverses in 1858.

Fun factEvery 3D game. Each frame of a video game multiplies millions of position vectors by

4\times4

matrices to rotate, move and project the scene onto your screen.

Section 3 of 14

Key formulae

Determinant (2×2)

\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc

Inverse (2×2)

\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}

Determinant (3×3)

\det\mathbf A=a\begin{vmatrix}e&f\\h&i\end{vmatrix}-b\begin{vmatrix}d&f\\g&i\end{vmatrix}+c\begin{vmatrix}d&e\\g&h\end{vmatrix}

Product rules

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

TrickCheck with the identity. After finding

\mathbf A^{-1}

, multiply one row of

\mathbf A

by one column of your answer. You should get 1 or 0. It takes ten seconds and catches most sign slips.

Section 4 of 14

Worked examples

Example 1 · 2×2 inverse0 / 3 steps

Find the inverse of

\mathbf A=\begin{pmatrix}3&1\\5&2\end{pmatrix}

Example 2 · Singular matrix0 / 2 steps

Find

k

such that

\begin{pmatrix}1&2&0\\0&k&1\\3&1&2\end{pmatrix}

is singular.

Section 5 of 14

Explore the graph

Determinant as an area scale factor

Loading the interactive graph…

Try this

The unit square goes to the parallelogram with corners $(0,0)$ , $(a,c)$ , $(b,d)$ , $(a+b,c+d)$ . Its area is $|ad-bc|$ .
Make $ad-bc=0$ : the parallelogram flattens into a line, so no inverse can undo it.

$\det\mathbf A$ is the factor by which areas are multiplied.

In the real worldCryptography. The Hill cipher encodes blocks of letters by multiplying them by a matrix; the receiver decodes with the inverse matrix. If the determinant shares a factor with 26, the code can't be undone.

Section 6 of 14

Common mistakes

Incorrect

\mathbf{AB}=\mathbf{BA}

CorrectUsually

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

Order matters in matrix products.

Incorrect

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

Correct

\mathbf B^{-1}\mathbf A^{-1}

Undo the last operation first, like taking off socks then shoes.

IncorrectIgnoring the minus sign on the middle cofactor

CorrectSigns alternate

+,-,+

The middle term in the 3×3 expansion is subtracted.

Me: AB = BA, obviously

Matrices: we don't do that here.

Section 7 of 14

Exam tip

Exam tipShow the determinant. Even when asked only for an inverse, write the determinant line: it earns a method mark and proves the matrix is non-singular.

Section 8 of 14

Guided practice

SA-FP-MAT-G011 markL1 FoundationNon-calculatorDeterminants

Find the determinant of

\begin{pmatrix}4&3\\2&5\end{pmatrix}

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-FP-MAT-0012 marksL2 StandardNon-calculatorOperations

\mathbf A=\begin{pmatrix}1&2\\3&4\end{pmatrix}

\mathbf B=\begin{pmatrix}0&1\\5&2\end{pmatrix}

. Find the bottom-left entry of

\mathbf{AB}

Salam Academy Original Practice · Proposed mark scheme

SA-FP-MAT-0022 marksL2 StandardNon-calculatorInverses

\mathbf A=\begin{pmatrix}5&2\\7&3\end{pmatrix}

. Find the top-right entry of

\mathbf A^{-1}

Salam Academy Original Practice · Proposed mark scheme

SA-FP-MAT-0033 marksL3 ExamNon-calculatorDeterminants

Find

\det\begin{pmatrix}2&1&0\\1&3&1\\0&1&2\end{pmatrix}

Salam Academy Original Practice · Proposed mark scheme

SA-FP-MAT-0043 marksL3 ExamNon-calculatorDeterminants

Find the value of

k

for which

\begin{pmatrix}k&2\\3&k-1\end{pmatrix}

is singular, given

k>0

Salam Academy Original Practice · Proposed mark scheme

SA-FP-MAT-0053 marksL3 ExamNon-calculatorInverses

Use an inverse matrix to solve

3x+y=7

5x+2y=12

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-FP-MAT-0063 marksL4 ChallengingNon-calculatorInverses

\det\mathbf A=4

and

\det\mathbf B=-3

for

3\times3

matrices. Find

\det\left((\mathbf{AB})^{-1}\right)

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

Row-by-column multiplication; not commutative.
$\det$ = area scale factor; zero means singular.
$2\times2$ inverse: swap, negate, divide by det.
$(\mathbf{AB})^{-1}=\mathbf B^{-1}\mathbf A^{-1}$ .

Section 12 of 14

Quick quiz

Q1 · SA-FP-MAT-Q011 markL1 FoundationNon-calculatorOperations

\mathbf A

2\times3

and

\mathbf B

3\times4

. What is the order of

\mathbf{AB}

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: IAL Further Mathematics (YFM01) and A level Further Mathematics (9FM0)
Where: Core Pure · Matrices (IAL: WFM01 Matrix algebra)
Expected: Operations, determinants, inverses of 2×2 and 3×3 matrices, and solving three linear equations; geometric meaning of consistent and inconsistent systems.
Calculator: Permitted; show enough working to earn the method marks.
Typical style: Edexcel often opens with a routine part and builds to a "show that" or "prove" part.
Open in: YFM01 · WFM01 9FM0 · Paper 1 9FM0 · Paper 2

Section 14 of 14

Past paper questions

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Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Pearson Edexcel YFM01, 2018 specification (official specification).