# Matrices

> CIE A-Level Further Mathematics · Further Pure 1
> Source: https://www.owlsprep.com/study/cie-9231-u1-matrices/

This sub-topic covers core matrix operations for CIE A-Level Further Mathematics FP1, including calculating determinants, finding inverses, evaluating matrix rank, solving linear systems, and finding eigenvalues and eigenvectors. It forms the foundation for all further linear algebra topics.

**Prerequisites:** [Basic matrix operations from A-Level Mathematics](https://www.owlsprep.com/study/cie-9709-p1-matrices/)

## Learning objectives

- Calculate determinants of 3×3 matrices using cofactor expansion
- Find the inverse of non-singular 2×2 and 3×3 matrices
- Evaluate the rank of a 3×3 matrix and identify singular matrices
- Determine consistency and solve 3-variable systems of linear equations
- Find eigenvalues and eigenvectors for 2×2 and 3×3 square matrices

## Determinants of 3×3 Matrices

**Determinant** — 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).

*Notation:* \det(A) \text{ or } |A|

*Example:* For 2×2 matrix $A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}$, $\det(A) = ad - bc$.

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

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.

*Calculator:* forbidden

## Inverse Matrices and Matrix Rank

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

*Notation:* \text{rank}(A)

*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: $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$, where $\text{adj}(A)$ (the adjugate) is the transpose of the cofactor matrix of A.

**Worked example:** Find $A^{-1}$ for $A = \begin{pmatrix} 1 & 2 & 0 \\ 3 & 1 & -1 \\ -2 & 0 & 3 \end{pmatrix}$

1. We already know $\det(A) = -11 ≠ 0$, 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 $AA^{-1}$ 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.

*Calculator:* forbidden

## Solving Systems of Linear Equations

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

1. If $\det(A) ≠ 0$: rank(A) = 3, unique solution $\mathbf{x} = A^{-1}\mathbf{b}$
2. 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: $x + 2y = 5$, $3x + y - z = 0$, $-2x + 3z = -11$

1. Write the system in matrix form $A\mathbf{x} = \mathbf{b}$, where A is the matrix from earlier examples, and $\mathbf{b} = \begin{pmatrix}5 \\ 0 \\ -11\end{pmatrix}$.
2. We know $\det(A) = -11 ≠ 0$, so there is a unique solution. Use $\mathbf{x} = A^{-1}\mathbf{b}$:
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 $k≠0$.

*Calculator:* forbidden

## Eigenvalues and Eigenvectors

**Eigenvalue and Eigenvector** — For a square matrix A, a non-zero vector $\mathbf{v}$ is an eigenvector if $A\mathbf{v} = \lambda \mathbf{v}$ for some scalar $\lambda$, which is the corresponding eigenvalue. To find eigenvalues, solve the characteristic equation $\det(A - \lambda I) = 0$.

*Notation:* $\lambda$ (eigenvalue), $\mathbf{v}$ (eigenvector)

**Worked example:** Find the eigenvalues and eigenvectors of $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$

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 $\lambda=1$ by solving $(A - I)\mathbf{v} = \mathbf{0}$:
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 $\lambda=1$ are all non-zero scalar multiples of $\begin{pmatrix}1 \\ -1\end{pmatrix}$. Repeat for $\lambda=3$:
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.

*Calculator:* forbidden

## Common pitfalls

- **Wrong:** Forgetting the $(-1)^{i+j}$ sign when calculating cofactors
  - Why it fails: Sign errors are the most common mistake in determinant and inverse calculations
  - Correct: Write the cofactor sign matrix $\begin{pmatrix}+ & - & + \\ - & + & - \\ + & - & +\end{pmatrix}$ next to your work to check each entry's sign
- **Wrong:** Forgetting to transpose the cofactor matrix to get the adjugate
  - Why it fails: Many students use the cofactor matrix directly as the adjugate, leading to an incorrect inverse
  - Correct: Always transpose the cofactor matrix before dividing by the determinant to get $A^{-1}$
- **Wrong:** Claiming $\det(A) = 0$ means the system has no solutions
  - Why it fails: A zero determinant only means there is no unique solution, not that there are no solutions
  - Correct: Check the rank of the augmented matrix: if rank(A) < rank(augmented), no solution; if ranks are equal, infinitely many solutions
- **Wrong:** Accepting the zero vector as an eigenvector
  - Why it fails: The definition of an eigenvector requires it to be non-zero, and zero satisfies $A\mathbf{0} = \lambda \mathbf{0}$ for any $\lambda$
  - Correct: Always state that eigenvectors are non-zero scalar multiples of your basis vector
- **Wrong:** Expanding a 3x3 determinant along a row with no zeros
  - Why it fails: This increases the number of calculations you need to do, raising the chance of arithmetic error and wasting time
  - Correct: Always pick the row or column with the most zeros to expand the determinant

## Cheatsheet

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

## 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.

- [Polar coordinates](https://www.owlsprep.com/study/cie-9231-u1-polar-coordinates/)
- [Vectors](https://www.owlsprep.com/study/cie-9231-u1-vectors/)
- [Further Pure 2](https://www.owlsprep.com/study/cie-9231-u2-overview/)

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