Vectors & Matrices HL

IB Mathematics: Applications & Interpretation · Topic 3: Geometry & Trigonometry / Topic 1: Number & Algebra

JMaths

www.jmaths.xyz

1

Vector Operations

A vector has magnitude and direction. Written as a (bold) or \( \vec{a} \) (arrow notation). Components in 2D: \( (a_1,\ a_2) \) or 3D: \( (a_1,\ a_2,\ a_3) \).

Addition / Subtraction

\( a + b = (a_1+b_1,\ a_2+b_2,\ a_3+b_3) \)

Add/subtract corresponding components

Scalar multiplication

\( k\mathbf{a} = (ka_1,\ ka_2,\ ka_3) \)

Scales magnitude by \( |k| \); reverses direction if \( k < 0 \)

Magnitude

\( |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)

Unit vector

\( \hat{a} = \dfrac{\mathbf{a}}{|\mathbf{a}|} \)

Magnitude = 1, same direction

2

Dot Product (Scalar Product)

Component form

\( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

Geometric form

\( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}| \cos \theta \)

Finding the angle between two vectors

\( \cos \theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} \)

If \( \mathbf{a} \cdot \mathbf{b} = 0 \), the vectors are perpendicular

Worked Example

Find the angle between \( \mathbf{a} = (2, 1, -3) \) and \( \mathbf{b} = (4, -2, 1) \).

\( \mathbf{a} \cdot \mathbf{b} = 2(4) + 1(-2) + (-3)(1) = 8 - 2 - 3 = 3 \)

\( |\mathbf{a}| = \sqrt{4+1+9} = \sqrt{14}, \quad |\mathbf{b}| = \sqrt{16+4+1} = \sqrt{21} \)

\( \cos \theta = \dfrac{3}{\sqrt{14} \times \sqrt{21}} = \dfrac{3}{\sqrt{294}} = 0.17496\ldots \)

Answer: \( \theta = 79.9^\circ \) (3 s.f.)

Angle between two lines

\( \cos \theta = \dfrac{|\mathbf{d_1} \cdot \mathbf{d_2}|}{|\mathbf{d_1}||\mathbf{d_2}|} \)

Use the lines' direction vectors \( \mathbf{d_1},\ \mathbf{d_2} \); the absolute value gives the acute angle

3

Cross Product (Vector Product)

Cross product (3D only)

\( \mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2,\ \ a_3b_1 - a_1b_3,\ \ a_1b_2 - a_2b_1) \)

Magnitude of cross product

\( |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}| \sin \theta \)

Area of parallelogram

\( \text{Area} = |\mathbf{a} \times \mathbf{b}| \)

Triangle area = \( \frac{1}{2}|\mathbf{a} \times \mathbf{b}| \)

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Direction: \( \mathbf{a} \times \mathbf{b} \) is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \) (right-hand rule). Also \( \mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a}) \).

4

Vector Equation of a Line

Line through point a in direction b

\( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \)

\( \mathbf{a} \) = position vector of a point on the line, \( \mathbf{b} \) = direction vector, \( \lambda \in \mathbb{R} \)

▶

Parallel lines: direction vectors are scalar multiples. Intersecting lines: solve for \( \lambda \) and \( \mu \) — if all components are consistent, they intersect. Skew lines: not parallel and do not intersect (3D only).

Vectors & Matrices HL

IB Mathematics: Applications & Interpretation · Topic 3: Geometry & Trigonometry

JMaths

www.jmaths.xyz

5

Vector Kinematics (3.12)

A moving object's position is a vector function of time \( \mathbf{r}(t) \). Velocity and displacement are found component by component.

Constant velocity

\( \mathbf{r}(t) = \mathbf{r_0} + t\,\mathbf{v} \)

\( \mathbf{r_0} \) = start position, \( \mathbf{v} \) = velocity. Speed = \( |\mathbf{v}| \)

Variable velocity

\( \mathbf{v}(t) = \dfrac{d\mathbf{r}}{dt}, \quad \mathbf{r} = \int \mathbf{v}\,dt \)

Differentiate / integrate each component separately

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Two objects meet when \( \mathbf{r_1}(t) = \mathbf{r_2}(t) \): the same value of \( t \) must satisfy every component equation. If no common \( t \) works, the paths may cross but the objects never collide.

Worked Example

A particle starts at \( (1, 2) \) with constant velocity \( \mathbf{v} = (3, -1) \). Find its position at \( t = 4 \) and its speed.

\( \mathbf{r}(4) = (1, 2) + 4(3, -1) = (1+12,\ 2-4) = (13, -2) \)

Speed \( = |\mathbf{v}| = \sqrt{3^2 + (-1)^2} = \sqrt{10} \)

Answer: position \( (13, -2) \); speed \( \sqrt{10} \approx 3.16 \) (3 s.f.)

6

Matrix Transformations (3.9)

A \( 2 \times 2 \) matrix transforms points in the plane. Apply by multiplying: \( \text{image} = M\,\mathbf{p} \), where \( \mathbf{p} \) is the point as a column vector.

Rotation by \( \theta \) anticlockwise (about O)

\( \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \)

Reflection in \( y = x \)

\( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \)

Reflection in the \( x \)-axis

\( \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \)

Enlargement, scale factor \( k \)

\( \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} \)

Translation by vector t

\( \text{image} = \mathbf{p} + \mathbf{t} \)

A pure translation is vector addition, not a \( 2 \times 2 \) matrix. Combined: \( \text{image} = M\,\mathbf{p} + \mathbf{t} \)

▶

Determinant meaning: \( |\det(M)| \) is the area scale factor of the transformation. If \( \det(M) < 0 \), the transformation reverses orientation (e.g. a reflection).

✗

Common error: For combined transformations, order matters. “Reflect then rotate” means \( \text{image} = R\,M\,\mathbf{p} \) (the rightmost matrix acts first).

Vectors & Matrices HL

IB Mathematics: Applications & Interpretation · Topic 1: Number & Algebra

JMaths

www.jmaths.xyz

7

Matrix Operations (1.14)

Matrices are rectangular arrays of numbers. Multiplication is not commutative: \( AB \neq BA \) in general.

Matrix multiplication (row × column)

\( (AB)_{ij} = (\text{row } i \text{ of } A) \cdot (\text{column } j \text{ of } B) \)

Conformable only if \( \text{cols}(A) = \text{rows}(B) \); result is \( \text{rows}(A) \times \text{cols}(B) \). Identity \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \) satisfies \( AI = IA = A \).

Determinant \( (2 \times 2) \)

\( \det(A) = |A| = ad - bc \)

For \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)

Inverse \( (2 \times 2) \)

\( A^{-1} = \dfrac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \)

Exists only if \( \det(A) \neq 0 \)

▶

Solving systems: \( A\mathbf{x} = \mathbf{b} \ \Rightarrow\ \mathbf{x} = A^{-1}\mathbf{b} \). Use the GDC for \( 3 \times 3 \) systems — finding inverses by hand is not expected (only \( 2 \times 2 \) by hand).

GDC: Matrix arithmetic, determinant, inverse & systems

TI-84 Plus CE

[2ND][x⁻¹] (MATRIX) → EDIT → enter dimensions and values
Multiply: \( [A] \times [B] \) Inverse: \( [A] \)[x⁻¹]
Determinant: [2ND][x⁻¹] → MATH → det([A])
Systems: enter \( [A] \) and \( [B] \), compute \( [A]^{-1}[B] \)

Note: there is no standalone MATRIX key — it is the 2nd function on the \( x^{-1} \) key.

TI-Nspire CX II

Define: a := [[2,3][1,4]] (use the matrix template, or type)
Inverse: \( a^{-1} \) Determinant: det(a)
Solve system: simultEqn() or rref() on the augmented matrix

Casio fx-CG50

[MENU] → Run-Matrix → [F3] (MAT) to enter matrices
Multiply: \( \text{Mat A} \times \text{Mat B} \) Inverse: Mat A[x⁻¹]
Determinant: det(Mat A) via [OPTN] → MAT/VCT → Det

Worked Example

Multiply \( \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix} \).

Top-left: \( 1(5)+2(7)=19 \); top-right: \( 1(6)+2(8)=22 \)

Bottom-left: \( 3(5)+4(7)=43 \); bottom-right: \( 3(6)+4(8)=50 \)

Answer: \( \begin{pmatrix} 19 & 22 \\ 43 & 50 \end{pmatrix} \)

Vectors & Matrices HL

IB Mathematics: Applications & Interpretation · Topic 1: Number & Algebra

JMaths

www.jmaths.xyz

8

Eigenvalues & Eigenvectors (1.15)

If \( A\mathbf{v} = \lambda \mathbf{v} \) (with \( \mathbf{v} \neq \mathbf{0} \)), then \( \lambda \) is an eigenvalue and \( \mathbf{v} \) is the corresponding eigenvector.

Finding eigenvalues

\( \det(A - \lambda I) = 0 \)

For \( 2 \times 2 \): \( (a - \lambda)(d - \lambda) - bc = 0 \ \Rightarrow\ \) solve the quadratic for \( \lambda \)

Steps: (1) Solve \( \det(A - \lambda I) = 0 \) for the eigenvalues. (2) For each \( \lambda \), solve \( (A - \lambda I)\mathbf{v} = \mathbf{0} \) for the eigenvector.

Worked Example — eigenvalues

Find the eigenvalues of \( A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix} \).

\( \det(A - \lambda I) = (3-\lambda)(2-\lambda) - (1)(0) = 0 \)

\( \lambda^2 - 5\lambda + 6 = 0 \ \Rightarrow\ (\lambda - 2)(\lambda - 3) = 0 \)

Answer: \( \lambda = 2 \) and \( \lambda = 3 \)

Worked Example — eigenvectors

Find an eigenvector for each \( \lambda \) of \( A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix} \).

\( \lambda = 3 \): \( (A - 3I)\mathbf{v} = \begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}\mathbf{v} = \mathbf{0} \Rightarrow v_2 = 0 \), so \( \mathbf{v} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \)

\( \lambda = 2 \): \( (A - 2I)\mathbf{v} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\mathbf{v} = \mathbf{0} \Rightarrow v_1 + v_2 = 0 \), so \( \mathbf{v} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \)

Answer: \( \begin{pmatrix} 1 \\ 0 \end{pmatrix} \) for \( \lambda = 3 \); \( \begin{pmatrix} 1 \\ -1 \end{pmatrix} \) for \( \lambda = 2 \)

TI-84 Plus CE — eigenvalues

No native eigenvalue command. Find the characteristic polynomial by hand (above), then solve it — or use the PlySmlt2 app (Poly Root Finder) on the quadratic \( \lambda^2 - (a+d)\lambda + \det A = 0 \).

TI-Nspire CX II — eigenvalues

Eigenvalues: eigVl(a) Eigenvectors: eigVc(a)
Menu → Matrix & Vector → Advanced → Eigenvalues / Eigenvectors

Casio fx-CG50 — eigenvalues

No eigenvalue command. Use the by-hand method: form \( \det(A - \lambda I) = 0 \) and solve the quadratic for \( \lambda \).

9

Matrix Powers & Diagonalization (1.15)

Diagonalization (distinct real eigenvalues)

\( A = PDP^{-1}, \quad A^n = PD^nP^{-1} \)

\( D = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} \), \( P \) has the eigenvectors as columns; \( D^n = \begin{pmatrix} \lambda_1^n & 0 \\ 0 & \lambda_2^n \end{pmatrix} \)

▶

Use for large powers / long-term behaviour. On the GDC you can also just compute \( A^n \) directly (e.g. \( [A]^{20} \) on TI-84, \( a^{20} \) on Nspire). Diagonalization is the by-hand route when a formula for \( A^n \) is required.

Vectors & Matrices HL

IB Mathematics: Applications & Interpretation · Topic 4: Statistics & Probability

JMaths

www.jmaths.xyz

10

Transition (Markov) Matrices & Steady State (4.19)

A transition matrix \( T \) gives the probabilities of moving between states in one step. Each column sums to 1. The state after \( n \) steps is found by repeated multiplication.

State after \( n \) steps

\( \mathbf{s}_n = T^n\,\mathbf{s}_0 \)

\( \mathbf{s}_0 \) = initial state column (entries sum to 1)

Steady (long-term) state

\( T\mathbf{s} = \mathbf{s}, \quad \textstyle\sum s_i = 1 \)

Solve this linear system, OR compute \( T^n \) for large \( n \)

✗

Method note: the steady state is found by solving \( T\mathbf{s} = \mathbf{s} \) with \( \sum s_i = 1 \) (a linear system) or by raising \( T \) to a high power on the GDC. Do not treat it as an eigenvalue/eigenvector exercise — that is a separate topic.

Worked Example

Find the steady state of \( T = \begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix} \).

Let \( \mathbf{s} = \begin{pmatrix} x \\ y \end{pmatrix} \). Then \( T\mathbf{s} = \mathbf{s} \Rightarrow 0.8x + 0.3y = x \Rightarrow 0.3y = 0.2x \Rightarrow y = \tfrac{2}{3}x \)

Constraint \( x + y = 1 \Rightarrow x + \tfrac{2}{3}x = 1 \Rightarrow \tfrac{5}{3}x = 1 \Rightarrow x = 0.6 \)

Answer: \( x = 0.6,\ y = 0.4 \), i.e. \( \mathbf{s} = \begin{pmatrix} 0.6 \\ 0.4 \end{pmatrix} \)

GDC: Transition matrices & powers

TI-84 Plus CE

Enter \( T \) and \( \mathbf{s_0} \) via [2ND][x⁻¹] (MATRIX) → EDIT
State after \( n \) steps: \( [T]^n [S_0] \) Long-term: compute \( [T]^{20} \) and read off a column

TI-Nspire CX II

Define t and s0; compute \( t^n \cdot s0 \) for the state after \( n \) steps
Long-term: evaluate \( t^{20} \) (columns converge to the steady state)

Casio fx-CG50

Run-Matrix → enter Mat T and Mat S
State after \( n \) steps: \( \text{Mat T}^n \times \text{Mat S} \) Long-term: high power of Mat T

✗

Common error: Confusing the dot product (scalar result) with matrix multiplication (matrix result). \( \mathbf{a} \cdot \mathbf{b} \) gives a number; \( AB \) gives a matrix.

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Formula booklet: the dot product, cross product, magnitude, and vector line equation are given. The eigenvalue method, transformation matrices, and the steady-state condition are not in the booklet — learn the process.