Matrices

Matrices

Matrices provide a compact algebraic framework for representing and manipulating systems of linear Equations, geometric transformations, and — at a more advanced level — quantum mechanical states and Data structures. This topic develops the algebra of matrices and their interpretation as linear Transformations of the plane and space.

Board Coverage

Board	Paper	Notes
AQA	Paper 1	3D transformations, eigenvalues and eigenvectors
Edexcel	FP1	2D transformations, $3\times3$ matrices, determinants
OCR (A)	Paper 1	2D transformations, $3\times3$ matrices
CIE	P1	2D and basic 3D transformations, inverses, determinants

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1. Matrix Notation and Operations

Definition. An $m \times n$ matrix is a rectangular array of numbers arranged in $m$ rows and $n$ columns. We write $\mathbf{A} = (a_{ij})$ where $a_{ij}$ is the entry in row $i$Column $j$. The set of all $m \times n$ matrices with real entries is denoted $M_{m\times n}(\mathbb{R})$.

A matrix with a single row is a row vector, and a matrix with a single column is a column Vector.

1.1 Matrix Addition and Scalar Multiplication

For matrices $\mathbf{A}, \mathbf{B} \in M_{m \times n}(\mathbb{R})$ and scalar $k \in \mathbb{R}$:

$$\begin{aligned} (\mathbf{A} + \mathbf{B})_{ij} &= a_{ij} + b_{ij} \\ (k\mathbf{A})_{ij} &= k \cdot a_{ij} \end{aligned}$$

Matrix addition is commutative ($\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}$) and Associative.

1.2 Matrix Multiplication

Definition. If $\mathbf{A} \in M_{m \times p}(\mathbb{R})$ and $\mathbf{B} \in M_{p \times n}(\mathbb{R})$The product $\mathbf{AB} \in M_{m \times n}(\mathbb{R})$ is defined by:

$\boxed{(\mathbf{AB})_{ij} = \sum_{k=1}^{p} a_{ik}\, b_{kj}}$

Matrix multiplication is associative but not commutative : $\mathbf{AB} \neq \mathbf{BA}$.

The $n \times n$ identity matrix $\mathbf{I}_n$ satisfies $\mathbf{A}\mathbf{I}_n = \mathbf{I}_n\mathbf{A} = \mathbf{A}$ for any $\mathbf{A} \in M_{m \times n}(\mathbb{R})$.

:::caution warning “apply $\mathbf{B}$ first, then $\mathbf{A}$.” When composing transformations, The rightmost matrix is applied first. :::

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2. Determinants

2.1 The $2 \times 2$ Determinant

Definition. For $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$:

$\boxed{\det(\mathbf{A}) = ad - bc}$

2.2 The $3 \times 3$ Determinant

Definition. For $\mathbf{A} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$ The determinant is computed by cofactor expansion along any row or column:

$\boxed{\det(\mathbf{A}) = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}}$

The signs alternate $+$, $-$, $+$ along the first row (following the checkerboard pattern).

2.3 Properties of Determinants

For $n \times n$ matrices $\mathbf{A}, \mathbf{B}$ and scalar $k$:

1. $\det(\mathbf{I}_n) = 1$
2. $\det(\mathbf{A}^T) = \det(\mathbf{A})$
3. $\det(k\mathbf{A}) = k^n \det(\mathbf{A})$
4. Swapping two rows (or columns) multiplies the determinant by $-1$
5. A matrix with a zero row (or column) has $\det = 0$
6. $\det(\mathbf{A}) = 0$ if and only if $\mathbf{A}$ is singular (non-invertible)

Proof of $\det(\mathbf{AB}) = \det(\mathbf{A})\det(\mathbf{B})$ (for $2\times2$ matrices)

Let $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix} e & f \\ g & h \end{pmatrix}$.

$\mathbf{AB} = \begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix}$

$$\begin{aligned} \det(\mathbf{AB}) &= (ae + bg)(cf + dh) - (af + bh)(ce + dg) \\ &= acef + adeh + bcfg + bdgh - acef - adfg - bceh - bdgh \\ &= adeh + bcfg - adfg - bceh \\ &= ad(eh - fg) - bc(eh - fg) \\ &= (ad - bc)(eh - fg) \\ &= \det(\mathbf{A}) \cdot \det(\mathbf{B}) \quad \square \end{aligned}$$

Intuition. The determinant measures how a matrix scales area (in 2D) or volume (in 3D). Composing two transformations multiplies their area/volume scaling factors, which is why Determinants multiply.

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3. Inverse Matrices

Definition. The inverse of a square matrix $\mathbf{A}$Written $\mathbf{A}^{-1}$Is the Unique matrix satisfying:

$\mathbf{A}\mathbf{A}^{-1} = \mathbf{A}^{-1}\mathbf{A} = \mathbf{I}$

An inverse exists if and only if $\det(\mathbf{A}) \neq 0$. Such a matrix is called non-singular.

3.1 Inverse of a $2 \times 2$ Matrix

For $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with $\det(\mathbf{A}) = ad - bc \neq 0$:

$\boxed{\mathbf{A}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}}$

3.2 The Adjugate Method ($3 \times 3$)

For a $3 \times 3$ matrix $\mathbf{A}$:

$\boxed{\mathbf{A}^{-1} = \frac◆LB◆1◆RB◆◆LB◆\det(\mathbf{A})◆RB◆\operatorname{adj}(\mathbf{A})}$

Where the adjugate (or adjoint) matrix $\operatorname{adj}(\mathbf{A})$ is the transpose of the cofactor matrix.

Definition. The cofactor $C_{ij}$ of entry $a_{ij}$ is $(-1)^{i+j}$ times the determinant of The submatrix obtained by deleting row $i$ and column $j$. The cofactor matrix has entries $C_{ij}$And $\operatorname{adj}(\mathbf{A}) = (C_{ij})^T$.

Proof that the inverse is unique

Suppose $\mathbf{B}$ and $\mathbf{C}$ are both inverses of $\mathbf{A}$. Then:

$\mathbf{B} = \mathbf{B}\mathbf{I} = \mathbf{B}(\mathbf{AC}) = (\mathbf{BA})\mathbf{C} = \mathbf{IC} = \mathbf{C} \quad \square$

:::tip To verify your inverse, always check that $\mathbf{A}\mathbf{A}^{-1} = \mathbf{I}$. This Catches sign errors and arithmetic mistakes immediately. :::

Worked Example: $3\times3$ inverse

Find $\mathbf{A}^{-1}$ where $\mathbf{A} = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 1 & 0 & 1 \end{pmatrix}$.

$\det(\mathbf{A}) = 1\begin{vmatrix} 1 & 3 \\ 0 & 1 \end{vmatrix} - 2\begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} + 0 = 1(1) - 2(-3) = 7$.

Cofactors:

$$\begin{aligned} C_{11} &= +\begin{vmatrix} 1 & 3 \\ 0 & 1 \end{vmatrix} = 1, \quad C_{12} &= -\begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} = 3, \quad C_{13} &= +\begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} = -1 \\ C_{21} &= -\begin{vmatrix} 2 & 0 \\ 0 & 1 \end{vmatrix} = -2, \quad C_{22} &= +\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = 1, \quad C_{23} &= -\begin{vmatrix} 1 & 2 \\ 1 & 0 \end{vmatrix} = 2 \\ C_{31} &= +\begin{vmatrix} 2 & 0 \\ 1 & 3 \end{vmatrix} = 6, \quad C_{32} &= -\begin{vmatrix} 1 & 0 \\ 0 & 3 \end{vmatrix} = -3, \quad C_{33} &= +\begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 \end{aligned}$$

$\operatorname{adj}(\mathbf{A}) = \begin{pmatrix} 1 & -2 & 6 \\ 3 & 1 & -3 \\ -1 & 2 & 1 \end{pmatrix}$

$\mathbf{A}^{-1} = \frac{1}{7}\begin{pmatrix} 1 & -2 & 6 \\ 3 & 1 & -3 \\ -1 & 2 & 1 \end{pmatrix}$

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4. Solving Systems of Linear Equations

A system of $n$ linear equations in $n$ unknowns can be written as $\mathbf{A}\mathbf{x} = \mathbf{b}$ where $\mathbf{A}$ is the coefficient matrix, $\mathbf{x}$ is The column vector of unknowns, and $\mathbf{b}$ is the column vector of constants.

If $\mathbf{A}$ is non-singular, the unique solution is:

$\boxed{\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}}$

4.1 Geometric Interpretation (2D)

For a $2 \times 2$ system:

• $\det(\mathbf{A}) \neq 0$: the two lines intersect at a unique point.
• $\det(\mathbf{A}) = 0$ and the equations are consistent: the lines are coincident (infinitely many solutions).
• $\det(\mathbf{A}) = 0$ and the equations are inconsistent: the lines are parallel (no solutions).

4.2 Cramer’s Rule

For a system $\mathbf{A}\mathbf{x} = \mathbf{b}$ where $\det(\mathbf{A}) \neq 0$:

$x_i = \frac◆LB◆\det(\mathbf{A}_i)◆RB◆◆LB◆\det(\mathbf{A})◆RB◆$

Where $\mathbf{A}_i$ is $\mathbf{A}$ with column $i$ replaced by $\mathbf{b}$.

Worked Example: Solving a $3\times3$ system

Solve: $\begin{cases} x + 2y = 4 \\ y + 3z = 5 \\ x + z = 2 \end{cases}$

In matrix form: $\begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 1 & 0 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 5 \\ 2 \end{pmatrix}$.

Using the inverse from the previous example:

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{7}\begin{pmatrix} 1 & -2 & 6 \\ 3 & 1 & -3 \\ -1 & 2 & 1 \end{pmatrix}\begin{pmatrix} 4 \\ 5 \\ 2 \end{pmatrix} = \frac{1}{7}\begin{pmatrix} 4 - 10 + 12 \\ 12 + 5 - 6 \\ -4 + 10 + 2 \end{pmatrix} = \frac{1}{7}\begin{pmatrix} 6 \\ 11 \\ 8 \end{pmatrix} = \begin{pmatrix} 6/7 \\ 11/7 \\ 8/7 \end{pmatrix}$

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5. Matrix Transformations in 2D

A $2 \times 2$ matrix $\mathbf{T}$ represents a linear transformation of $\mathbb{R}^2$: the point $(x, y)$ is mapped to $(x', y')$ where $\begin{pmatrix} x' \\ y' \end{pmatrix} = \mathbf{T}\begin{pmatrix} x \\ y \end{pmatrix}$.

5.1 Reflection

Reflection in the $x$-axis: $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

Reflection in the $y$-axis: $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$

Reflection in the line $y = x$: $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

Reflection in the line $y = \tan\theta\, x$:

$\boxed{\mathbf{R} = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}}$

Proof of the reflection matrix

The reflection of a vector in a line through the origin making angle $\theta$ with the $x$-axis can Be decomposed: first rotate by $-\theta$ to align the mirror with the $x$-axis, reflect in the $x$-axis, then rotate back by $\theta$.

$\mathbf{R} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$

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

$= \begin{pmatrix} \cos^2\theta - \sin^2\theta & \cos\theta\sin\theta + \sin\theta\cos\theta \\ \sin\theta\cos\theta + \cos\theta\sin\theta & \sin^2\theta - \cos^2\theta \end{pmatrix} = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix} \quad \square$

5.2 Rotation

Rotation anticlockwise by angle $\theta$ about the origin:

$\boxed{\mathbf{R}_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}}$

Note: $\det(\mathbf{R}_\theta) = \cos^2\theta + \sin^2\theta = 1$.

5.3 Enlargement

Enlargement by scale factor $k$ about the origin: $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$

$\det = k^2$Confirming the area is scaled by $k^2$.

5.4 Shear

Shear parallel to the $x$-axis by factor $k$: $\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$

Shear parallel to the $y$-axis by factor $k$: $\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}$

Note: $\det = 1$ for shears, so area is preserved.

5.5 Combining Transformations

If transformation $\mathbf{A}$ is followed by transformation $\mathbf{B}$The combined Transformation is represented by $\mathbf{BA}$ (rightmost applied first).

:::caution Order matters. A rotation followed by a reflection generally produces a different result From a reflection followed by a rotation. The combined matrix is $\mathbf{BA}$ (not $\mathbf{AB}$) When $\mathbf{A}$ is applied first. :::

:::tip To find the matrix of a combined transformation, multiply the matrices in reverse order of Application. If the question says “reflect then rotate,” compute $\mathbf{R}_{\mathrm{rot}} \times \mathbf{R}_{\mathrm{ref}}$. :::

Worked Example: Combined transformation

Find the matrix representing a rotation of $90^\circ$ anticlockwise about the origin followed by a Reflection in the line $y = x$.

Rotation by $90^\circ$: $\mathbf{R} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$

Reflection in $y = x$: $\mathbf{S} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

Combined (reflection applied after rotation): $\mathbf{S}\mathbf{R} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

This is a reflection in the $x$-axis. $\det = -1$Consistent with an orientation-reversing Transformation.

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6. Matrix Transformations in 3D

:::info 3D transformations are required by AQA and appear on CIE P1. Edexcel and OCR focus primarily On 2D but may include basic $3\times3$ determinant and inverse calculations. :::

6.1 Rotations in 3D

Rotation about the $x$-axis by angle $\theta$:

$\mathbf{R}_x = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{pmatrix}$

Rotation about the $y$-axis by angle $\theta$:

$\mathbf{R}_y = \begin{pmatrix} \cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\ -\sin\theta & 0 & \cos\theta \end{pmatrix}$

Rotation about the $z$-axis by angle $\theta$:

$\mathbf{R}_z = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}$

Each has $\det = 1$ and represents a rigid motion preserving distances and orientation.

6.2 Reflections in 3D

Reflection in the plane $x = 0$ (the $yz$-plane): $\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

Reflection in the plane $y = 0$ (the $xz$-plane): $\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

Reflection in the plane $z = 0$ (the $xy$-plane): $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$

Each has $\det = -1$Confirming orientation reversal.

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7. Eigenvalues and Eigenvectors

:::info info require this topic at A Level. ::: Definition. Let $\mathbf{A}$ be an $n \times n$ matrix. A scalar $\lambda$ is an eigenvalue of $\mathbf{A}$ if there exists a non-zero vector $\mathbf{v}$ such that:

$\boxed{\mathbf{A}\mathbf{v} = \lambda\mathbf{v}}$

The vector $\mathbf{v}$ is called an eigenvector corresponding to $\lambda$.

7.1 The Characteristic Equation

$\mathbf{A}\mathbf{v} = \lambda\mathbf{v} \iff (\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0}$.

For a non-trivial solution ($\mathbf{v} \neq \mathbf{0}$), we require $\det(\mathbf{A} - \lambda\mathbf{I}) = 0$.

$\boxed{\det(\mathbf{A} - \lambda\mathbf{I}) = 0}$

This is the characteristic equation of $\mathbf{A}$. Its roots are the eigenvalues.

7.2 Finding Eigenvectors

For each eigenvalue $\lambda$Solve $(\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0}$ by Row reduction.

7.3 Diagonalisation

Definition. A matrix $\mathbf{A}$ is diagonalisable if there exists an invertible matrix $\mathbf{P}$ and a diagonal matrix $\mathbf{D}$ such that:

$\boxed{\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}}$

The columns of $\mathbf{P}$ are the eigenvectors of $\mathbf{A}$And the diagonal entries of $\mathbf{D}$ are the corresponding eigenvalues.

A matrix is diagonalisable if and only if it has $n$ linearly independent eigenvectors (always true For $n$ distinct eigenvalues).

Intuition. Diagonalisation changes to a coordinate system where the transformation acts Independently on each axis (stretching by eigenvalues). In this basis, the matrix takes its simplest Possible form.

Worked Example: Eigenvalues and eigenvectors

Find the eigenvalues and eigenvectors of $\mathbf{A} = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$.

Characteristic equation: $\det\!\begin{pmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{pmatrix} = 0$

$(4 - \lambda)(3 - \lambda) - 2 = \lambda^2 - 7\lambda + 10 = (\lambda - 5)(\lambda - 2) = 0$

Eigenvalues: $\lambda_1 = 5$, $\lambda_2 = 2$.

For $\lambda_1 = 5$: $\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies -x + y = 0 \implies y = x$.

Eigenvector: $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ (or any non-zero scalar Multiple).

For $\lambda_2 = 2$: $\begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies 2x + y = 0 \implies y = -2x$.

Eigenvector: $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$.

Diagonalisation: $\mathbf{P} = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}$ $\mathbf{D} = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}$Giving $\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$.

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8. Summary of Key Results

$\boxed{\det(\mathbf{AB}) = \det(\mathbf{A})\det(\mathbf{B})}$

$\boxed{\mathbf{A}^{-1} = \frac◆LB◆1◆RB◆◆LB◆\det(\mathbf{A})◆RB◆\operatorname{adj}(\mathbf{A}) \quad \mathrm{when } \det(\mathbf{A}) \neq 0}$

$\boxed{\mathrm{Rotation by } \theta: \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}}$

$\boxed{\mathrm{Reflection in } y = (\tan\theta)x: \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}}$

$\boxed{\det(\mathbf{A} - \lambda\mathbf{I}) = 0 \implies \mathrm{eigenvalues of } \mathbf{A}}$

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Problems

Problem 1. Given $\mathbf{A} = \begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix} 1 & 5 \\ -2 & 0 \end{pmatrix}$Compute $\mathbf{AB} - \mathbf{BA}$.

Hint

Compute both products separately and subtract. They will not be equal.

Answer

$\mathbf{AB} = \begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} 1 & 5 \\ -2 & 0 \end{pmatrix} = \begin{pmatrix} 5 & 15 \\ -6 & 10 \end{pmatrix}$

$\mathbf{BA} = \begin{pmatrix} 1 & 5 \\ -2 & 0 \end{pmatrix}\begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 13 & 19 \\ -6 & 2 \end{pmatrix}$

$\mathbf{AB} - \mathbf{BA} = \begin{pmatrix} 5 & 15 \\ -6 & 10 \end{pmatrix} - \begin{pmatrix} 13 & 19 \\ -6 & 2 \end{pmatrix} = \begin{pmatrix} -8 & -4 \\ 0 & 8 \end{pmatrix}$

This confirms $\mathbf{AB} \neq \mathbf{BA}$.

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Problem 2. Find the determinant and inverse of $\mathbf{A} = \begin{pmatrix} 2 & 1 & 3 \\ 0 & -1 & 2 \\ 1 & 0 & 1 \end{pmatrix}$.

Hint

Expand the determinant along the first row. Then compute cofactors for the adjugate.

Answer

$\det(\mathbf{A}) = 2\begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} - 1\begin{vmatrix} 0 & 2 \\ 1 & 1 \end{vmatrix} + 3\begin{vmatrix} 0 & -1 \\ 1 & 0 \end{vmatrix}$

$= 2(-1) - 1(-2) + 3(1) = -2 + 2 + 3 = 3$

Cofactors:

$$\begin{aligned} C_{11} = +(-1) = -1, \quad C_{12} = -(0 \cdot 1 - 2 \cdot 1) = 2, \quad C_{13} = +(0 \cdot 0 - (-1) \cdot 1) = 1 \\ C_{21} = -(1 \cdot 1 - 3 \cdot 0) = -1, \quad C_{22} = +(2 \cdot 1 - 3 \cdot 1) = -1, \quad C_{23} = -(2 \cdot 0 - 1 \cdot 1) = 1 \\ C_{31} = +(2 + 2) = 4, \quad C_{32} = -(4 - 0) = -4, \quad C_{33} = +(-2 - 0) = -2 \end{aligned}$$

$\operatorname{adj}(\mathbf{A}) = \begin{pmatrix} -1 & -1 & 4 \\ 2 & -1 & -4 \\ 1 & 1 & -2 \end{pmatrix}$

$\mathbf{A}^{-1} = \frac{1}{3}\begin{pmatrix} -1 & -1 & 4 \\ 2 & -1 & -4 \\ 1 & 1 & -2 \end{pmatrix}$

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Problem 3. Find the matrix representing a reflection in the line $y = \sqrt{3}\,x$And verify That $\mathbf{M}^2 = \mathbf{I}$.

Hint

The line makes angle $\theta$ with the $x$-axis where $\tan\theta = \sqrt{3}$. A reflection applied Twice is the identity.

Answer

$\tan\theta = \sqrt{3} \implies \theta = \dfrac◆LB◆\pi◆RB◆◆LB◆3◆RB◆$.

$\mathbf{M} = \begin{pmatrix} \cos\frac◆LB◆2\pi◆RB◆◆LB◆3◆RB◆ & \sin\frac◆LB◆2\pi◆RB◆◆LB◆3◆RB◆ \\ \sin\frac◆LB◆2\pi◆RB◆◆LB◆3◆RB◆ & -\cos\frac◆LB◆2\pi◆RB◆◆LB◆3◆RB◆ \end{pmatrix} = \begin{pmatrix} -\frac{1}{2} & \frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆ \\ \frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆ & \frac{1}{2} \end{pmatrix}$

Verification: $\mathbf{M}^2 = \begin{pmatrix} -\frac{1}{2} & \frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆ \\ \frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆ & \frac{1}{2} \end{pmatrix}\begin{pmatrix} -\frac{1}{2} & \frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆ \\ \frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆ & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} \frac{1}{4} + \frac{3}{4} & -\frac◆LB◆\sqrt{3}◆RB◆◆LB◆4◆RB◆ + \frac◆LB◆\sqrt{3}◆RB◆◆LB◆4◆RB◆ \\ -\frac◆LB◆\sqrt{3}◆RB◆◆LB◆4◆RB◆ + \frac◆LB◆\sqrt{3}◆RB◆◆LB◆4◆RB◆ & \frac{3}{4} + \frac{1}{4} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbf{I} \quad \square$

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Problem 4. The triangle with vertices (0, 0)$$(2, 0)$$(0, 1) is transformed by the matrix $\mathbf{T} = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$. Find the coordinates of the vertices of The image, and verify that the area scales by $|\det(\mathbf{T})|$.

Hint

Apply $\mathbf{T}$ to each vertex. The original triangle has area 1.

Answer

(0, 0) \mapsto (0, 0)$$(2, 0) \mapsto (6, 0)$$(0, 1) \mapsto (1, 2).

Image vertices: (0, 0)$$(6, 0)$$(1, 2).

Original area: $\dfrac{1}{2} \times 2 \times 1 = 1$.

Image area using the determinant formula: $\dfrac{1}{2}\left|6 \cdot 2 - 0 \cdot 1\right| = \dfrac{1}{2} \times 12 = 6$.

$\det(\mathbf{T}) = 6$And $|\det(\mathbf{T})| \times \mathrm{original area} = 6 \times 1 = 6$. ✓

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Problem 5. Solve the system of equations using matrices:

$\begin{cases} 2x + y - z = 3 \\ x - y + 2z = 1 \\ 3x + 2y + z = 10 \end{cases}$

Hint

Write as $\mathbf{A}\mathbf{x} = \mathbf{b}$ and compute $\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}$.

Answer

$\mathbf{A} = \begin{pmatrix} 2 & 1 & -1 \\ 1 & -1 & 2 \\ 3 & 2 & 1 \end{pmatrix}$ $\mathbf{b} = \begin{pmatrix} 3 \\ 1 \\ 10 \end{pmatrix}$.

$\det(\mathbf{A}) = 2\begin{vmatrix} -1 & 2 \\ 2 & 1 \end{vmatrix} - 1\begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} + (-1)\begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix}$

$= 2(-5) - 1(-5) - 1(5) = -10 + 5 - 5 = -10$

Cofactors: C_{11} = -5$$C_{12} = 5$$C_{13} = 5$$C_{21} = -3$$C_{22} = 5$$C_{23} = -1 C_{31} = 1$$C_{32} = -5$$C_{33} = -3.

$\mathbf{A}^{-1} = -\frac{1}{10}\begin{pmatrix} -5 & -3 & 1 \\ 5 & 5 & -5 \\ 5 & -1 & -3 \end{pmatrix}$

$\mathbf{x} = -\frac{1}{10}\begin{pmatrix} -5 & -3 & 1 \\ 5 & 5 & -5 \\ 5 & -1 & -3 \end{pmatrix}\begin{pmatrix} 3 \\ 1 \\ 10 \end{pmatrix} = -\frac{1}{10}\begin{pmatrix} -15 - 3 + 10 \\ 15 + 5 - 50 \\ 15 - 1 - 30 \end{pmatrix} = -\frac{1}{10}\begin{pmatrix} -8 \\ -30 \\ -16 \end{pmatrix} = \begin{pmatrix} 4/5 \\ 3 \\ 8/5 \end{pmatrix}$

So x = \dfrac{4}{5}$$y = 3$$z = \dfrac{8}{5}.

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Problem 6. Find the single $2 \times 2$ matrix that represents an enlargement by scale factor 2 About the origin followed by a rotation of $90^\circ$ anticlockwise. Show that this is equivalent to A single rotation of $90^\circ$ combined with an enlargement by factor 2.

Hint

The enlargement matrix is $2\mathbf{I}$ and the rotation is $\mathbf{R}_{\pi/2}$. Since $2\mathbf{I}$ commutes with all matrices, the order doesn’t matter.

Answer

Enlargement by 2: $\mathbf{E} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$.

Rotation by $90^\circ$: $\mathbf{R} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.

Enlargement then rotation: $\mathbf{R}\mathbf{E} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}$.

Rotation then enlargement: $\mathbf{E}\mathbf{R} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}$.

Both give the same result: $\begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix} = 2\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.

This is a rotation by $90^\circ$ combined with an enlargement by factor 2, and the order is Irrelevant because scalar multiples of the identity commute with all matrices.

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Problem 7. AQA only. Find the eigenvalues and corresponding eigenvectors of $\mathbf{A} = \begin{pmatrix} 5 & 4 \\ 1 & 2 \end{pmatrix}$And write down a matrix $\mathbf{P}$ And diagonal matrix $\mathbf{D}$ such that $\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$.

Hint

Solve $\det(\mathbf{A} - \lambda\mathbf{I}) = 0$ for $\lambda$Then find eigenvectors.

Answer

Characteristic equation: $\det\begin{pmatrix} 5 - \lambda & 4 \\ 1 & 2 - \lambda \end{pmatrix} = (5 - \lambda)(2 - \lambda) - 4 = \lambda^2 - 7\lambda + 6 = (\lambda - 1)(\lambda - 6) = 0$.

Eigenvalues: \lambda_1 = 6$$\lambda_2 = 1.

For $\lambda_1 = 6$: $\begin{pmatrix} -1 & 4 \\ 1 & -4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{0} \implies -x + 4y = 0 \implies x = 4y$.

Eigenvector: $\mathbf{v}_1 = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$.

For $\lambda_2 = 1$: $\begin{pmatrix} 4 & 4 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{0} \implies x + y = 0 \implies x = -y$.

Eigenvector: $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

$\mathbf{P} = \begin{pmatrix} 4 & 1 \\ 1 & -1 \end{pmatrix}, \quad \mathbf{D} = \begin{pmatrix} 6 & 0 \\ 0 & 1 \end{pmatrix}$

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Problem 8. AQA only. The matrix $\mathbf{A} = \begin{pmatrix} 3 & -2 \\ 1 & 0 \end{pmatrix}$ has Eigenvalues 1 and 2. Use this to compute $\mathbf{A}^5$ without multiplying matrices five times.

Hint

Diagonalise $\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$Then $\mathbf{A}^5 = \mathbf{P}\mathbf{D}^5\mathbf{P}^{-1}$.

Answer

For $\lambda = 1$: $\begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies x = y$. Eigenvector $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

For $\lambda = 2$: $\begin{pmatrix} 1 & -2 \\ 1 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies x = 2y$. Eigenvector $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

$\mathbf{P} = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}$ $\mathbf{D} = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$.

$\det(\mathbf{P}) = 1 - 2 = -1$So $\mathbf{P}^{-1} = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}$.

$\mathbf{D}^5 = \begin{pmatrix} 1 & 0 \\ 0 & 32 \end{pmatrix}$.

$\mathbf{A}^5 = \mathbf{P}\mathbf{D}^5\mathbf{P}^{-1} = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 32 \end{pmatrix}\begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 64 \\ 1 & 32 \end{pmatrix}\begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 63 & -62 \\ 31 & -30 \end{pmatrix}$

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Problem 9. Find the $3 \times 3$ matrix representing a rotation of $90^\circ$ anticlockwise About the $z$-axis. Verify that this matrix has determinant 1 and that it maps $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ to $\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$.

Hint

Use the standard formula for $\mathbf{R}_z$ with $\theta = \pi/2$.

Answer

$\mathbf{R}_z = \begin{pmatrix} \cos\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ & -\sin\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ & 0 \\ \sin\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ & \cos\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

$\det(\mathbf{R}_z) = 0 \cdot (0 - 0) - (-1)(1 - 0) + 0 = 1$. ✓

$\mathbf{R}_z\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \quad \checkmark$

The $x$-axis is correctly rotated to the $y$-axis by a $90^\circ$ anticlockwise rotation about $z$.

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Problem 10. Prove that if $\mathbf{A}$ and $\mathbf{B}$ are non-singular $n \times n$ matrices, Then $(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$.

Hint

Show that $\mathbf{B}^{-1}\mathbf{A}^{-1}$ satisfies the definition of the inverse of $\mathbf{AB}$ And invoke uniqueness.

Answer

We need to show that $(\mathbf{AB})(\mathbf{B}^{-1}\mathbf{A}^{-1}) = \mathbf{I}$ and $(\mathbf{B}^{-1}\mathbf{A}^{-1})(\mathbf{AB}) = \mathbf{I}$.

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

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

Since the inverse is unique, $(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$. $\square$

Intuition. The order reverses, just like putting on and taking off socks and shoes. To undo “A Then B,” you must undo B first, then undo A.

:::

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8. Advanced Worked Examples

Example 8.1: Diagonalisation of a 3×3 matrix

Problem. Diagonalise $\mathbf{A} = \begin{pmatrix}2&1&0\\0&2&0\\0&1&3\end{pmatrix}$.

Solution. Find eigenvalues: $\det(\mathbf{A}-\lambda\mathbf{I}) = 0$.

$\det\begin{pmatrix}2-\lambda&1&0\\0&2-\lambda&0\\0&1&3-\lambda\end{pmatrix} = (2-\lambda)^2(3-\lambda) = 0$

$\lambda_1 = 2$ (repeated), $\lambda_2 = 3$.

For $\lambda = 2$: $(\mathbf{A}-2\mathbf{I})\mathbf{v} = \mathbf{0}$ gives $\begin{pmatrix}0&1&0\\0&0&0\\0&1&1\end{pmatrix}\mathbf{v} = \mathbf{0}$So $v_2 = 0$ and $v_3 = 0$With $v_1$ free. Only one eigenvector: $(1,0,0)$. Since the geometric multiplicity (1) is Less than the algebraic multiplicity (2), $\mathbf{A}$ is not diagonalisable.

Example 8.2: Finding $\mathbf{A}^n$ using Cayley—Hamilton

Problem. For $\mathbf{A} = \begin{pmatrix}1&2\\0&3\end{pmatrix}$Find $\mathbf{A}^5$.

Solution. By Cayley—Hamilton, $\mathbf{A}^2 - 4\mathbf{A} + 3\mathbf{I} = \mathbf{O}$So $\mathbf{A}^2 = 4\mathbf{A} - 3\mathbf{I}$.

$\mathbf{A}^3 = \mathbf{A}(4\mathbf{A}-3\mathbf{I}) = 4(4\mathbf{A}-3\mathbf{I}) - 3\mathbf{A} = 13\mathbf{A} - 12\mathbf{I}$.

$\mathbf{A}^4 = 13(4\mathbf{A}-3\mathbf{I}) - 12\mathbf{A} = 40\mathbf{A} - 39\mathbf{I}$.

$\mathbf{A}^5 = 40(4\mathbf{A}-3\mathbf{I}) - 39\mathbf{A} = 121\mathbf{A} - 120\mathbf{I}$.

$= 121\begin{pmatrix}1&2\\0&3\end{pmatrix} - 120\begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}121&242\\0&363\end{pmatrix} - \begin{pmatrix}120&0\\0&120\end{pmatrix} = \boxed{\begin{pmatrix}1&242\\0&243\end{pmatrix}}$

Example 8.3: Invariant points and invariant lines

Problem. $\mathbf{A} = \begin{pmatrix}3&1\\0&2\end{pmatrix}$. Find all invariant points and Invariant lines of the transformation $\mathbf{x} \mapsto \mathbf{Ax}$.

Solution. Invariant points: $\mathbf{Ax} = \mathbf{x} \implies (\mathbf{A}-\mathbf{I})\mathbf{x} = \mathbf{0}$.

$\begin{pmatrix}2&1\\0&1\end{pmatrix}\mathbf{x} = \mathbf{0} \implies x_2 = 0, \; 2x_1 = 0$

Only the origin $(0,0)$ is an invariant point.

Invariant lines through the origin: These are the eigenspaces. Eigenvalues: $(3-\lambda)(2-\lambda) = 0$So $\lambda = 3$ and $\lambda = 2$.

For $\lambda = 3$: $\begin{pmatrix}0&1\\0&-1\end{pmatrix}\mathbf{v} = \mathbf{0} \implies v_2 = 0$. Line: $y = 0$ (the $x$-axis).

For $\lambda = 2$: $\begin{pmatrix}1&1\\0&0\end{pmatrix}\mathbf{v} = \mathbf{0} \implies v_1 + v_2 = 0$. Line: $y = -x$.

Example 8.4: Determinant as a scaling factor

Problem. The matrix $\mathbf{T} = \begin{pmatrix}2&1\\-1&3\end{pmatrix}$ represents a Transformation. A triangle has vertices $(0,0)$, $(1,0)$, $(0,1)$. Find the area of its image.

Solution. $\det(\mathbf{T}) = 6 - (-1) = 7$.

Original area $= \dfrac{1}{2}$. Image area $= |\det(\mathbf{T})| \times \text{original area} = 7 \times \dfrac{1}{2} = \boxed{3.5}$.

Example 8.5: Commutator and non-commuting matrices

Problem. For $\mathbf{A} = \begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix}0&0\\1&0\end{pmatrix}$Compute the commutator $[\mathbf{A}, \mathbf{B}] = \mathbf{AB} - \mathbf{BA}$.

Solution. $\mathbf{AB} = \begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix} = \begin{pmatrix}1&0\\0&0\end{pmatrix}$.

$\mathbf{BA} = \begin{pmatrix}0&0\\1&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix} = \begin{pmatrix}0&0\\0&1\end{pmatrix}$.

$\boxed{[\mathbf{A},\mathbf{B}] = \begin{pmatrix}1&0\\0&-1\end{pmatrix}}$

Since $[\mathbf{A},\mathbf{B}] \neq \mathbf{O}$, $\mathbf{A}$ and $\mathbf{B}$ do not commute.

Example 8.6: Matrix representing successive transformations

Problem. Transformation $R$ is a reflection in the line $y = x\sqrt{3}$. Transformation $S$ is a Rotation by $90°$ anticlockwise about the origin. Find the matrix representing $RS$.

Solution. The line $y = x\sqrt{3}$ makes angle $60°$ with the $x$-axis.

$R = \begin{pmatrix}\cos 120°&\sin 120°\\\sin 120°&-\cos 120°\end{pmatrix} = \begin{pmatrix}-\frac{1}{2}&\frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆\\\frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆&\frac{1}{2}\end{pmatrix}$.

$S = \begin{pmatrix}0&-1\\1&0\end{pmatrix}$.

$RS = \begin{pmatrix}-\frac{1}{2}&\frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆\\\frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆&\frac{1}{2}\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix} = \begin{pmatrix}\frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆&\frac{1}{2}\\\frac{1}{2}&-\frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆\end{pmatrix}$.

$\det(RS) = -\dfrac{3}{4} - \dfrac{1}{4} = -1$ and $\text{tr}(RS) = 0$Confirming this is a Reflection.

Example 8.7: Finding the inverse of a 3×3 matrix

Problem. Find $\mathbf{A}^{-1}$ where $\mathbf{A} = \begin{pmatrix}1&0&2\\0&1&-1\\1&1&0\end{pmatrix}$.

Solution. $\det(\mathbf{A}) = 1(0+1) - 0 + 2(0-1) = 1 - 2 = -1 \neq 0$.

$\mathbf{A}^{-1} = \dfrac{1}{-1}\begin{pmatrix}1&2&-2\\-1&-2&1\\-1&-1&1\end{pmatrix} = \boxed{\begin{pmatrix}-1&-2&2\\1&2&-1\\1&1&-1\end{pmatrix}}$

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9. Common Pitfalls

Pitfall	Correct Approach
Assuming all matrices are diagonalisable	Check geometric multiplicity equals algebraic multiplicity for each eigenvalue
Forgetting that $(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$	The order reverses
Computing $\det(\mathbf{A}+\mathbf{B}) = \det(\mathbf{A})+\det(\mathbf{B})$	, $\det(\mathbf{A}+\mathbf{B}) \neq \det(\mathbf{A})+\det(\mathbf{B})$
Mixing up row and column operations	Column operations change the determinant differently from row operations

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10. Additional Exam-Style Questions

Question 8

Find the eigenvalues and eigenvectors of $\mathbf{A} = \begin{pmatrix}4&-1\\2&1\end{pmatrix}$. Hence Write down $\mathbf{P}$ and $\mathbf{D}$ such that $\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \mathbf{D}$.

Solution

$\det(\mathbf{A}-\lambda\mathbf{I}) = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+6 = 0$. $\lambda = 2, 3$.

$\lambda = 2$: $\begin{pmatrix}2&-1\\2&-1\end{pmatrix}\mathbf{v}=\mathbf{0} \implies 2v_1-v_2=0 \implies \mathbf{v}=(1,2)$.

$\lambda = 3$: $\begin{pmatrix}1&-1\\2&-2\end{pmatrix}\mathbf{v}=\mathbf{0} \implies v_1=v_2 \implies \mathbf{v}=(1,1)$.

$\mathbf{P} = \begin{pmatrix}1&1\\2&1\end{pmatrix}$ $\mathbf{D} = \begin{pmatrix}2&0\\0&3\end{pmatrix}$.

Question 9

Prove that if $\lambda$ is an eigenvalue of $\mathbf{A}$ with eigenvector $\mathbf{v}$Then $\lambda^2$ is an eigenvalue of $\mathbf{A}^2$ with the same eigenvector.

Solution

$\mathbf{Av} = \lambda\mathbf{v}$.

$\mathbf{A}^2\mathbf{v} = \mathbf{A}(\mathbf{Av}) = \mathbf{A}(\lambda\mathbf{v}) = \lambda(\mathbf{Av}) = \lambda(\lambda\mathbf{v}) = \lambda^2\mathbf{v}$.

Therefore $\lambda^2$ is an eigenvalue of $\mathbf{A}^2$ with eigenvector $\mathbf{v}$. $\blacksquare$

Question 10

The transformation represented by $\mathbf{M} = \begin{pmatrix}a&b\\c&d\end{pmatrix}$ maps the unit Square to a parallelogram of area 6. Given $a+d = 5$ and $ad-bc = 6$Find the eigenvalues of $\mathbf{M}$.

Solution

The characteristic equation: $\lambda^2 - (a+d)\lambda + \det(\mathbf{M}) = \lambda^2 - 5\lambda + 6 = 0$.

$(\lambda-2)(\lambda-3) = 0$.

$\boxed{\lambda = 2 \text{ and } \lambda = 3}$

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11. Connections to Other Topics

11.1 Matrices and complex numbers

Complex eigenvalues lead to rotation-scaling transformations. See Complex Numbers.

11.2 Matrices and vectors

The cross product can be written as a matrix multiplication. See Vectors in 3D.

11.3 Matrices and further algebra

Cayley—Hamilton connects matrices to polynomial algebra. See Further Algebra.

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12. Key Results Summary

| Result | Formula/Condition | | -------------------- | --------------------------------------------------------------------------- | ---------------- | ------------------------------------ | | Invertibility | $\mathbf{A}$ is invertible $\iff$ $\det(\mathbf{A}) \neq 0$ | | $(\mathbf{AB})^{-1}$ | $\mathbf{B}^{-1}\mathbf{A}^{-1}$ | | $(\mathbf{AB})^T$ | $\mathbf{B}^T\mathbf{A}^T$ | | $\det(\mathbf{AB})$ | $\det(\mathbf{A})\det(\mathbf{B})$ | | Trace of product | $\text{tr}(\mathbf{AB}) = \text{tr}(\mathbf{BA})$ | | Cayley—Hamilton | $\mathbf{A}$ satisfies its own characteristic equation | | Diagonalisability | All eigenvalues must have geometric multiplicity $=$ algebraic multiplicity | | Area scaling | $| \det(\mathbf{T}) | \times$ original area $=$ image area |

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13. Further Exam-Style Questions

Question 11

Find the matrix representing a stretch of scale factor 3 parallel to the $y$-axis followed by a Reflection in the $x$-axis.

Solution

Stretch: $S = \begin{pmatrix}1&0\\0&3\end{pmatrix}$. Reflection: $R = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$.

Combined: $RS = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1&0\\0&3\end{pmatrix} = \begin{pmatrix}1&0\\0&-3\end{pmatrix}$.

$\boxed{\begin{pmatrix}1&0\\0&-3\end{pmatrix}}$

Question 12

Prove that $\det(\mathbf{A}^T) = \det(\mathbf{A})$ for any square matrix $\mathbf{A}$.

Solution

The determinant can be computed by cofactor expansion along any row or column. Expanding $\det(\mathbf{A})$ along row $i$ and $\det(\mathbf{A}^T)$ along column $i$ (which is row $i$ of $\mathbf{A}$) gives the same expression, since the cofactors are the same.

Formally: this follows from the permutation definition of the determinant. $\blacksquare$

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14. Advanced Topics

14.1 Eigenvalues and the characteristic polynomial

The characteristic polynomial of $\mathbf{A}$ is $p(\lambda) = \det(\mathbf{A}-\lambda\mathbf{I})$.

Properties:

• The sum of eigenvalues equals the trace: $\sum \lambda_i = \mathrm{tr}(\mathbf{A})$
• The product of eigenvalues equals the determinant: $\prod \lambda_i = \det(\mathbf{A})$
• $\mathbf{A}$ is invertible iff no eigenvalue is zero

14.2 Jordan normal form

Not all matrices are diagonalisable. The Jordan normal form is a generalisation where the diagonal Matrix $\mathbf{D}$ may have 1s on the superdiagonal (Jordan blocks).

For example, if $\mathbf{A}$ has a repeated eigenvalue $\lambda$ with only one eigenvector:

$\mathbf{P}^{-1}\mathbf{AP} = \begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$

14.3 Orthogonal diagonalisation

A real symmetric matrix $\mathbf{A}$ can always be diagonalised by an orthogonal matrix: $\mathbf{A} = \mathbf{Q}\mathbf{D}\mathbf{Q}^T$ where $\mathbf{Q}^T = \mathbf{Q}^{-1}$.

The Spectral Theorem states that $\mathbf{A} = \sum_{i=1}^{n} \lambda_i \mathbf{q}_i\mathbf{q}_i^T$ Where $\lambda_i$ are eigenvalues and $\mathbf{q}_i$ are orthonormal eigenvectors.

14.4 Matrix norms

The Frobenius norm: $\|\mathbf{A}\|_F = \sqrt◆LB◆\sum_{i,j} a_{ij}^2◆RB◆ = \sqrt◆LB◆\mathrm{tr}(\mathbf{A}^T\mathbf{A})◆RB◆$.

The spectral norm: $\|\mathbf{A}\|_2 = \sigma_{\max}$ (largest singular value).

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15. Further Exam-Style Questions

Question 14

Find the eigenvalues and eigenvectors of $\mathbf{A} = \begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$.

Solution

$\det(\mathbf{A}-\lambda\mathbf{I}) = (1-\lambda)^3 - 3(1-\lambda) - 2 + 3(1-\lambda) = (1-\lambda)^3 - 3(1-\lambda) + 3(1-\lambda) - 2$.

Wait, let me compute directly. $\det\begin{pmatrix}1-\lambda&1&1\\1&1-\lambda&1\\1&1&1-\lambda\end{pmatrix}$.

$= (1-\lambda)[(1-\lambda)^2-1] - 1[(1-\lambda)-1] + 1[1-(1-\lambda)]$

$= (1-\lambda)[1-2\lambda+\lambda^2-1] + \lambda + \lambda = (1-\lambda)(\lambda^2-2\lambda) + 2\lambda$

$= (1-\lambda)\lambda(\lambda-2) + 2\lambda = \lambda[(1-\lambda)(\lambda-2)+2] = \lambda[\lambda-2-\lambda^2+2\lambda+2] = \lambda[-\lambda^2+3\lambda] = \lambda^2(3-\lambda)$.

Eigenvalues: $\lambda = 0$ (double) and $\lambda = 3$.

$\lambda = 0$: $\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}\mathbf{v}=\mathbf{0} \implies v_1+v_2+v_3=0$. Two Independent eigenvectors: $(1,-1,0)$ and $(1,0,-1)$.

$\lambda = 3$: $\begin{pmatrix}-2&1&1\\1&-2&1\\1&1&-2\end{pmatrix}\mathbf{v}=\mathbf{0} \implies v_1=v_2=v_3$. Eigenvector: $(1,1,1)$.

Question 15

Prove that similar matrices have the same eigenvalues.

Solution

If $\mathbf{B} = \mathbf{P}^{-1}\mathbf{AP}$Then:

$\det(\mathbf{B}-\lambda\mathbf{I}) = \det(\mathbf{P}^{-1}\mathbf{AP}-\lambda\mathbf{I}) = \det(\mathbf{P}^{-1}(\mathbf{A}-\lambda\mathbf{I})\mathbf{P})$

$= \det(\mathbf{P}^{-1})\det(\mathbf{A}-\lambda\mathbf{I})\det(\mathbf{P}) = \det(\mathbf{A}-\lambda\mathbf{I})$.

Since the characteristic polynomials are identical, the eigenvalues are the same. $\blacksquare$

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16. Further Advanced Topics

16.1 LU decomposition

Any square matrix $\mathbf{A}$ can be decomposed as $\mathbf{A} = \mathbf{L}\mathbf{U}$ where $\mathbf{L}$ is lower triangular and $\mathbf{U}$ is upper triangular. This is used for efficient Numerical solution of systems $\mathbf{Ax} = \mathbf{b}$.

16.2 The Cayley—Hamilton theorem — applications

Since $\mathbf{A}$ satisfies $p(\mathbf{A}) = \mathbf{O}$ where $p$ is the characteristic Polynomial:

• $\mathbf{A}^{-1}$ can be computed from $\mathbf{A}^n$ terms
• $\mathbf{A}^n$ for large $n$ can be reduced using the recurrence

16.3 Singular Value Decomposition (SVD)

Any $m \times n$ matrix $\mathbf{A}$ can be written as $\mathbf{A} = \mathbf{U}\boldsymbol{\Sigma}\mathbf{V}^T$ where $\mathbf{U}$ and $\mathbf{V}$ are Orthogonal and $\boldsymbol{\Sigma}$ is diagonal with non-negative singular values.

16.4 Positive definite matrices

A symmetric matrix $\mathbf{A}$ is positive definite if $\mathbf{x}^T\mathbf{A}\mathbf{x} > 0$ for All $\mathbf{x} \neq \mathbf{0}$.

Equivalent conditions: all eigenvalues positive, all leading principal minors positive.

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17. Further Exam-Style Questions

Question 16

Find $\det(\mathbf{A})$ where $\mathbf{A} = \begin{pmatrix}1&2&3\\4&5&6\\7&8&0\end{pmatrix}$.

Solution

$\det(\mathbf{A}) = 1(0-48) - 2(0-42) + 3(32-35) = -48 + 84 - 9 = \boxed{27}$.

Question 17

Prove that the trace of a matrix equals the sum of its eigenvalues (with multiplicity).

Solution

The characteristic polynomial is $p(\lambda) = (-1)^n[\lambda^n - (\text{tr}\,\mathbf{A})\lambda^{n-1} + \cdots + (-1)^n\det(\mathbf{A})]$.

By Vieta’s formulae, the coefficient of $\lambda^{n-1}$ equals $-(\lambda_1 + \lambda_2 + \cdots + \lambda_n)$.

Therefore $\text{tr}\,\mathbf{A} = \lambda_1 + \lambda_2 + \cdots + \lambda_n$. $\blacksquare$

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