Why is "Forgetting to transpose the cofactor matrix when forming the $3\times3$ adjugate" a common mistake in Inverse of a matrix?

Why it happens: Students compute the cofactor matrix and divide straight by the determinant, skipping the adjugate step. How to avoid it: Write 'adj = C^T' explicitly and reflect the entries across the leading diagonal BEFORE dividing by A. Source: 9231 Paper 1 Examiner Reports — recurring comment

Why is "Sign errors in the cofactor checkerboard (e.g. taking $C_{12}$ as $+M_{12}$)" a common mistake in Inverse of a matrix?

Why it happens: Forgetting the (-1)^i+j sign pattern, or mis-counting which entries are negative. How to avoid it: Sketch the pmatrix + & - & + - & + & - + & - & + pmatrix grid in the margin first and read signs off it. Source: 9231 Paper 1 mark schemes

Why is "Attempting to invert a singular matrix (dividing by a zero determinant)" a common mistake in Inverse of a matrix?

Why it happens: Jumping into the inverse recipe without computing A first. How to avoid it: Always find A before anything else. If it is 0, state 'singular — no inverse exists' (that statement scores). Source: 9231 Paper 1 Examiner Reports

Why is "Writing $(AB)^{-1} = A^{-1}B^{-1}$ instead of $B^{-1}A^{-1}$" a common mistake in Inverse of a matrix?

Why it happens: Assuming the inverse distributes in the same order as the product, as it would for commuting numbers. How to avoid it: Remember the order REVERSES: (AB)^-1 = B^-1A^-1 ('socks and shoes'). Matrix multiplication is not commutative. Source: 9231 Paper 1 Examiner Reports

Why is "Solving $A\mathbf{x} = \mathbf{b}$ by computing $\mathbf{b}A^{-1}$ (post-multiplying)" a common mistake in Inverse of a matrix?

Why it happens: Treating matrix multiplication as commutative; not realising bA^-1 may not even be defined. How to avoid it: Pre-multiply both sides on the LEFT: A^-1Ax = A^-1b, giving x = A^-1b. Source: 9231 Paper 1 mark schemes

Why is "Swapping and negating the $2\times2$ entries but forgetting to divide by the determinant" a common mistake in Inverse of a matrix?

Why it happens: Treating the pmatrix d & -b -c & a pmatrix rearrangement as the whole answer. How to avoid it: The rearranged matrix is only the adjugate; the inverse is 1/ad-bc times it. Keep the 1/ factor visible throughout. Source: 9231 Paper 1 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Inverse of a matrix

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Detailed Study Notes

Detailed notes on Matrices (paper 1) for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Inverse of a Matrix — Cambridge International AS & A Level Further Mathematics 9231 Paper 1 (2026-2027 syllabus)

The inverse $A^{-1}$ undoes a matrix: $AA^{-1} = A^{-1}A = I$ . Master the lightning-fast $2\times2$ recipe, the cofactor route for $3\times3$ , and the single most useful application — solving $A\mathbf{x} = \mathbf{b}$ by $\mathbf{x} = A^{-1}\mathbf{b}$ .

At a glance

$2\times2$ recipe: swap the leading diagonal, negate the off-diagonal, divide by $\det = ad - bc$ .
$3\times3$ route: matrix of cofactors $\rightarrow$ transpose to get the adjugate $\rightarrow$ divide by $\det A$ .
No inverse exists when $\det A = 0$ — the matrix is then singular.
Reverse-order law $(AB)^{-1} = B^{-1}A^{-1}$ is NOT in MF19 — you MUST recall it.
Killer application: $A\mathbf{x} = \mathbf{b} \Rightarrow \mathbf{x} = A^{-1}\mathbf{b}$ solves simultaneous equations in one step.

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

1.4a — Find the inverse of a non-singular $2\times2$ matrix using the standard formula, and of a $3\times3$ matrix using the cofactor/adjugate method.
1.4b — Use the properties $(A^{-1})^{-1} = A$ and $(AB)^{-1} = B^{-1}A^{-1}$ , and recognise when a matrix is singular ( $\det = 0$ ).
1.4c — Solve a system of linear equations by writing it as $A\mathbf{x} = \mathbf{b}$ and computing $\mathbf{x} = A^{-1}\mathbf{b}$ .

What an inverse is — and when one exists

The inverse undoes the matrix; it exists only when the determinant is non-zero.

The key idea. Just as $5 \times \frac{1}{5} = 1$ for ordinary numbers, the inverse $A^{-1}$ of a square matrix satisfies

$AA^{-1} = A^{-1}A = I$

where $I$ is the identity matrix (1s on the leading diagonal, 0s elsewhere). Multiplying by $A^{-1}$ "cancels" a multiplication by $A$ — exactly what you need to make $A$ disappear from an equation.

When does an inverse exist? Only when the determinant is non-zero, $\det A \neq 0$ . Look at the $2\times2$ formula below: it divides by $\det A$ , so a zero determinant means dividing by zero — impossible. A matrix with $\det A = 0$ is called singular and has no inverse. A matrix with $\det A \neq 0$ is non-singular (or invertible).

Why the determinant matters geometrically. The determinant measures how the matrix scales area (in 2D) or volume (in 3D). If $\det A = 0$ the transformation collapses everything onto a line or plane — information is lost, and there is no way to undo it. That is precisely why no inverse can exist.

Two properties you must know (the second is not in MF19):

$(A^{-1})^{-1} = A$ — undoing the undo gives you back the original.
$(AB)^{-1} = B^{-1}A^{-1}$ — the inverse of a product reverses the order (think "socks then shoes" to undress: shoes off first, then socks).

Multiplying a non-singular matrix by its inverse collapses to the identity I — the matrix equivalent of "1".

Cambridge tip. Before you start any inverse, compute $\det A$ first. If it is zero, stop and state "the matrix is singular, so no inverse exists" — that statement itself earns a mark.

$AA^{-1} = A^{-1}A = I$ defines the inverse.
An inverse exists only when $\det A \neq 0$ (non-singular).
$(A^{-1})^{-1} = A$ and $(AB)^{-1} = B^{-1}A^{-1}$ (reverse order).

See the full worked example for inverse of a matrix →

The 2×2 inverse — swap, negate, divide

Swap the leading diagonal, negate the other two, divide by ad − bc.

For a $2\times2$ matrix the inverse is a three-step recipe that you should be able to do in seconds.

If $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ then

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

The recipe in words:

Swap the two entries on the leading diagonal ( $a \leftrightarrow d$ ).
Negate the two off-diagonal entries ( $b \to -b$ , $c \to -c$ ).
Divide every entry by the determinant $ad - bc$ .

Swap the green diagonal, negate the red off-diagonal, then divide the whole matrix by det = ad − bc.

Worked sketch. For $A = \begin{pmatrix} 4 & 3 \\ 5 & 2 \end{pmatrix}$ : $\det A = (4)(2) - (3)(5) = 8 - 15 = -7$ , so

$A^{-1} = \frac{1}{-7}\begin{pmatrix} 2 & -3 \\ -5 & 4 \end{pmatrix} = \begin{pmatrix} -\frac{2}{7} & \frac{3}{7} \\[2pt] \frac{5}{7} & -\frac{4}{7} \end{pmatrix}.$

Always verify. Multiply back: $AA^{-1}$ should give $I$ . This catches almost every slip and the check is fast.

Cambridge tip. Keep the $\frac{1}{\det}$ outside the matrix as long as you can — it keeps the working clean and is exactly how mark schemes present the answer. Only multiply it in at the very end if a numerical matrix is demanded.

$A^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ .
Swap the leading diagonal, negate the off-diagonal, divide by the determinant.
Keep $\frac{1}{\det}$ as a factor outside the matrix until the last step.

See the full worked example for inverse of a matrix →

The 3×3 inverse — cofactors, adjugate, divide

Cofactor matrix → transpose for the adjugate → divide by the determinant.

The $3\times3$ inverse follows the same "divide by the determinant" idea, but the numerator is built from cofactors. This whole method is not in MF19 — you must know every step.

Step 1 — the matrix of cofactors. For each entry, cross out its row and column, take the determinant of the remaining $2\times2$ block (this is the minor $M_{ij}$ ), then attach the sign from the checkerboard pattern:

$\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix} \qquad \text{cofactor } C_{ij} = (-1)^{i+j}M_{ij}.$

Step 2 — transpose to get the adjugate. The adjugate (also "adjoint") is the transpose of the cofactor matrix: $\operatorname{adj}(A) = C^{\mathsf{T}}$ . Reflect the cofactor matrix across its leading diagonal.

Step 3 — divide by the determinant.

$A^{-1} = \frac{1}{\det A}\operatorname{adj}(A) = \frac{1}{\det A}\,C^{\mathsf{T}}.$

The determinant can itself be found by expanding along the top row using those same cofactors: $\det A = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$ .

A → cofactors → transpose (adjugate) → divide by det A. Skipping the transpose is the most common error.

Cambridge tip. The single biggest $3\times3$ trap is forgetting the transpose — students compute the cofactor matrix and divide by the determinant without transposing. Write "adj $= C^{\mathsf{T}}$ " explicitly so you never miss it.

Cofactor $C_{ij} = (-1)^{i+j} \times$ (minor): checkerboard signs.
Adjugate $=$ transpose of the cofactor matrix.
$A^{-1} = \frac{1}{\det A}\operatorname{adj}(A)$ — divide last.

See the full worked example for inverse of a matrix →

Solving Ax = b with the inverse

Write the equations as Ax = b, then x = A⁻¹b in one matrix product.

The most valuable application of inverses is solving simultaneous equations. A system such as

$\begin{cases} 2x + y = 5 \\ 3x - 2y = 4 \end{cases}$

can be packaged as a single matrix equation $A\mathbf{x} = \mathbf{b}$ :

$\underbrace{\begin{pmatrix} 2 & 1 \\ 3 & -2 \end{pmatrix}}_{A}\underbrace{\begin{pmatrix} x \\ y \end{pmatrix}}_{\mathbf{x}} = \underbrace{\begin{pmatrix} 5 \\ 4 \end{pmatrix}}_{\mathbf{b}}.$

The solution. Multiply both sides on the left by $A^{-1}$ :

$A^{-1}A\mathbf{x} = A^{-1}\mathbf{b} \;\Rightarrow\; I\mathbf{x} = A^{-1}\mathbf{b} \;\Rightarrow\; \boxed{\mathbf{x} = A^{-1}\mathbf{b}}.$

The order matters: matrix multiplication is not commutative, so you must pre-multiply ( $A^{-1}\mathbf{b}$ ), never post-multiply ( $\mathbf{b}A^{-1}$ , which is not even defined here).

Pre-multiplying by A⁻¹ turns A x = b into x = A⁻¹b. Always multiply on the LEFT.

The recipe:

Write the system as $A\mathbf{x} = \mathbf{b}$ , reading off the coefficient matrix $A$ and the constant vector $\mathbf{b}$ .
Find $A^{-1}$ (and if $\det A = 0$ , the system has no unique solution).
Compute $\mathbf{x} = A^{-1}\mathbf{b}$ — the entries of $\mathbf{x}$ are your answers.

This scales directly to three equations in three unknowns using the $3\times3$ inverse.

Cambridge tip. The whole point of the method is that $A^{-1}$ depends only on the left-hand-side coefficients. If a question gives several different right-hand sides $\mathbf{b}_1, \mathbf{b}_2, \ldots$ with the same $A$ , find $A^{-1}$ once and reuse it.

Package the equations as $A\mathbf{x} = \mathbf{b}$ .
Solution is $\mathbf{x} = A^{-1}\mathbf{b}$ — pre-multiply on the LEFT.
$\det A = 0$ means no unique solution.

See the full worked example for inverse of a matrix →

Quick recap

$A^{-1}$ satisfies $AA^{-1} = A^{-1}A = I$ ; it exists only when $\det A \neq 0$ .
$2\times2$ : swap diagonal, negate off-diagonal, divide by $ad - bc$ .
$3\times3$ : cofactor matrix $\rightarrow$ transpose (adjugate) $\rightarrow$ divide by $\det A$ .
$(A^{-1})^{-1} = A$ and $(AB)^{-1} = B^{-1}A^{-1}$ (reverse order — not in MF19).
$A\mathbf{x} = \mathbf{b} \Rightarrow \mathbf{x} = A^{-1}\mathbf{b}$ (pre-multiply on the left).
Always verify by checking $AA^{-1} = I$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$2\times2$ inverse: $A^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ .
Singular $\Leftrightarrow \det A = 0$ (no inverse).
$3\times3$ method: cofactors $\rightarrow$ adjugate $= C^{\mathsf{T}} \rightarrow$ divide by $\det A$ .
Reverse-order law $(AB)^{-1} = B^{-1}A^{-1}$ (not in MF19).
$A\mathbf{x} = \mathbf{b} \Rightarrow \mathbf{x} = A^{-1}\mathbf{b}$ (left-multiply).

How it’s examined

Topic 1.4 appears in most Paper 1 sessions, often worth 5-9 marks. A $2\times2$ inverse is usually a quick warm-up; the $3\times3$ cofactor/adjugate method and the $A\mathbf{x}=\mathbf{b}$ solve carry the bigger tariffs. Note that neither the $3\times3$ cofactor method nor the reverse-order law $(AB)^{-1}=B^{-1}A^{-1}$ is in MF19, so secure recall is essential. Marks are lost through (i) forgetting to transpose when forming the adjugate, (ii) sign slips in the cofactor checkerboard, and (iii) post-multiplying instead of pre-multiplying when solving systems. A single line stating 'the matrix is singular, so no inverse exists' earns credit when $\det A = 0$ .

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 1); 9231/11, 9231/12, 9231/13 Further Pure Mathematics 1 — past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Inverse of a matrix

Step-by-step solutions to past-paper-style questions on inverse of a matrix, written exactly the way a tutor would explain them at the board.

Question type:

1Inverse of a 2×2 matrix — and verifying it
Getting startedDirect calculation• 2x2, inverse, verify
Question
Find the inverse of $A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$ and verify your answer by computing $AA^{-1}$ .
Step-by-step solution
1. Step 1
  Determinant first. With $a=3, b=1, c=5, d=2$ , compute $\det A = ad - bc$ .
  $\det A = (3)(2) - (1)(5) = 6 - 5 = 1$
2. Step 2
  Swap the leading diagonal, negate the off-diagonal, divide by $\det A = 1$ .
  $A^{-1} = \frac{1}{1}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$
3. Step 3
  Verify $AA^{-1} = I$ . Multiply the original by the inverse.
  $AA^{-1} = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 6-5 & -3+3 \\ 10-10 & -5+6 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
Answer
$A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$ , confirmed by $AA^{-1} = I$ .
Examiner tip
A determinant of $1$ makes this clean, but always show the $\frac{1}{\det}$ step explicitly — the method mark depends on it. The $AA^{-1} = I$ check is the fastest way to catch a sign slip.
2Inverse of a 2×2 matrix with a fractional result
Getting startedDirect calculation• 2x2, inverse, fractions
Question
Find the inverse of $B = \begin{pmatrix} 2 & 4 \\ 1 & 3 \end{pmatrix}$ .
Step-by-step solution
1. Step 1
  Determinant. $a=2, b=4, c=1, d=3$ .
  $\det B = (2)(3) - (4)(1) = 6 - 4 = 2$
2. Step 2
  Apply the recipe: swap diagonal, negate off-diagonal, keep $\frac{1}{2}$ outside.
  $B^{-1} = \frac{1}{2}\begin{pmatrix} 3 & -4 \\ -1 & 2 \end{pmatrix}$
3. Step 3
  Multiply in the scalar to give the explicit matrix.
  $B^{-1} = \begin{pmatrix} \tfrac{3}{2} & -2 \\ -\tfrac{1}{2} & 1 \end{pmatrix}$
Answer
$B^{-1} = \frac{1}{2}\begin{pmatrix} 3 & -4 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{2} & -2 \\ -\frac{1}{2} & 1 \end{pmatrix}$ .
Examiner tip
Leaving the answer as $\frac{1}{2}\begin{pmatrix}\cdots\end{pmatrix}$ is fully acceptable and how mark schemes present it. Only multiply in when an explicit numerical matrix is demanded.
3Recognising a singular matrix
Building confidenceIdentify & classify• singular, determinant
Question
Determine whether $C = \begin{pmatrix} 2 & 6 \\ 1 & 3 \end{pmatrix}$ has an inverse. Justify your answer.
Step-by-step solution
1. Step 1
  Compute the determinant before attempting any inverse.
  $\det C = (2)(3) - (6)(1) = 6 - 6 = 0$
2. Step 2
  Interpret. The $2\times2$ inverse formula divides by $\det C$ . Since $\det C = 0$ , division is impossible, so no inverse exists.
3. Step 3
  State the conclusion in the language the mark scheme wants.
Answer
$\det C = 0$ , so $C$ is singular and has no inverse.
Examiner tip
The single sentence ' $\det = 0$ , so the matrix is singular and has no inverse' is itself worth a mark. Notice the second row is exactly half the first — a hallmark of a singular matrix.
4Solving simultaneous equations with x = A⁻¹b
Building confidenceMulti-step problem• simultaneous, Ax=b, 2x2
Question
Use a matrix inverse to solve $\begin{cases} 2x + y = 5 \\ 3x - 2y = 4 \end{cases}$ .
Step-by-step solution
1. Step 1
  Write as $A\mathbf{x} = \mathbf{b}$ .
  $\begin{pmatrix} 2 & 1 \\ 3 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}$
2. Step 2
  Find $A^{-1}$ . $\det A = (2)(-2) - (1)(3) = -4 - 3 = -7$ .
  $A^{-1} = \frac{1}{-7}\begin{pmatrix} -2 & -1 \\ -3 & 2 \end{pmatrix} = \frac{1}{7}\begin{pmatrix} 2 & 1 \\ 3 & -2 \end{pmatrix}$
3. Step 3
  Pre-multiply $\mathbf{b}$ by $A^{-1}$ : $\mathbf{x} = A^{-1}\mathbf{b}$ .
  $\mathbf{x} = \frac{1}{7}\begin{pmatrix} 2 & 1 \\ 3 & -2 \end{pmatrix}\begin{pmatrix} 5 \\ 4 \end{pmatrix} = \frac{1}{7}\begin{pmatrix} 14 \\ 7 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$
Answer
$x = 2$ , $\quad y = 1$ .
Examiner tip
Check by substitution: $2(2)+1 = 5$ ✓ and $3(2)-2(1) = 4$ ✓. Note $A^{-1}$ must pre-multiply $\mathbf{b}$ — writing $\mathbf{b}A^{-1}$ is not even a valid product here.
5Inverse of a 3×3 matrix via cofactors and the adjugate
StretchMulti-step problem• 3x3, cofactors, adjugate
Question
Find the inverse of $M = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$ .
Step-by-step solution
1. Step 1
  Determinant (expand along the top row using minors).
  $\det M = 1(1\cdot 0 - 4\cdot 6) - 2(0\cdot 0 - 4\cdot 5) + 3(0\cdot 6 - 1\cdot 5) = -24 + 40 - 15 = 1$
2. Step 2
  Matrix of cofactors $C_{ij} = (-1)^{i+j}M_{ij}$ (checkerboard signs).
  $C = \begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{pmatrix}$
3. Step 3
  Adjugate = transpose of $C$ . Reflect across the leading diagonal.
  $\operatorname{adj}(M) = C^{\mathsf{T}} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}$
4. Step 4
  Divide by $\det M = 1$ and verify with one row of $MM^{-1}$ .
  $M^{-1} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix},\quad \text{row 1 of } MM^{-1}: (1)(-24)+(2)(20)+(3)(-5)=1\;\checkmark$
Answer
$M^{-1} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}$ (since $\det M = 1$ ).
Examiner tip
The transpose step is where most marks are lost — note how the cofactor entry $20$ in position (2,1) moves to (1,2) in the adjugate. A quick $MM^{-1}$ row-check (here giving $1$ on the diagonal) confirms the whole calculation.
6The reverse-order law (AB)⁻¹ = B⁻¹A⁻¹
StretchShow that / prove• reverse-order, product, proof
Question
Given square invertible matrices $A$ and $B$ , prove that $(AB)^{-1} = B^{-1}A^{-1}$ .
Step-by-step solution
1. Step 1
  Strategy. To show some matrix $X$ is the inverse of $AB$ , show $(AB)X = I$ . Take $X = B^{-1}A^{-1}$ .
2. Step 2
  Multiply $(AB)(B^{-1}A^{-1})$ and use associativity (drop the brackets).
  $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = A\,I\,A^{-1} = AA^{-1} = I$
3. Step 3
  Check the other side $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I$ . Both products give $I$ , so $B^{-1}A^{-1}$ is the inverse of $AB$ .
Answer
$(AB)(B^{-1}A^{-1}) = I$ and $(B^{-1}A^{-1})(AB) = I$ , so $(AB)^{-1} = B^{-1}A^{-1}$ .
Examiner tip
The order reversal comes from the inner $BB^{-1}$ collapsing first. Writing $(AB)^{-1} = A^{-1}B^{-1}$ (wrong order) earns nothing — this law is not in MF19, so the proof technique '(AB)X = I' is worth memorising.

Model Answers — Inverse of a matrix

High-scoring sample answers for inverse of a matrix on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
Find the inverse of $A = \begin{pmatrix} 4 & 7 \\ 2 & 6 \end{pmatrix}$ .
Model answer
Determinant: $\det A = (4)(6) - (7)(2) = 24 - 14 = 10$

Apply the $2\times2$ inverse formula (swap diagonal, negate off-diagonal, divide by the determinant): $A^{-1} = \frac{1}{10}\begin{pmatrix} 6 & -7 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} \frac{3}{5} & -\frac{7}{10} \\[2pt] -\frac{1}{5} & \frac{2}{5} \end{pmatrix}$
Why this scores
M1 for $\det A = 10$ , M1 for the swap-and-negate matrix $\begin{pmatrix} 6 & -7 \\ -2 & 4 \end{pmatrix}$ , A1 for the correct $\frac{1}{10}$ factor. Leaving the answer in $\frac{1}{10}(\cdots)$ form is fully credited.
Question 2
3 marks
The matrix $D = \begin{pmatrix} k & 3 \\ 4 & 6 \end{pmatrix}$ is singular. Find the value of $k$ .
Model answer
A matrix is singular when its determinant is zero. $\det D = 6k - 12 = 0$

Solve for $k$ : $6k = 12 \;\Rightarrow\; k = 2$
Why this scores
M1 for setting $\det D = 6k - 12$ , M1 for equating to $0$ , A1 for $k = 2$ . The keyword 'singular' must be linked to ' $\det = 0$ ' — a definition mark candidates often miss.
Question 3
6 marks
(a) Write the system $\begin{cases} 3x + 2y = 12 \\ x + 4y = 14 \end{cases}$ in the form $A\mathbf{x} = \mathbf{b}$ . (b) Find $A^{-1}$ and hence solve the system.
Model answer
(a) $\begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 12 \\ 14 \end{pmatrix}$

(b) Determinant: $\det A = (3)(4) - (2)(1) = 10$ , so $A^{-1} = \frac{1}{10}\begin{pmatrix} 4 & -2 \\ -1 & 3 \end{pmatrix}$

Pre-multiply $\mathbf{b}$ by $A^{-1}$ : $\mathbf{x} = A^{-1}\mathbf{b} = \frac{1}{10}\begin{pmatrix} 4 & -2 \\ -1 & 3 \end{pmatrix}\begin{pmatrix} 12 \\ 14 \end{pmatrix} = \frac{1}{10}\begin{pmatrix} 48 - 28 \\ -12 + 42 \end{pmatrix} = \frac{1}{10}\begin{pmatrix} 20 \\ 30 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$

So $x = 2$ , $y = 3$ .
Why this scores
(a) B1 for the correct matrix equation. (b) M1 $\det A = 10$ , A1 $A^{-1}$ , M1 for the product $A^{-1}\mathbf{b}$ (must pre-multiply), A1 for $x=2, y=3$ . Check: $3(2)+2(3)=12$ ✓, $2+4(3)=14$ ✓.
Question 4
6 marks
Given $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$ , verify that $(AB)^{-1} = B^{-1}A^{-1}$ .
Model answer
Compute $AB$ : $AB = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}$

$\det(AB) = (7)(1) - (2)(3) = 1$ , so $(AB)^{-1} = \begin{pmatrix} 1 & -2 \\ -3 & 7 \end{pmatrix}$

Now $B^{-1}A^{-1}$ . $\det A = 1$ and $\det B = 1$ , so $A^{-1} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}, \qquad B^{-1} = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}$

$B^{-1}A^{-1} = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ -3 & 7 \end{pmatrix}$

The two results agree, so $(AB)^{-1} = B^{-1}A^{-1}$ . ∎
Why this scores
M1 $AB$ , A1 $(AB)^{-1}$ ; M1 for both $A^{-1}, B^{-1}$ , M1 for the product $B^{-1}A^{-1}$ (correct order), A1 for the matching matrix. A common loss: computing $A^{-1}B^{-1}$ (wrong order), which gives $\begin{pmatrix} 7 & -2 \\ -3 & 1 \end{pmatrix} \neq (AB)^{-1}$ .
Question 5
7 marks
Find the inverse of $P = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}$ .
Model answer
Determinant (expand along row 1): $\det P = 2(2\cdot2 - 1\cdot1) - 1(1\cdot2 - 1\cdot1) + 1(1\cdot1 - 2\cdot1) = 2(3) - 1(1) + 1(-1) = 4$

Matrix of cofactors ( $C_{ij} = (-1)^{i+j}M_{ij}$ ): $C = \begin{pmatrix} 3 & -1 & -1 \\ -1 & 3 & -1 \\ -1 & -1 & 3 \end{pmatrix}$

Adjugate = transpose of $C$ . (Here $C$ is symmetric, so $\operatorname{adj}(P) = C$ .)

Divide by $\det P = 4$ : $P^{-1} = \frac{1}{4}\begin{pmatrix} 3 & -1 & -1 \\ -1 & 3 & -1 \\ -1 & -1 & 3 \end{pmatrix}$

Check (row 1 of $PP^{-1}$ ): $\frac{1}{4}\big[(2)(3)+(1)(-1)+(1)(-1)\big] = \frac{4}{4} = 1$ ✓.
Why this scores
M1 $\det P = 4$ , M1 for a correct cofactor (any), A1 full cofactor matrix, M1 transpose (stated, even though $C$ is symmetric here), A1 $\frac{1}{4}$ factor, A1 final matrix, plus B1 for an explicit check. State the transpose step even when $C$ is symmetric — examiners want the method shown.
Question 6
8 marks
Solve the system $\begin{cases} x + y + z = 6 \\ 2x + 3y + z = 11 \\ x + 2y + 3z = 14 \end{cases}$ using a matrix inverse, given that the coefficient matrix has determinant $3$ .
Model answer
Matrix form $A\mathbf{x} = \mathbf{b}$ : $A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & 3 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 6 \\ 11 \\ 14 \end{pmatrix}, \quad \det A = 3$

Cofactor matrix of $A$ : $C = \begin{pmatrix} 7 & -5 & 1 \\ -1 & 2 & -1 \\ -2 & 1 & 1 \end{pmatrix}$

Adjugate = transpose: $\operatorname{adj}(A) = \begin{pmatrix} 7 & -1 & -2 \\ -5 & 2 & 1 \\ 1 & -1 & 1 \end{pmatrix}, \qquad A^{-1} = \frac{1}{3}\begin{pmatrix} 7 & -1 & -2 \\ -5 & 2 & 1 \\ 1 & -1 & 1 \end{pmatrix}$

Solve $\mathbf{x} = A^{-1}\mathbf{b}$ : $\mathbf{x} = \frac{1}{3}\begin{pmatrix} 7 & -1 & -2 \\ -5 & 2 & 1 \\ 1 & -1 & 1 \end{pmatrix}\begin{pmatrix} 6 \\ 11 \\ 14 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 42 - 11 - 28 \\ -30 + 22 + 14 \\ 6 - 11 + 14 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 3 \\ 6 \\ 9 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$

So $x = 1$ , $y = 2$ , $z = 3$ .
Why this scores
M1 matrix form; M1 A1 cofactors; M1 transpose to adjugate; A1 $A^{-1}$ ; M1 product $A^{-1}\mathbf{b}$ ; A1 $x=1,y=2,z=3$ . Check: $1+2+3=6$ ✓, $2+6+3=11$ ✓, $1+4+9=14$ ✓. The transpose must be done before forming $A^{-1}$ — omit it and every $x,y,z$ is wrong.

Key Formulae — Inverse of a matrix

The formulae you need to memorise for inverse of a matrix on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Inverse of a 2×2 matrix
$A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}, \quad A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$
$a, b, c, d$
the four entries of $A$
$ad - bc$
the determinant $\det A$ (must be non-zero)
When to use
Any $2\times2$ inverse. Swap the leading diagonal, negate the off-diagonal, divide by $\det A$ .
Determinant of a 2×2 matrix
$\det A = ad - bc$
$a, b, c, d$
entries of $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$
When to use
Compute FIRST. If $\det A = 0$ the matrix is singular and has no inverse.
Inverse of a 3×3 matrix (adjugate method)
$A^{-1} = \frac{1}{\det A}\operatorname{adj}(A) = \frac{1}{\det A}\,C^{\mathsf{T}}$
$C$
matrix of cofactors $C_{ij} = (-1)^{i+j}M_{ij}$
$C^{\mathsf{T}}$
the adjugate — transpose of the cofactor matrix
$\det A$
determinant (non-zero)
When to use
For $3\times3$ inverses. Build cofactors, transpose to the adjugate, divide by $\det A$ . Not in MF19.
Reverse-order law for inverses
$(AB)^{-1} = B^{-1}A^{-1}$
$A, B$
invertible square matrices of the same size
When to use
Inverting a product. The order reverses. Not in MF19 — memorise it.
Inverse of an inverse
$(A^{-1})^{-1} = A$
$A$
any invertible matrix
When to use
Whenever you undo an inverse — you recover the original matrix.
Solving a linear system by inverse
$A\mathbf{x} = \mathbf{b} \;\Rightarrow\; \mathbf{x} = A^{-1}\mathbf{b}$
$A$
coefficient matrix (non-singular)
$\mathbf{x}$
column of unknowns
$\mathbf{b}$
column of constants
When to use
Simultaneous equations. Pre-multiply (left-multiply) by $A^{-1}$ , never post-multiply.

Key Definitions and Keywords — Inverse of a matrix

Definitions to memorise and the exact keywords mark schemes credit for inverse of a matrix answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Inverse matrix $A^{-1}$
Examiner keyword
The unique matrix satisfying $AA^{-1} = A^{-1}A = I$ . It 'undoes' the action of $A$ and exists only when $A$ is non-singular ( $\det A \neq 0$ ).
Determinant $\det A$
Examiner keyword
A scalar computed from a square matrix ( $ad - bc$ for $2\times2$ ). The inverse divides by it, so $\det A \neq 0$ is the condition for an inverse to exist.
Singular matrix
Examiner keyword
A square matrix with $\det A = 0$ . It has NO inverse. A matrix with $\det A \neq 0$ is called non-singular or invertible.
Cofactor $C_{ij}$
Examiner keyword
The minor $M_{ij}$ (determinant of the block left after deleting row $i$ and column $j$ ) multiplied by the sign $(-1)^{i+j}$ from the checkerboard pattern.
Adjugate (adjoint) $\operatorname{adj}(A)$
Examiner keyword
The transpose of the cofactor matrix, $C^{\mathsf{T}}$ . The $3\times3$ inverse is $\frac{1}{\det A}\operatorname{adj}(A)$ .
Identity matrix $I$
The square matrix with $1$ s on the leading diagonal and $0$ s elsewhere. It is the matrix analogue of the number $1$ : $AI = IA = A$ .

Common Mistakes and Misconceptions — Inverse of a matrix

The traps other students keep falling into on inverse of a matrix questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Forgetting to transpose the cofactor matrix when forming the $3\times3$ adjugate
9231 Paper 1 Examiner Reports — recurring comment
Why it happens
Students compute the cofactor matrix and divide straight by the determinant, skipping the adjugate step.
How to avoid it
Write 'adj $= C^{\mathsf{T}}$ ' explicitly and reflect the entries across the leading diagonal BEFORE dividing by $\det A$ .
✕Sign errors in the cofactor checkerboard (e.g. taking $C_{12}$ as $+M_{12}$ )
9231 Paper 1 mark schemes
Why it happens
Forgetting the $(-1)^{i+j}$ sign pattern, or mis-counting which entries are negative.
How to avoid it
Sketch the $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$ grid in the margin first and read signs off it.
✕Attempting to invert a singular matrix (dividing by a zero determinant)
9231 Paper 1 Examiner Reports
Why it happens
Jumping into the inverse recipe without computing $\det A$ first.
How to avoid it
Always find $\det A$ before anything else. If it is $0$ , state 'singular — no inverse exists' (that statement scores).
✕Writing $(AB)^{-1} = A^{-1}B^{-1}$ instead of $B^{-1}A^{-1}$
9231 Paper 1 Examiner Reports
Why it happens
Assuming the inverse distributes in the same order as the product, as it would for commuting numbers.
How to avoid it
Remember the order REVERSES: $(AB)^{-1} = B^{-1}A^{-1}$ ('socks and shoes'). Matrix multiplication is not commutative.
✕Solving $A\mathbf{x} = \mathbf{b}$ by computing $\mathbf{b}A^{-1}$ (post-multiplying)
9231 Paper 1 mark schemes
Why it happens
Treating matrix multiplication as commutative; not realising $\mathbf{b}A^{-1}$ may not even be defined.
How to avoid it
Pre-multiply both sides on the LEFT: $A^{-1}A\mathbf{x} = A^{-1}\mathbf{b}$ , giving $\mathbf{x} = A^{-1}\mathbf{b}$ .
✕Swapping and negating the $2\times2$ entries but forgetting to divide by the determinant
9231 Paper 1 Examiner Reports
Why it happens
Treating the $\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ rearrangement as the whole answer.
How to avoid it
The rearranged matrix is only the adjugate; the inverse is $\frac{1}{ad-bc}$ times it. Keep the $\frac{1}{\det}$ factor visible throughout.

Inverse of a matrix

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Inverse of a 2×2 matrix — and verifying it

2Inverse of a 2×2 matrix with a fractional result

3Recognising a singular matrix

4Solving simultaneous equations with x = A⁻¹b

5Inverse of a 3×3 matrix via cofactors and the adjugate

6The reverse-order law (AB)⁻¹ = B⁻¹A⁻¹

Inverse of a 2×2 matrix

Determinant of a 2×2 matrix

Inverse of a 3×3 matrix (adjugate method)

Reverse-order law for inverses

Inverse of an inverse

Solving a linear system by inverse

Inverse matrix A−1A^{-1}A−1

Determinant det⁡A\det AdetA

Singular matrix

Cofactor CijC_{ij}Cij​

Adjugate (adjoint) adj⁡(A)\operatorname{adj}(A)adj(A)

Identity matrix III

✕Forgetting to transpose the cofactor matrix when forming the 3×33\times33×3 adjugate

✕Sign errors in the cofactor checkerboard (e.g. taking C12C_{12}C12​ as +M12+M_{12}+M12​)

✕Attempting to invert a singular matrix (dividing by a zero determinant)

✕Writing (AB)−1=A−1B−1(AB)^{-1} = A^{-1}B^{-1}(AB)−1=A−1B−1 instead of B−1A−1B^{-1}A^{-1}B−1A−1

✕Solving Ax=bA\mathbf{x} = \mathbf{b}Ax=b by computing bA−1\mathbf{b}A^{-1}bA−1 (post-multiplying)

✕Swapping and negating the 2×22\times22×2 entries but forgetting to divide by the determinant

Inverse matrix $A^{-1}$

Determinant $\det A$

Cofactor $C_{ij}$

Adjugate (adjoint) $\operatorname{adj}(A)$

Identity matrix $I$

✕Forgetting to transpose the cofactor matrix when forming the $3\times3$ adjugate

✕Sign errors in the cofactor checkerboard (e.g. taking $C_{12}$ as $+M_{12}$ )

✕Writing $(AB)^{-1} = A^{-1}B^{-1}$ instead of $B^{-1}A^{-1}$

✕Solving $A\mathbf{x} = \mathbf{b}$ by computing $\mathbf{b}A^{-1}$ (post-multiplying)

✕Swapping and negating the $2\times2$ entries but forgetting to divide by the determinant