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	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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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.

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.

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 $\mathbf{A} \in M_{m \times n}(\mathbb{R})$.

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

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

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).

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)

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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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.

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

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

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

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

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 $\operatorname{adj}(\mathbf{A}) = (C_{ij})^T$.

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

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

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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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:

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

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).

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

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

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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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 $\begin{pmatrix} x' \\ y' \end{pmatrix} = \mathbf{T}\begin{pmatrix} x \\ y \end{pmatrix}$.

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 $y = \tan\theta\, x$:

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

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 $\theta$.

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

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

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

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

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 $k^2$.

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.

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

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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### 6.1 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.

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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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 $\lambda$.

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

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.

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

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

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

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

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

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

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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

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

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

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

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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

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

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

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

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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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.

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 $\mathbf{A}^2 = 4\mathbf{A} - 3\mathbf{I}$.

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

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

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

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

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: $y = -x$.

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

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.

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.

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

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

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

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

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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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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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

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

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

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

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

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 $\mathbf{v}$. $\blacksquare$

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

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

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

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

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

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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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

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:

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

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.

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

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

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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: $(1,1,1)$.

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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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 $\mathbf{Ax} = \mathbf{b}$.

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

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.

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

Solution

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

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

$$$$

• Complex Numbers — Complex eigenvalues lead to rotation-scaling transformations, connecting matrix theory to the geometry of complex arithmetic.
• Further Algebra — The Cayley-Hamilton theorem links matrices to polynomial algebra through the characteristic equation.
• Further Calculus — Matrix methods underpin systems of differential equations solved using eigenvalues and eigenvectors.
• Vectors in 3D — The cross product can be written as a matrix multiplication, connecting vector algebra to linear transformations.