Matrices

CIE A-Level Further Mathematics· Unit 1: Further Pure 1, Topic 4: Matrices· 15 min read

1. Determinants of 3×3 Matrices★★☆☆☆⏱ 5 min

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📘 Definition

Determinant

\det(A) \text{ or } |A|

A scalar value calculated from a square matrix. A non-zero determinant means the matrix is invertible, while a determinant of zero means the matrix is singular (non-invertible).

Example:

For 2×2 matrix $A = (a c b d)$ , $det (A) = a d - b c$ .

To calculate the determinant of a 3×3 matrix, we use cofactor expansion. For any entry $a_{ij}$ , the cofactor is $C_{ij} = (- 1)^{i + j} M_{ij}$ , where $M_{ij}$ (the minor) is the determinant of the 2×2 submatrix obtained by deleting row $i$ and column $j$ . We can expand along any row or column.

📐 Worked Example

Find $det (A)$ for $A = 13 - 2 210 0 - 1 3$

1
Expand along the first row, which has a zero to reduce computation:
2
$\det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} \\ C_{11} = (-1)^{1+1} \begin{vmatrix} 1 & -1 \\ 0 & 3 \end{vmatrix} = 1(3 - 0) = 3 \\ C_{12} = (-1)^{1+2} \begin{vmatrix} 3 & -1 \\ -2 & 3 \end{vmatrix} = -1(9 - 2) = -7 \\ C_{13} = (-1)^{1+3} \begin{vmatrix} 3 & 1 \\ -2 & 0 \end{vmatrix} = 1(0 + 2) = 2$
3
Substitute back to get the final determinant:
4
$\det(A) = (1)(3) + (2)(-7) + (0)(2) = 3 - 14 = -11$

Exam tip:

Always expand along the row or column with the most zeros to reduce arithmetic and cut down on calculation time.

2. Inverse Matrices and Matrix Rank★★★☆☆⏱ 8 min

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📘 Definition

Rank of a Matrix

\text{rank}(A)

The maximum number of linearly independent rows (or columns) of A. For a 3×3 matrix: rank = 3 if $det (A) \neq = 0$ , rank = 2 if $det (A) = 0$ but at least one non-zero 2×2 minor, rank = 1 if all 2×2 minors are zero.

Example:

A 3×3 singular matrix has rank less than 3.

To find the inverse of a 3×3 matrix, we use the adjugate method: $A^{- 1} = \frac{1}{d e t ( A )} adj (A)$ , where $adj (A)$ (the adjugate) is the transpose of the cofactor matrix of A.

📐 Worked Example

Find $A^{- 1}$ for $A = 13 - 2 210 0 - 1 3$

1
We already know $det (A) = - 11 \neq = 0$ , so the inverse exists. First, construct the full cofactor matrix C:
2
$C = \begin{pmatrix} 3 & -7 & 2 \\ -6 & 3 & -4 \\ -2 & 1 & -5 \end{pmatrix}$
3
The adjugate matrix is the transpose of C:
4
$\text{adj}(A) = C^T = \begin{pmatrix} 3 & -6 & -2 \\ -7 & 3 & 1 \\ 2 & -4 & -5 \end{pmatrix}$
5
Divide by $det (A)$ to get the inverse:
6
$A^{-1} = \frac{1}{-11} \begin{pmatrix} 3 & -6 & -2 \\ -7 & 3 & 1 \\ 2 & -4 & -5 \end{pmatrix} = \begin{pmatrix} -\frac{3}{11} & \frac{6}{11} & \frac{2}{11} \\ \frac{7}{11} & -\frac{3}{11} & -\frac{1}{11} \\ -\frac{2}{11} & \frac{4}{11} & \frac{5}{11} \end{pmatrix}$
7
Verify by multiplying $A A^{- 1}$ to confirm you get the identity matrix, to catch errors.

Exam tip:

Always transpose the cofactor matrix to get the adjugate. This is one of the most common errors in inverse matrix calculations.

3. Solving Systems of Linear Equations★★★☆☆⏱ 7 min

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Any system of 3 linear equations in 3 unknowns can be written as $A x = b$ , where A is the coefficient matrix, $x$ is the vector of unknowns, and $b$ is the constant vector. The number of solutions depends on the determinant and rank of A and the augmented matrix $(A ∣ b)$ .

If $det (A) \neq = 0$ : rank(A) = 3, unique solution $x = A^{- 1} b$
If $det (A) = 0$ : either no solution (inconsistent, rank(A) < rank(augmented)) or infinitely many solutions (rank(A) = rank(augmented) < 3)

📐 Worked Example

Find the solution to the system: $x + 2 y = 5$ , $3 x + y - z = 0$ , $- 2 x + 3 z = - 11$

1
Write the system in matrix form $A x = b$ , where A is the matrix from earlier examples, and $b = 50 - 11$ .
2
We know $det (A) = - 11 \neq = 0$ , so there is a unique solution. Use $x = A^{- 1} b$ :
3
$\mathbf{x} = \begin{pmatrix} -\frac{3}{11} & \frac{6}{11} & \frac{2}{11} \\ \frac{7}{11} & -\frac{3}{11} & -\frac{1}{11} \\ -\frac{2}{11} & \frac{4}{11} & \frac{5}{11} \end{pmatrix} \begin{pmatrix}5 \\ 0 \\ -11\end{pmatrix} = \begin{pmatrix} -\frac{37}{11} \\ \frac{46}{11} \\ -\frac{65}{11} \end{pmatrix}$
4
The unique solution is $x = - \frac{37}{11}$ , $y = \frac{46}{11}$ , $z = - \frac{65}{11}$ .

Exam tip:

If asked to show a system is inconsistent, eliminate variables and look for a contradiction of the form $0 = k$ where $k \neq = 0$ .

4. Eigenvalues and Eigenvectors★★★☆☆⏱ 6 min

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📘 Definition

Eigenvalue and Eigenvector

$λ$ (eigenvalue), $v$ (eigenvector)

For a square matrix A, a non-zero vector $v$ is an eigenvector if $A v = λ v$ for some scalar $λ$ , which is the corresponding eigenvalue. To find eigenvalues, solve the characteristic equation $det (A - λ I) = 0$ .

📐 Worked Example

Find the eigenvalues and eigenvectors of $A = (2112)$

1
Form the characteristic equation:
2
$\det(A - \lambda I) = \begin{vmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{vmatrix} = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = 0$
3
Factor the quadratic to find eigenvalues:
4
$(\lambda - 1)(\lambda - 3) = 0 \implies \lambda = 1 \text{ and } \lambda = 3$
5
Find eigenvector for $λ = 1$ by solving $(A - I) v = 0$ :
6
$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies y = -x$
7
Eigenvectors for $λ = 1$ are all non-zero scalar multiples of $(1 - 1)$ . Repeat for $λ = 3$ :
8
$(A - 3I)\mathbf{v} = \mathbf{0} \implies y = x \implies \text{eigenvectors} = k\begin{pmatrix}1 \\ 1\end{pmatrix}, k≠0$

Exam tip:

Never accept the zero vector as an eigenvector. Always state that the scalar multiple must be non-zero in your answer.

5. Common Pitfalls

Wrong move:

Forgetting the $(- 1)^{i + j}$ sign when calculating cofactors

Why:

Sign errors are the most common mistake in determinant and inverse calculations

Correct move:

Write the cofactor sign matrix $+ - + - + - + - +$ next to your work to check each entry's sign

Wrong move:

Forgetting to transpose the cofactor matrix to get the adjugate

Why:

Many students use the cofactor matrix directly as the adjugate, leading to an incorrect inverse

Correct move:

Always transpose the cofactor matrix before dividing by the determinant to get $A^{- 1}$

Wrong move:

Claiming $det (A) = 0$ means the system has no solutions

Why:

A zero determinant only means there is no unique solution, not that there are no solutions

Correct move:

Check the rank of the augmented matrix: if rank(A) < rank(augmented), no solution; if ranks are equal, infinitely many solutions

Wrong move:

Accepting the zero vector as an eigenvector

Why:

The definition of an eigenvector requires it to be non-zero, and zero satisfies $A 0 = λ 0$ for any $λ$

Correct move:

Always state that eigenvectors are non-zero scalar multiples of your basis vector

Wrong move:

Expanding a 3x3 determinant along a row with no zeros

Why:

This increases the number of calculations you need to do, raising the chance of arithmetic error and wasting time

Correct move:

Always pick the row or column with the most zeros to expand the determinant

6. Quick Reference Cheatsheet

Concept	Key Rule/Formula
3×3 Determinant	Expand along row/column with most zeros: $det (A) = \sum a_{ij} C_{ij}$
Inverse Matrix	$A^{- 1} = \frac{1}{d e t ( A )} adj (A)$ , $adj (A) = C^{T}$
Singular Matrix	$det (A) = 0$ , no inverse exists
Linear System $A x = b$	$det (A) \neq = 0 \Rightarrow$ unique solution $x = A^{- 1} b$ ; $det (A) = 0 \Rightarrow$ check rank for no/infinitely many solutions
Characteristic Equation	$det (A - λ I) = 0$ , solve for eigenvalues
Eigenvector	Solve $(A - λ I) v = 0$ , $v \neq = 0$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
3x3 inverse, system of linear equations
2023 · 1
Eigenvalues of 2x2 matrix
2024 · 1
Rank and consistency of linear system

Going deeper

syllabusOfficial CIE 9231 SyllabusFull syllabus and assessment objectives

What's Next

Matrices are the foundational building block for all linear algebra topics in CIE A-Level Further Mathematics. The skills you master here — calculating determinants, finding inverses, solving linear systems, and working with eigenvalues — are applied repeatedly in advanced topics across all Further Pure units. Understanding core matrix properties is essential to scoring full marks on linear transformation, diagonalization, and differential system questions later in your course. Building fluency with matrix arithmetic now will reduce cognitive load for more complex topics in future units.