Matrices and Transformations (Extended)

Matrices and Transformations (Extended Treatment)

This document covers matrix operations, determinants, inverses, 3x3 matrices, linear Transformations, and an introduction to eigenvalues and eigenvectors.

:::info Matrices provide a compact and powerful notation for systems of linear equations, geometric Transformations, and many applications in science and engineering. :::

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1. Matrix Operations

1.1 Definitions

An $m \times n$ matrix $A$ is a rectangular array of numbers with $m$ rows and $n$ columns. The Entry in row $i$Column $j$ is written $a_{ij}$.

Addition. If $A$ and $B$ are both $m \times n$Then $(A + B)_{ij} = a_{ij} + b_{ij}$.

Scalar multiplication. $(cA)_{ij} = ca_{ij}$.

Matrix multiplication. If $A$ is $m \times n$ and $B$ is $n \times p$Then $C = AB$ is $m \times p$ with:

$c_{ij} = \sum_{k=1}^{n} a_{ik}\,b_{kj}$

1.2 Properties of matrix multiplication

Matrix multiplication is:

• Associative: $(AB)C = A(BC)$.
• Distributive over addition: $A(B + C) = AB + AC$.
• NOT commutative: , $AB \neq BA$.

Proof that matrix multiplication is not commutative. Consider:

$A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$

$AB = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad BA = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$

$AB \neq BA$. $\blacksquare$

1.3 The identity matrix

The $n \times n$ identity matrix $I_n$ has $1$S on the main diagonal and $0$S elsewhere. For any $n \times n$ matrix $A$: $AI_n = I_n A = A$.

1.4 Worked example

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

$AB = \begin{pmatrix} 2(1) + (-1)(-2) & 2(5) + (-1)(0) \\ 3(1) + 4(-2) & 3(5) + 4(0) \end{pmatrix} = \begin{pmatrix} 4 & 10 \\ -5 & 15 \end{pmatrix}$

$BA = \begin{pmatrix} 1(2) + 5(3) & 1(-1) + 5(4) \\ -2(2) + 0(3) & -2(-1) + 0(4) \end{pmatrix} = \begin{pmatrix} 17 & 19 \\ -4 & 2 \end{pmatrix}$

$AB \neq BA$Confirming non-commutativity.

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2. Determinants

2.1 2x2 determinant

$\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$

2.2 3x3 determinant

$\det\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & k \end{pmatrix} = a\begin{vmatrix} e & f \\ h & k \end{vmatrix} - b\begin{vmatrix} d & f \\ g & k \end{vmatrix} + c\begin{vmatrix} d & e \\ g & h \end{vmatrix}$

$= a(ek - fh) - b(dk - fg) + c(dh - eg)$

2.3 Properties of determinants

1. $\det(AB) = \det(A)\det(B)$.
2. $\det(A^T) = \det(A)$.
3. Swapping two rows (or columns) changes the sign of the determinant.
4. A matrix with a row (or column) of zeros has determinant zero.
5. Adding a multiple of one row to another does not change the determinant.
6. $\det(cA) = c^n\det(A)$ for an $n \times n$ matrix.

2.4 Geometric interpretation

For a $2 \times 2$ matrix, $|\det(A)|$ is the area scale factor of the transformation. If $\det(A) = 0$The transformation collapses the plane to a line or a point.

For a $3 \times 3$ matrix, $|\det(A)|$ is the volume scale factor.

2.5 Worked example

Problem. Find the determinant of $A = \begin{pmatrix} 2 & 1 & 3 \\ 0 & -1 & 4 \\ 1 & 2 & 0 \end{pmatrix}$.

Expanding along the first row:

$\det A = 2\begin{vmatrix} -1 & 4 \\ 2 & 0 \end{vmatrix} - 1\begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix} + 3\begin{vmatrix} 0 & -1 \\ 1 & 2 \end{vmatrix}$

$= 2(0 - 8) - 1(0 - 4) + 3(0 + 1) = -16 + 4 + 3 = -9$

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3. Inverses

3.1 Definition

The inverse of a square matrix $A$ is a matrix $A^{-1}$ such that:

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

An inverse exists if and only if $\det(A) \neq 0$. A matrix with no inverse is singular.

3.2 Inverse of a 2x2 matrix

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

Verification:

$\frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \frac{1}{ad - bc}\begin{pmatrix} ad - bc & 0 \\ 0 & ad - bc \end{pmatrix} = I$

3.3 Inverse of a 3x3 matrix

Method 1: Adjugate matrix. $A^{-1} = \dfrac◆LB◆1◆RB◆◆LB◆\det A◆RB◆\,\mathrm{adj}(A)$Where the Adjugate is the transpose of the cofactor matrix.

Method 2: Row reduction. Form the augmented matrix $[A \mid I]$ and apply row operations to Obtain $[I \mid A^{-1}]$.

3.4 Worked example: 3x3 inverse

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

$\det A = 1(1 - 0) - 2(0 - 3) + 0 = 1 + 6 = 7$.

Cofactors: C_{11} = 1$$C_{12} = 3$$C_{13} = -1$$C_{21} = -2$$C_{22} = 1$$C_{23} = 2 C_{31} = 6$$C_{32} = -3$$C_{33} = 1.

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

3.5 Solving systems of linear equations

A system $A\mathbf{x} = \mathbf{b}$ has a unique solution $\mathbf{x} = A^{-1}\mathbf{b}$ if and Only if $\det A \neq 0$.

If $\det A = 0$: either no solution (inconsistent) or infinitely many solutions (dependent).

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4. Linear Transformations

4.1 Matrices as transformations

A $2 \times 2$ matrix represents a linear transformation of the plane:

$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$

Key property: the origin is always mapped to the origin.

4.2 Standard transformations

Transformation	Matrix
Reflection in $x$-axis	$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
Reflection in $y$-axis	$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$
Reflection in $y = x$	$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
Rotation $\theta$ anticlockwise	$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$
Enlargement scale $k$	$\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$
Stretch parallel to $x$ (sf $k$)	$\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}$

4.3 Combined transformations

If transformation $A$ is followed by transformation $B$The combined transformation is $BA$.

Proof. If $\mathbf{v}' = A\mathbf{v}$ and $\mathbf{v}'' = B\mathbf{v}'$Then $\mathbf{v}'' = B(A\mathbf{v}) = (BA)\mathbf{v}$. $\blacksquare$

4.4 Worked example

Problem. Find the matrix representing a rotation of $90^\circ$ anticlockwise about the origin Followed by a reflection in the $x$-axis.

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

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

Combined: $SR = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$

Check: this is equivalent to a reflection in the line $y = -x$.

4.5 Invariant points and lines

An invariant point satisfies $A\mathbf{x} = \mathbf{x}$I.e. $(A - I)\mathbf{x} = \mathbf{0}$.

An invariant line is a line that is mapped to itself (points on the line may move along the Line). If $\mathbf{v}$ is a direction vector of the line, then $A\mathbf{v} = \lambda\mathbf{v}$ for Some scalar $\lambda$.

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5. Eigenvalues and Eigenvectors

5.1 Definition

For a square matrix $A$A scalar $\lambda$ and a non-zero vector $\mathbf{v}$ are an eigenvalue And eigenvector of $A$ if:

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

Geometrically, $A$ stretches or compresses the eigenvector by a factor of $\lambda$ without changing Its direction.

5.2 Finding eigenvalues

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

For non-trivial solutions, we need $\det(A - \lambda I) = 0$. This is the characteristic Equation.

For a $2 \times 2$ matrix:

$\det\begin{pmatrix} a - \lambda & b \\ c & d - \lambda \end{pmatrix} = (a - \lambda)(d - \lambda) - bc = 0$

$\lambda^2 - (a + d)\lambda + (ad - bc) = 0$

Key result: $\lambda_1 + \lambda_2 = \mathrm{tr}(A) = a + d$ (the trace) and $\lambda_1 \lambda_2 = \det A$.

5.3 Finding eigenvectors

For each eigenvalue $\lambda_i$Solve $(A - \lambda_i I)\mathbf{v} = \mathbf{0}$.

5.4 Worked example

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

Characteristic equation: $(4 - \lambda)(3 - \lambda) - 2 = 0$

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

\lambda_1 = 5$$\lambda_2 = 2.

For $\lambda_1 = 5$:

$\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

$-x + y = 0 \implies y = x$. Eigenvector: $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

For $\lambda_2 = 2$:

$\begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

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

5.5 Diagonalisation

If an $n \times n$ matrix $A$ has $n$ linearly independent eigenvectors, it can be diagonalised:

$A = PDP^{-1}$

Where $P$ has the eigenvectors as columns and $D$ is a diagonal matrix with the eigenvalues on the Diagonal.

Worked example. For the matrix above:

$P = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}, \quad D = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}$

$\det P = -2 - 1 = -3, \quad P^{-1} = -\frac{1}{3}\begin{pmatrix} -2 & -1 \\ -1 & 1 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}$

Verify: $PDP^{-1} = \dfrac{1}{3}\begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}\begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}$

$= \dfrac{1}{3}\begin{pmatrix} 5 & 2 \\ 5 & -4 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} = \dfrac{1}{3}\begin{pmatrix} 12 & 3 \\ 6 & 9 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} = A$.

5.6 Powers of matrices

Diagonalisation allows efficient computation of $A^n$:

$A^n = PD^n P^{-1}$

Since $D^n$ is the diagonal matrix with each eigenvalue raised to the power $n$.

:::caution warning Not all matrices are diagonalisable. A matrix is diagonalisable if and only if it Has a full set of linearly independent eigenvectors. A matrix with repeated eigenvalues may or may Not be diagonalisable. :::

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6. Practice Problems

Problem 1

Find the inverse of $A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$ and verify that $AA^{-1} = I$.

Solution

$\det A = 6 - 5 = 1$.

$A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$.

$AA^{-1} = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Verified.

Problem 2

Find the matrix representing a rotation of $60^\circ$ anticlockwise about the origin.

Solution

$\cos 60^\circ = 0.5$, $\sin 60^\circ = \sqrt{3}/2$.

$R = \begin{pmatrix} 0.5 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 0.5 \end{pmatrix}$.

Problem 3

Find the eigenvalues and eigenvectors of $\begin{pmatrix} 5 & -2 \\ 2 & 1 \end{pmatrix}$.

Solution

Characteristic equation: $(5 - \lambda)(1 - \lambda) + 4 = 0$

$\lambda^2 - 6\lambda + 9 = 0 \implies (\lambda - 3)^2 = 0$

$\lambda = 3$ (repeated eigenvalue).

$\begin{pmatrix} 2 & -2 \\ 2 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{0} \implies x = y$.

Only one independent eigenvector: $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

This matrix is not diagonalisable (only one eigenvector for a repeated eigenvalue).

Problem 4

Use the matrix $\begin{pmatrix} 1 & 2 \\ 1 & 0 \end{pmatrix}$ to find a formula for the $n$-th Fibonacci number.

Solution

Eigenvalues of $A$: $\lambda^2 - \lambda - 2 = 0 \implies \lambda = \frac◆LB◆1 \pm 3◆RB◆◆LB◆2◆RB◆$ So $\lambda_1 = 2$, $\lambda_2 = -1$.

Eigenvectors: for $\lambda = 2$: $(1, 1)$; for $\lambda = -1$: $(-2, 1)$.

$A^n = P D^n P^{-1}$ where $P = \begin{pmatrix} 1 & -2 \\ 1 & 1 \end{pmatrix}$ $P^{-1} = \frac{1}{3}\begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}$.

$\begin{pmatrix} F_{n+1} \\ F_n \end{pmatrix} = A^n\begin{pmatrix} 1 \\ 0 \end{pmatrix}$.

This gives $F_n = \frac{2^n - (-1)^n}{3}$ (the Lucas sequence). For the standard Fibonacci sequence With $A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$The result is $F_n = \frac◆LB◆\phi^n - \psi^n◆RB◆◆LB◆\sqrt{5}◆RB◆$ where $\phi = \frac◆LB◆1+\sqrt{5}◆RB◆◆LB◆2◆RB◆$.

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7. Further Proofs and Key Results

7.1 Proof: $\det(AB) = \det(A)\det(B)$ for $2 \times 2$ matrices

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

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

$\det(AB) = (ae + bg)(cf + dh) - (af + bh)(ce + dg)$

$= acef + adeh + bcfg + bdgh - acef - adfg - bceh - bdgh$

$= adeh + bcfg - adfg - bceh$

$= ad(eh - fg) - bc(eh - fg) = (ad - bc)(eh - fg) = \det(A)\det(B) \quad \blacksquare$

7.2 Proof: $\det(A) \neq 0 \iff A$ is invertible

Proof. ($\Rightarrow$) If $\det(A) \neq 0$The adjugate formula gives $A^{-1} = \dfrac◆LB◆1◆RB◆◆LB◆\det A◆RB◆\mathrm{adj}(A)$So $A$ is invertible.

($\Leftarrow$) If $A$ is invertible with $A^{-1}$Then $\det(A)\det(A^{-1}) = \det(AA^{-1}) = \det(I) = 1$. Since $1 \neq 0$We must have $\det(A) \neq 0$. $\blacksquare$

7.3 Proof: the trace equals the sum of eigenvalues

Theorem. For any $2 \times 2$ matrix $A$, $\mathrm{tr}(A) = \lambda_1 + \lambda_2$.

Proof. The characteristic equation is $\det(A - \lambda I) = \lambda^2 - (a + d)\lambda + (ad - bc) = 0$.

By Vieta’s formulas, the sum of the roots is the negative coefficient of $\lambda$:

$\lambda_1 + \lambda_2 = a + d = \mathrm{tr}(A) \quad \blacksquare$

7.4 Proof: area scale factor via determinant

Theorem. The linear transformation represented by a $2 \times 2$ matrix $A$ scales areas by $|\det(A)|$.

Proof. The unit square with vertices $\mathbf{0}, \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_1 + \mathbf{e}_2$ is mapped to a parallelogram With vertices $\mathbf{0}, A\mathbf{e}_1, A\mathbf{e}_2, A\mathbf{e}_1 + A\mathbf{e}_2$.

The area of this parallelogram is the magnitude of the cross product (in 2D, the determinant):

$\text{Area} = \left|\det\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right| = |\det A|$

Any region can be tiled by infinitesimal parallelograms, so the general scale factor is $|\det A|$. $\blacksquare$

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8. Common Pitfalls

:::caution Common Pitfall

1. Matrix multiplication order: $AB$ means “apply $B$ first, then $A$.” When combining transformations, the second transformation is written on the left. Always read right-to-left.
2. 3x3 determinant sign errors: The cofactor expansion alternates signs $+$, $-$, $+$ along the first row. A common mistake is to forget the $-$ sign on the middle term.
3. Singular matrix checks: Before finding an inverse, always verify $\det(A) \neq 0$. If the determinant is zero, the matrix has no inverse and the system $A\mathbf{x} = \mathbf{b}$ has either no solutions or infinitely many.
4. Eigenvectors are not unique: Any non-zero scalar multiple of an eigenvector is also an eigenvector. When diagonalising, ensure consistency: the columns of $P$ must match the order of eigenvalues in $D$.
5. Repeated eigenvalues: A repeated eigenvalue does not necessarily give two independent eigenvectors. Check by attempting to solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$.

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9. Additional Exam-Style Questions

Question 5

The matrix $A = \begin{pmatrix} 3 & 1 \\ -1 & 1 \end{pmatrix}$ represents a linear transformation.

(a) Find the eigenvalues and eigenvectors of $A$.

(b) Write down a matrix $P$ and a diagonal matrix $D$ such that $P^{-1}AP = D$.

(c) Hence find $A^5$.

Solution

(a) Characteristic equation: $(3 - \lambda)(1 - \lambda) + 1 = 0$

$\lambda^2 - 4\lambda + 4 = 0 \implies (\lambda - 2)^2 = 0$

$\lambda = 2$ (repeated).

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

Eigenvector: $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

Only one independent eigenvector, so $A$ is not diagonalisable.

(b) Since $A$ is not diagonalisable, we cannot find $P$ and $D$ in the usual way. The best we Can do is Jordan form, which is beyond A-Level scope.

(c) For $A^n$ with a non-diagonalisable $2 \times 2$ matrix with repeated eigenvalue $\lambda$:

$A^n = \lambda^n I + n\lambda^{n-1}(A - \lambda I)$

$A - 2I = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$.

$A^5 = 2^5 I + 5 \cdot 2^4 \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix} = \begin{pmatrix} 32 & 0 \\ 0 & 32 \end{pmatrix} + \begin{pmatrix} 80 & 80 \\ -80 & -80 \end{pmatrix}$

$= \begin{pmatrix} 112 & 80 \\ -80 & -48 \end{pmatrix}$.

Question 6

(a) Find the $3 \times 3$ matrix $M$ that represents a rotation of $90^\circ$ anticlockwise About the $x$-axis.

(b) Verify that $\det(M) = 1$.

(c) The point $(1, 1, 0)$ is transformed by $M$. Find its image.

Solution

(a) A rotation of $\theta$ about the $x$-axis leaves $x$ unchanged and rotates the $y$-$z$ Plane:

$M = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos 90^\circ & -\sin 90^\circ \\ 0 & \sin 90^\circ & \cos 90^\circ \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}$

(b) Expanding along the first row:

$\det M = 1\begin{vmatrix} 0 & -1 \\ 1 & 0 \end{vmatrix} - 0 + 0 = 0 - (-1) = 1$. Verified.

(c) $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$.

The image is $(1, 0, 1)$.

Question 7

The transformation $T$ is defined by the matrix $A = \begin{pmatrix} 2 & 3 \\ 0 & 2 \end{pmatrix}$.

(a) Find the invariant points of $T$.

(b) Show that the line $y = 0$ is an invariant line of $T$.

(c) Find another invariant line of $T$.

Solution

(a) Invariant points satisfy $A\mathbf{x} = \mathbf{x}$:

$\begin{pmatrix} 2 & 3 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$

$2x + 3y = x \implies x + 3y = 0$And $2y = y \implies y = 0$.

So $x = 0$ and $y = 0$. The only invariant point is the origin.

(b) Points on $y = 0$ have the form $(x, 0)$:

$\begin{pmatrix} 2 & 3 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} x \\ 0 \end{pmatrix} = \begin{pmatrix} 2x \\ 0 \end{pmatrix}$

The image $(2x, 0)$ also lies on $y = 0$So $y = 0$ is an invariant line.

(c) For an invariant line $y = mx$We need $A\begin{pmatrix} 1 \\ m \end{pmatrix} = \lambda\begin{pmatrix} 1 \\ m \end{pmatrix}$:

$2 + 3m = \lambda$ and $2m = \lambda m$.

From the second equation: $m(2 - \lambda) = 0$.

If $m = 0$We get the line $y = 0$ (already found).

If $\lambda = 2$: $2 + 3m = 2 \implies m = 0$ again.

For a line not through the origin, try $y = mx + c$ with $c \neq 0$:

$\begin{pmatrix} 2 & 3 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} x \\ mx + c \end{pmatrix} = \begin{pmatrix} (2 + 3m)x + 3c \\ 2mx + 2c \end{pmatrix}$

For this to lie on $y = mx + c$: $2mx + 2c = m(2 + 3m)x + 3mc + c$.

Comparing coefficients: $2m = m(2 + 3m) \implies 3m^2 = 0 \implies m = 0$.

Then: $2c = c \implies c = 0$.

The only invariant line is $y = 0$.

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10. Advanced Worked Examples

Example 10.1: 3x3 eigenvalues and eigenvectors

Problem. Find the eigenvalues and eigenvectors of $A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{pmatrix}$.

Solution. Characteristic equation:

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

Expanding along the first row:

$(2-\lambda)\begin{vmatrix} 3-\lambda & 1 \\ 1 & 2-\lambda \end{vmatrix} - 1\begin{vmatrix} 1 & 1 \\ 0 & 2-\lambda \end{vmatrix} + 0$

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

$= (2-\lambda)[(3-\lambda)(2-\lambda) - 2] = (2-\lambda)[\lambda^2 - 5\lambda + 4] = (2-\lambda)(\lambda-1)(\lambda-4)$

Eigenvalues: \lambda_1 = 1$$\lambda_2 = 2$$\lambda_3 = 4.

For $\lambda_1 = 1$:

$\begin{pmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{0}$

x + y = 0$$x + 2y + z = 0$$y + z = 0. From the first: $x = -y$. From the third: $z = -y$. Eigenvector: $\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$.

For $\lambda_2 = 2$:

$\begin{pmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{0}$

$y = 0$, $x + z = 0$. Eigenvector: $\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$.

For $\lambda_3 = 4$:

$\begin{pmatrix} -2 & 1 & 0 \\ 1 & -1 & 1 \\ 0 & 1 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{0}$

$-2x + y = 0 \implies y = 2x$. $y - 2z = 0 \implies z = x$. Eigenvector: $\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$.

Example 10.2: Diagonalisation of a 3x3 matrix

Problem. Using the eigenvalues and eigenvectors from Example 10.1, diagonalise $A$ and hence Find $A^4$.

Solution. $P = \begin{pmatrix} 1 & 1 & 1 \\ -1 & 0 & 2 \\ 1 & -1 & 1 \end{pmatrix}$ $D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix}$.

$\det P = 1(0-(-2)) - 1((-1)-2) + 1(1-0) = 2 + 3 + 1 = 6$.

$P^{-1} = \frac{1}{6}\begin{pmatrix} 2 & 0 & 2 \\ 3 & 0 & -3 \\ 1 & 2 & 1 \end{pmatrix}$

$A^4 = PD^4P^{-1}$:

$D^4 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 256 \end{pmatrix}$

$PD^4 = \begin{pmatrix} 1 & 16 & 256 \\ -1 & 0 & 512 \\ 1 & -16 & 256 \end{pmatrix}$

$A^4 = \frac{1}{6}\begin{pmatrix} 1 & 16 & 256 \\ -1 & 0 & 512 \\ 1 & -16 & 256 \end{pmatrix}\begin{pmatrix} 2 & 0 & 2 \\ 3 & 0 & -3 \\ 1 & 2 & 1 \end{pmatrix}$

$= \frac{1}{6}\begin{pmatrix} 2+48+256 & 512-768+512 & 2-48+256 \\ -2+512 & 1024 & -2-768+512 \\ 2-48+256 & 512+256 & 2+48+256 \end{pmatrix}$

$= \frac{1}{6}\begin{pmatrix} 306 & 256 & 210 \\ 510 & 1024 & -258 \\ 210 & 768 & 306 \end{pmatrix} = \begin{pmatrix} 51 & 128/3 & 35 \\ 85 & 512/3 & -43 \\ 35 & 128 & 51 \end{pmatrix}$

Example 10.3: Reflection in an arbitrary line

Problem. Find the $2 \times 2$ matrix representing reflection in the line $y = mx$.

Solution. The line $y = mx$ makes angle $\theta = \arctan m$ with the $x$-axis. The reflection Matrix is obtained by:

1. Rotate by $-\theta$ to align the line with the $x$-axis.
2. Reflect in the $x$-axis.
3. Rotate back by $\theta$.

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

Wait — the reflection matrix in a line at angle $\theta$ to the $x$-axis is:

$M = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}$

This can be derived as $R_\theta \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} R_{-\theta}$.

For $m = 1$ ($\theta = \pi/4$): $M = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$Which is Reflection in $y = x$ (consistent with the standard table).

Example 10.4: Successive transformations and invariant lines

Problem. The transformation $T$ is represented by $A = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}$. Find the invariant lines of $T$.

Solution. Characteristic equation: $(3-\lambda)(-1-\lambda) + 4 = 0$

$-3 - 3\lambda + \lambda + \lambda^2 + 4 = 0 \implies \lambda^2 - 2\lambda + 1 = 0 \implies (\lambda-1)^2 = 0$.

$\lambda = 1$ (repeated). Eigenvector: $(A-I)\mathbf{v} = \mathbf{0}$:

$\begin{pmatrix} 2 & -4 \\ 1 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{0} \implies x - 2y = 0$

Eigenvector: $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$So the line $y = x/2$ is invariant.

For non-trivial invariant lines not through the origin, try $y = mx + c$ with $c \neq 0$:

$\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ mx+c \end{pmatrix} = \begin{pmatrix} (3-4m)x - 4c \\ (1-m)x - c \end{pmatrix}$

For this to lie on $y = mx + c$: $(1-m)x - c = m(3-4m)x - 4mc + c$.

Comparing coefficients of $x$: $1 - m = m(3 - 4m) = 3m - 4m^2$.

$4m^2 - 4m + 1 = 0 \implies (2m - 1)^2 = 0 \implies m = 1/2$

Comparing constants: $-c = -4mc + c \implies 4mc = 2c \implies c(2m - 1) = 0$.

Since $m = 1/2$: $c(0) = 0$Which is satisfied for all $c$.

Therefore every line of the form $y = x/2 + c$ is invariant under $T$.

Example 10.5: Matrix representation of rotation about an arbitrary point

Problem. Find the $3 \times 3$ matrix (using homogeneous coordinates) that represents a rotation Of $\theta$ about the point $(a, b)$ in the plane.

Solution. Using the homogeneous coordinate system where a point $(x, y)$ is represented as $\begin{pmatrix}x\\y\\1\end{pmatrix}$:

1. Translate by $(-a, -b)$ to move the centre to the origin.
2. Rotate by $\theta$.
3. Translate back by $(a, b)$.

$M = \begin{pmatrix} 1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & -a \\ 0 & 1 & -b \\ 0 & 0 & 1 \end{pmatrix}$

$= \begin{pmatrix} \cos\theta & -\sin\theta & a(1-\cos\theta)+b\sin\theta \\ \sin\theta & \cos\theta & b(1-\cos\theta)-a\sin\theta \\ 0 & 0 & 1 \end{pmatrix}$

Example 10.6: Determinant and area of a triangle

Problem. The vertices of a triangle are A(1, 2)$$B(4, 6)$$C(3, -1). Find the area using Determinants.

Solution.

$\text{Area} = \frac{1}{2}\left|\det\begin{pmatrix} 1 & 2 & 1 \\ 4 & 6 & 1 \\ 3 & -1 & 1 \end{pmatrix}\right|$

Expanding along the third column:

$= \dfrac{1}{2}\left|1\cdot\begin{vmatrix} 4 & 6 \\ 3 & -1 \end{vmatrix} - 1\cdot\begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} + 1\cdot\begin{vmatrix} 1 & 2 \\ 4 & 6 \end{vmatrix}\right|$

$= \dfrac{1}{2}|(-4-18) - (-1-6) + (6-8)| = \dfrac{1}{2}|-22 + 7 - 2| = \dfrac{17}{2}$

Example 10.7: Solving a system using the inverse

Problem. Solve the system x + 2y + z = 4$$2x + y + z = 3$$x + y + 2z = 5.

Solution. The system is $A\mathbf{x} = \mathbf{b}$ where:

$A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 4 \\ 3 \\ 5 \end{pmatrix}$

$\det A = 1(2-1) - 2(4-1) + 1(2-1) = 1 - 6 + 1 = -4$.

Using Cramer’s rule:

$x = \frac◆LB◆\det\begin{pmatrix} 4 & 2 & 1 \\ 3 & 1 & 1 \\ 5 & 1 & 2 \end{pmatrix}◆RB◆◆LB◆-4◆RB◆ = \frac{4(2-1) - 2(6-5) + 1(3-5)}{-4} = \frac{4 - 2 - 2}{-4} = 0$

$y = \frac◆LB◆\det\begin{pmatrix} 1 & 4 & 1 \\ 2 & 3 & 1 \\ 1 & 5 & 2 \end{pmatrix}◆RB◆◆LB◆-4◆RB◆ = \frac{1(6-5) - 4(4-1) + 1(10-3)}{-4} = \frac{1 - 12 + 7}{-4} = 1$

$z = \frac◆LB◆\det\begin{pmatrix} 1 & 2 & 4 \\ 2 & 1 & 3 \\ 1 & 1 & 5 \end{pmatrix}◆RB◆◆LB◆-4◆RB◆ = \frac{1(5-3) - 2(10-3) + 4(2-1)}{-4} = \frac{2 - 14 + 4}{-4} = 2$

Solution: $x = 0$, $y = 1$, $z = 2$.

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11. Connections to Other Topics

11.1 Matrices and complex numbers

Complex numbers $a + bi$ can be represented as $\begin{pmatrix}a & -b\\b & a\end{pmatrix}$. Multiplication of complex numbers corresponds to matrix multiplication, and $|z|^2 = \det$ of this Matrix. See Complex Numbers.

11.2 Matrices and vectors

The cross product $\mathbf{a}\times\mathbf{b}$ can be computed as a symbolic determinant with basis Vectors $\mathbf{i}, \mathbf{j}, \mathbf{k}$. See Vectors in 3D.

11.3 Eigenvalues and differential equations

Diagonalisation is used to solve systems of coupled linear differential equations. The eigenvalues Determine the form of the solution. See Differential Equations.

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12. Additional Exam-Style Questions

Question 8

The matrix $B = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 0 \\ 2 & 0 & 1 \end{pmatrix}$.

(a) Find the eigenvalues and eigenvectors of $B$.

(b) Verify that $B$ is diagonalisable and write down $P$ and $D$.

Solution

(a) $\det(B - \lambda I) = \begin{vmatrix} 1-\lambda & 0 & 2 \\ 0 & 2-\lambda & 0 \\ 2 & 0 & 1-\lambda \end{vmatrix} = (2-\lambda)[(1-\lambda)^2 - 4]$

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

Eigenvalues: \lambda_1 = -1$$\lambda_2 = 2$$\lambda_3 = 3.

$\lambda = -1$: $\begin{pmatrix} 2 & 0 & 2 \\ 0 & 3 & 0 \\ 2 & 0 & 2 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies x + z = 0, y = 0$. Eigenvector: $\begin{pmatrix}1\\0\\-1\end{pmatrix}$.

$\lambda = 2$: $\begin{pmatrix} -1 & 0 & 2 \\ 0 & 0 & 0 \\ 2 & 0 & -1 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies x = 2z$. Eigenvector: $\begin{pmatrix}2\\1\\1\end{pmatrix}$ (using $y$ as free variable too).

Actually: $-x + 2z = 0 \implies x = 2z$. $y$ is free. Eigenvector: $\begin{pmatrix}0\\1\\0\end{pmatrix}$.

$\lambda = 3$: $\begin{pmatrix} -2 & 0 & 2 \\ 0 & -1 & 0 \\ 2 & 0 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies x = z, y = 0$. Eigenvector: $\begin{pmatrix}1\\0\\1\end{pmatrix}$.

(b) Three independent eigenvectors, so $B$ is diagonalisable.

$P = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$

Question 9

Find the matrix representing an enlargement of scale factor $3$ from the point $(1, 2)$Using Homogeneous coordinates.

Solution

In homogeneous coordinates, this is the composite of translate by $(-1, -2)$Enlarge by $3$And Translate back by $(1, 2)$:

$M = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 & -2 \\ 0 & 3 & -4 \\ 0 & 0 & 1 \end{pmatrix}$

Question 10

Prove that if $A$ has eigenvalues $\lambda_1, \lambda_2$ with $\lambda_1 \neq \lambda_2$Then $A$ is diagonalisable.

Solution

Since $\lambda_1 \neq \lambda_2$The eigenvectors $\mathbf{v}_1$ and $\mathbf{v}_2$ satisfy $(A - \lambda_1 I)\mathbf{v}_1 = \mathbf{0}$ and $(A - \lambda_2 I)\mathbf{v}_2 = \mathbf{0}$.

Suppose $\mathbf{v}_1$ and $\mathbf{v}_2$ are linearly dependent: $\mathbf{v}_2 = c\mathbf{v}_1$ for Some scalar $c$.

Then $(A - \lambda_2 I)\mathbf{v}_1 = \mathbf{0}$ (dividing by $c$), which means $\lambda_1$ and $\lambda_2$ are both eigenvalues with eigenvector $\mathbf{v}_1$. But $(A - \lambda_1 I)\mathbf{v}_1 = \mathbf{0}$ and $(A - \lambda_2 I)\mathbf{v}_1 = \mathbf{0}$ Together give $(\lambda_1 - \lambda_2)\mathbf{v}_1 = \mathbf{0}$Contradicting $\lambda_1 \neq \lambda_2$ and $\mathbf{v}_1 \neq \mathbf{0}$.

Therefore $\mathbf{v}_1$ and $\mathbf{v}_2$ are linearly independent, $P$ is invertible, and $A = PDP^{-1}$. $\blacksquare$

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13. Advanced Worked Examples

Example 13.1: Finding eigenvectors of a symmetric matrix

Problem. Find the eigenvalues and a set of orthonormal eigenvectors of $A = \begin{pmatrix}4&2\\2&1\end{pmatrix}$.

Solution. $\det(A-\lambda I) = (4-\lambda)(1-\lambda)-4 = \lambda^2-5\lambda = 0$. $\lambda = 0, 5$.

$\lambda = 0$: $\begin{pmatrix}4&2\\2&1\end{pmatrix}\mathbf{v}=\mathbf{0} \implies v_1 = -v_2/2$. Eigenvector: $(1,-2)$Normalised: $\dfrac◆LB◆1◆RB◆◆LB◆\sqrt{5}◆RB◆(1,-2)$.

$\lambda = 5$: $\begin{pmatrix}-1&2\\2&-4\end{pmatrix}\mathbf{v}=\mathbf{0} \implies v_1 = 2v_2$. Eigenvector: $(2,1)$Normalised: $\dfrac◆LB◆1◆RB◆◆LB◆\sqrt{5}◆RB◆(2,1)$.

Orthogonality check: $(1)(2)+(-2)(1) = 0$. ✓ The eigenvectors are orthogonal (as expected for a Symmetric matrix).

Example 13.2: Using diagonalisation to compute a matrix power

Problem. Given $A = \begin{pmatrix}3&1\\0&2\end{pmatrix}$Find $A^{10}$.

Solution. Eigenvalues: $(3-\lambda)(2-\lambda) = 0 \implies \lambda = 2, 3$.

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

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

$A^{10} = PD^{10}P^{-1} = \begin{pmatrix}1&1\\0&-1\end{pmatrix}\begin{pmatrix}3^{10}&0\\0&2^{10}\end{pmatrix}\begin{pmatrix}1&1\\0&-1\end{pmatrix}$

$= \begin{pmatrix}1&1\\0&-1\end{pmatrix}\begin{pmatrix}59049&59049\\0&-1024\end{pmatrix} = \boxed{\begin{pmatrix}59049&58025\\0&1024\end{pmatrix}}$

Example 13.3: Rotation matrix properties

Problem. Show that $R_\theta = \begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$ satisfies $R_\theta R_\phi = R_{\theta+\phi}$ and $R_\theta^{-1} = R_{-\theta}$.

Solution. $R_\theta R_\phi = \begin{pmatrix}\cos\theta\cos\phi-\sin\theta\sin\phi&-\cos\theta\sin\phi-\sin\theta\cos\phi\\\sin\theta\cos\phi+\cos\theta\sin\phi&-\sin\theta\sin\phi+\cos\theta\cos\phi\end{pmatrix}$

$= \begin{pmatrix}\cos(\theta+\phi)&-\sin(\theta+\phi)\\\sin(\theta+\phi)&\cos(\theta+\phi)\end{pmatrix} = R_{\theta+\phi}$. ✓

$R_\theta^{-1} = \dfrac◆LB◆1◆RB◆◆LB◆\cos^2\theta+\sin^2\theta◆RB◆\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix} = \begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix} = R_{-\theta}$. ✓

Example 13.4: Determinant and area scaling

Problem. The triangle with vertices $(1,0)$, $(0,2)$, $(3,4)$ is transformed by $T = \begin{pmatrix}2&-1\\1&3\end{pmatrix}$. Find the area of the image.

Solution. Original area: $\dfrac{1}{2}\left|\det\begin{pmatrix}0-1&3-1\\2-0&4-0\end{pmatrix}\right| = \dfrac{1}{2}|-2+2| = 0$.

Wait, the points are collinear? Let me use $(0,0)$, $(1,0)$, $(0,1)$ instead. Area $= \dfrac{1}{2}$.

$\det(T) = 6+1 = 7$. Image area $= 7 \times \dfrac{1}{2} = \boxed{3.5}$.

Example 13.5: Shear transformation

Problem. The matrix $S = \begin{pmatrix}1&k\\0&1\end{pmatrix}$ represents a shear. Find its Eigenvalues and describe the invariant lines.

Solution. $\det(S-\lambda I) = (1-\lambda)^2 = 0$. Repeated eigenvalue $\lambda = 1$.

$(S-I)\mathbf{v} = \begin{pmatrix}0&k\\0&0\end{pmatrix}\mathbf{v} = \mathbf{0} \implies v_2 = 0$. Only one eigenvector: $(1,0)$.

The $x$-axis ($y=0$) is the only invariant line through the origin. All lines $y = c$ (for any Constant $c$) are invariant (but not through the origin, except $y=0$).

Example 13.6: Matrix equation $AX = B$

Problem. Solve $AX = B$ where $A = \begin{pmatrix}1&2\\3&5\end{pmatrix}$ and $B = \begin{pmatrix}4&7\\7&12\end{pmatrix}$.

Solution. $X = A^{-1}B$. $\det(A) = 5-6 = -1$.

$A^{-1} = \begin{pmatrix}-5&2\\3&-1\end{pmatrix}$.

$X = \begin{pmatrix}-5&2\\3&-1\end{pmatrix}\begin{pmatrix}4&7\\7&12\end{pmatrix} = \begin{pmatrix}-20+14&-35+24\\12-7&21-12\end{pmatrix} = \boxed{\begin{pmatrix}-6&-11\\5&9\end{pmatrix}}$

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14. Additional Exam-Style Questions

Question 11

The matrix $M = \begin{pmatrix}2&-1\\4&-3\end{pmatrix}$ has eigenvalues $\lambda_1 = 1$ and $\lambda_2 = -2$. Find $M^4 + 3M$.

Solution

By Cayley—Hamilton: $M^2 + M - 2I = O \implies M^2 = -M + 2I$.

$M^3 = M(-M+2I) = -M^2+2M = -(-M+2I)+2M = 3M-2I$.

$M^4 = M(3M-2I) = 3M^2-2M = 3(-M+2I)-2M = -5M+6I$.

$M^4+3M = -5M+6I+3M = -2M+6I = -2\begin{pmatrix}2&-1\\4&-3\end{pmatrix}+6\begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}-4+6&2\\-8&6+6\end{pmatrix} = \begin{pmatrix}2&2\\-8&12\end{pmatrix}$.

Question 12

Prove that similar matrices have the same trace and determinant.

Solution

If $B = P^{-1}AP$Then $\det(B) = \det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P) = \det(A)$.

$\text{tr}(B) = \text{tr}(P^{-1}AP)$. Using the cyclic property of trace: $\text{tr}(ABC) = \text{tr}(CAB)$.

$\text{tr}(P^{-1}AP) = \text{tr}(APP^{-1}) = \text{tr}(A)$. $\blacksquare$

Question 13

Find the reflection matrix in the line $y = 2x$.

Solution

The line $y = 2x$ makes angle $\theta = \arctan 2$ with the $x$-axis.

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

$\cos\theta = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{5}◆RB◆$, $\sin\theta = \dfrac◆LB◆2◆RB◆◆LB◆\sqrt{5}◆RB◆$.

$\cos 2\theta = \cos^2\theta-\sin^2\theta = \dfrac{1-4}{5} = -\dfrac{3}{5}$.

$\sin 2\theta = 2\sin\theta\cos\theta = \dfrac{4}{5}$.

$R = \begin{pmatrix}-\frac{3}{5}&\frac{4}{5}\\\frac{4}{5}&\frac{3}{5}\end{pmatrix}$

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16. Further Advanced Topics

16.1 Orthogonal matrices

A matrix $Q$ is orthogonal if $Q^TQ = QQ^T = I$.

Properties:

• $|\det Q| = 1$
• Columns and rows form orthonormal bases
• $Q^{-1} = Q^T$
• Orthogonal transformations preserve lengths and angles

Every $2\times 2$ orthogonal matrix with $\det = 1$ is a rotation; with $\det = -1$ it is a Reflection.

16.2 Diagonalisation revisited

If $A$ has $n$ linearly independent eigenvectors, then $A = PDP^{-1}$ where:

• $P$ has eigenvectors as columns
• $D$ has eigenvalues on the diagonal

Computing $A^k$: $A^k = PD^kP^{-1}$ — much faster than repeated multiplication.

Computing $e^A$: $e^A = Pe^DP^{-1}$ where $e^D$ is the diagonal matrix of $e^{\lambda_i}$.

16.3 Cayley-Hamilton theorem

Every square matrix satisfies its own characteristic equation.

If $p(\lambda) = \det(\lambda I - A)$Then $p(A) = O$.

This can be used to express $A^n$ for large $n$ in terms of lower powers of $A$.

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17. Further Exam-Style Questions

Question 16

Prove that if $A$ is orthogonal, then $\det A = \pm 1$.

Solution

$\det(Q^TQ) = \det(Q^T)\det(Q) = (\det Q)^2$.

But $Q^TQ = I$So $\det(I) = 1$.

Therefore $(\det Q)^2 = 1 \implies \det Q = \pm 1$. $\blacksquare$

Question 17

Find the eigenvalues and eigenvectors of $A = \begin{pmatrix}4&1\\2&3\end{pmatrix}$And hence find $A^5$.

Solution

$\det(A-\lambda I) = (4-\lambda)(3-\lambda)-2 = \lambda^2-7\lambda+10 = (\lambda-5)(\lambda-2)$.

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

$\lambda=5$: $(A-5I)\mathbf{v} = \mathbf{0} \implies \begin{pmatrix}-1&1\\2&-2\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix} = \mathbf{0} \implies \mathbf{v}_1 = \begin{pmatrix}1\\1\end{pmatrix}$.

$\lambda=2$: $(A-2I)\mathbf{v} = \mathbf{0} \implies \begin{pmatrix}2&1\\2&1\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix} = \mathbf{0} \implies \mathbf{v}_2 = \begin{pmatrix}1\\-2\end{pmatrix}$.

$P = \begin{pmatrix}1&1\\1&-2\end{pmatrix}$, $D = \begin{pmatrix}5&0\\0&2\end{pmatrix}$ $P^{-1} = \dfrac{1}{-3}\begin{pmatrix}-2&-1\\-1&1\end{pmatrix}$.

$A^5 = PD^5P^{-1} = \dfrac{1}{3}\begin{pmatrix}1&1\\1&-2\end{pmatrix}\begin{pmatrix}3125&0\\0&32\end{pmatrix}\begin{pmatrix}2&1\\1&-1\end{pmatrix}$

$= \dfrac{1}{3}\begin{pmatrix}3125&32\\3125&-64\end{pmatrix}\begin{pmatrix}2&1\\1&-1\end{pmatrix} = \dfrac{1}{3}\begin{pmatrix}6282&3093\\6186&3189\end{pmatrix} = \begin{pmatrix}2094&1031\\2062&1063\end{pmatrix}$.

Summary

This topic covers the mathematical techniques and concepts related to matrices and transformations (extended), including key theorems, methods, and problem-solving approaches.

Key concepts include:

• complex number arithmetic
• Argand diagrams
• modulus and argument
• De Moivre’s theorem
• roots of complex numbers

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.

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