Unit 4  ·  Pre-Calculus
Unit 4

Functions Involving
Parameters, Vectors,
and Matrices

From core concepts to worked examples — everything you need to master this unit on your own.

Parametric Equations Vectors Matrices & Inverses Systems via Matrices
01 Parametric 02 Vectors 03 Matrices 04 Systems
Topic 01

Parametric Equations
& Their Graphs

Instead of writing y directly as a function of x, we express both x and y separately as functions of a third variable t — capturing path, direction, and time all at once.

Key Concept
What is a Parameter?
The variable t is called the parameter. Rather than linking x and y directly, both are expressed as functions of t.

Think of it this way: a ball in flight — as time t increases, x (horizontal) and y (vertical) both change.
Key Concept
Orientation of the Curve
As t increases, the point traces a path with a definite direction (orientation). Two parametric equations can produce the same curve but with opposite directions of travel.
Key Concept
Converting to Rectangular Form
Solve for t in one equation, then substitute into the other to eliminate t. The result is an x–y relationship. Watch out: the domain of t may restrict which part of the curve is shown.
Standard Form
x = f(t)    ← x expressed in terms of t
y = g(t)    ← y expressed in terms of t
a ≤ t ≤ b    ← domain of the parameter

Drag the slider or press Play to watch the point trace the curve as t increases.

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3-Step Process for Rectangular Conversion ① Solve for t from the simpler equation  →  ② Substitute into the other equation to eliminate t  →  ③ Check and apply any domain restrictions
Example 1 Eliminate the Parameter (Algebraic)
Problem. Eliminate the parameter t from x = t + 1, y = t² − 2 and write the rectangular equation. (t ∈ ℝ)
1
Solve for t using the simpler equation.
x = t + 1  ⟹  t = x − 1
2
Substitute into the y equation to eliminate t.
y = t² − 2 = (x − 1)² − 2
3
Expand.
y = x² − 2x + 1 − 2 = x² − 2x − 1
∴ y = x² − 2x − 1   (upward-opening parabola with vertex at (1, −2))
Example 2 Trigonometric Parameters → Ellipse
Problem. Identify and sketch the curve defined by x = 3cos t, y = 2sin t, for 0 ≤ t ≤ 2π.
1
Isolate the trig functions.
cos t = x/3,   sin t = y/2
2
Apply the Pythagorean identity cos²t + sin²t = 1.
(x/3)² + (y/2)² = 1
∴ x²/9 + y²/4 = 1   (ellipse: semi-major axis 3, semi-minor axis 2, traced counter-clockwise)
Topic 02

Vectors — Magnitude,
Direction & Operations

A vector has both magnitude (size) and direction. It's the language of physics, navigation, and computer graphics.

Notation
Writing a Vector
Vector v = ⟨a, b⟩  or  ai + bj
a = horizontal component, b = vertical component.

Drawn as an arrow from initial point to terminal point.
Magnitude
Length: |v|
By the Pythagorean theorem:
|v| = √(a² + b²)

Magnitude is always ≥ 0. It tells you "how long" the vector is, with no direction information.
Direction
Direction Angle θ
The angle measured counter-clockwise from the positive x-axis:
θ = arctan(b/a)

Always adjust for the correct quadrant!
x y a (horizontal) b (vertical) v = ⟨a, b⟩ θ |v| = √(a²+b²)
Vector Operations
// Addition & Subtraction — add component-wise
⟨a, b⟩ ± ⟨c, d⟩ = ⟨a±c, b±d⟩

// Scalar Multiplication — scale each component
k⟨a, b⟩ = ⟨ka, kb⟩

// Dot Product — multiply pairs, then sum
⟨a, b⟩ · ⟨c, d⟩ = ac + bd

// Unit Vector — same direction, length = 1
û = v / |v|
🧭
What the Dot Product Tells You u · v = |u||v|cosθ. If the dot product equals 0, the vectors are perpendicular (orthogonal). The sign of the dot product tells you whether the angle between the vectors is acute or obtuse.
Example 3 Magnitude, Direction Angle & Unit Vector
Problem. For vector v = ⟨−3, 4⟩, find the magnitude |v|, the direction angle θ, and the unit vector û.
1
Magnitude:
|v| = √((-3)² + 4²) = √(9 + 16) = √25 = 5
2
Direction angle: v is in Quadrant II (x < 0, y > 0)
Reference angle: arctan(4/3) ≈ 53.13°
Adjust for QII: θ = 180° − 53.13° ≈ 126.87°
3
Unit vector:
û = v / |v| = ⟨−3/5, 4/5⟩ = ⟨−0.6, 0.8⟩
|v| = 5  |  θ ≈ 126.87°  |  û = ⟨−3/5, 4/5⟩
Example 4 Dot Product and Angle Between Two Vectors
Problem. Given u = ⟨2, 3⟩ and v = ⟨−1, 4⟩, find the dot product u·v and the angle θ between them.
1
Dot product:
u·v = (2)(−1) + (3)(4) = −2 + 12 = 10
2
Magnitudes:
|u| = √(4 + 9) = √13,   |v| = √(1 + 16) = √17
3
cos θ = (u·v) / (|u||v|) = 10 / √221
θ = arccos(10/√221) ≈ 47.7°
u·v = 10  |  θ ≈ 47.7°   (acute angle — the vectors share a directional component)
Topic 03

Matrices — Operations
& Inverse Matrices

A matrix is a rectangular grid of numbers. We can add, subtract, and multiply them — and use inverse matrices to solve systems of equations efficiently.

Definition
Matrix Dimensions
An m × n matrix has m rows and n columns. Entry aij is in row i, column j.

Example: A 2×3 matrix has 2 rows, 3 columns, and 6 total entries.
Multiplication Rule
When Can We Multiply?
A(m×n) × B(n×p) = C(m×p)
Columns of A must equal rows of B.

Note: matrix multiplication is generally not commutative — AB ≠ BA.
Inverse Matrix
What Is A⁻¹?
A · A⁻¹ = A⁻¹ · A = I (identity matrix)

For a 2×2 matrix, A⁻¹ exists only when det(A) ≠ 0.
det(A) = ad − bc

2×2 Inverse Matrix Formula

A =
a
b
c
d
⟹
A⁻¹ = 1/(ad−bc) ·
d
−b
−c
a

Trick: swap the diagonal entries (a ↔ d), negate the off-diagonal entries (b and c)

Matrix Multiplication (2×2)
// Each entry = (row of A) · (column of B)
[a b] · [e f] = [ae+bg   af+bh]
[c d]   [g h]   [ce+dg   cf+dh]
⚠️
When No Inverse Exists If det(A) = ad − bc = 0, the matrix is called a singular matrix and has no inverse. This means the corresponding system of equations has either no solution or infinitely many solutions.
Example 5 Finding the Inverse of a 2×2 Matrix
Problem. Find the inverse of matrix A = [[3, 1], [5, 2]].
1
Compute the determinant:
det(A) = (3)(2) − (1)(5) = 6 − 5 = 1
Since det ≠ 0, the inverse exists.
2
Apply the formula (a=3, b=1, c=5, d=2):
A⁻¹ = (1/1) · [[2, −1], [−5, 3]]
3
Verify: A · A⁻¹ = I
[[3,1],[5,2]] · [[2,−1],[−5,3]] = [[6−5, −3+3],[10−10, −5+6]] = [[1,0],[0,1]] ✓
A⁻¹ = [[2, −1], [−5, 3]]
Example 6 Matrix Multiplication
Problem. Given A = [[1, 2], [3, 4]] and B = [[5, 0], [1, −1]], compute AB.
1
Entry (1,1): Row 1 of A · Col 1 of B = (1)(5)+(2)(1) = 7
2
Entry (1,2): Row 1 of A · Col 2 of B = (1)(0)+(2)(−1) = −2
3
Entry (2,1): Row 2 of A · Col 1 of B = (3)(5)+(4)(1) = 19
4
Entry (2,2): Row 2 of A · Col 2 of B = (3)(0)+(4)(−1) = −4
AB = [[7, −2], [19, −4]]
Topic 04

Solving Systems
Using Matrices

Rewrite a system of linear equations as a matrix equation AX = B, then solve in one step using the inverse: X = A⁻¹B.

Method 1
Inverse Matrix Method
Write the system as AX = B, then multiply both sides on the left by A⁻¹.
X = A⁻¹B

Requires A to be invertible (det ≠ 0).
Method 2
Gaussian Elimination
Form the augmented matrix [A|B] and apply row operations to reduce it.
Goal: [A|B] → [I|X]

Scales well for systems with many variables.
Number of Solutions
Three Possible Outcomes
Unique solution: det(A) ≠ 0
Infinitely many: dependent rows in A
No solution: inconsistent system

From System of Equations → Matrix Equation

2x + 3y = 8
x − y = 1
→
2
3
1
−1
x
y
=
8
1
A·X = B
Step-by-Step Solution Process
Step 1   Write the system in AX = B form
Step 2   Compute det(A) = ad − bc
Step 3   Find A⁻¹ = (1/det) · [[d, −b], [−c, a]]
Step 4   Calculate X = A⁻¹ · B
Step 5   Substitute back into the original equations to verify
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Why the Inverse Method Is Powerful When the same coefficient matrix A appears in multiple systems (only B changes), you compute A⁻¹ just once and reuse it for every right-hand side — a major efficiency gain in practice.
Example 7 Solving a System Using the Inverse Matrix
Problem. Use the inverse matrix method to solve:
  2x + 3y = 8
  x − y = 1
1
Determinant:
det(A) = (2)(−1) − (3)(1) = −2 − 3 = −5
2
Inverse matrix:
A⁻¹ = (1/−5) · [[−1, −3], [−1, 2]] = [[1/5, 3/5], [1/5, −2/5]]
3
Compute X = A⁻¹B:
x = (1/5)(8) + (3/5)(1) = 8/5 + 3/5 = 11/5
y = (1/5)(8) + (−2/5)(1) = 8/5 − 2/5 = 6/5
4
Verify:
2(11/5) + 3(6/5) = 22/5 + 18/5 = 40/5 = 8 ✓
11/5 − 6/5 = 5/5 = 1 ✓
x = 11/5 = 2.2  |  y = 6/5 = 1.2
Unit 4 — Quick Reference
Topic Key Formula Watch Out For
Parametric x=f(t), y=g(t) → eliminate t Domain restrictions on t may limit the curve
Vectors |v|=√(a²+b²)   u·v=|u||v|cosθ Adjust direction angle for the correct quadrant
Matrices A⁻¹ = (1/det)·[[d,−b],[−c,a]] If det = 0, no inverse exists (singular matrix)
Systems X = A⁻¹B Always substitute back in to verify your answer