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SPM WikiMathematicsChapter 12: Matrices

Chapter 12: Matrices

Master matrix operations, matrix algebra, and solving systems of equations using matrices.

Chapter 12: Matrices

Overview

Welcome to Chapter 12 of Form 5 Mathematics! This chapter introduces you to the powerful world of matrices and matrix algebra. You'll learn to represent information in matrix form, perform various matrix operations, find matrix inverses, and use matrices to solve systems of linear equations. Matrices are fundamental in mathematics, computer science, engineering, and many other fields.

What You'll Learn:

  • Represent information in matrix form
  • Perform matrix operations (addition, subtraction, multiplication)
  • Find matrix identity and inverse matrices
  • Use matrices to solve systems of linear equations

Learning Objectives

After completing this chapter, you will be able to:

  • Represent information in matrix form
  • Determine the order of matrices and specific elements
  • Perform addition, subtraction, and multiplication of matrices
  • Find the identity matrix and inverse of matrices
  • Use the matrix method to solve systems of linear equations

Key Concepts

Matrices

A matrix is a rectangular array of numbers arranged in rows and columns. Matrices are used to represent and organize data efficiently.

Notation:

  • Matrices are usually denoted by capital letters: A, B, C
  • Elements are denoted by lowercase letters: a, b, c
  • The element in row i, column j is denoted as aᵢⱼ

Matrix Structure:

Example:

A=(246135)A = \begin{pmatrix} 2 & 4 & 6 \\ 1 & 3 & 5 \end{pmatrix}
  • Matrix A has 2 rows and 3 columns
  • Order: 2 × 3
  • Element a1a_1₁ = 2 (row 1, column 1)
  • Element a2a_2₃ = 5 (row 2, column 3)

Matrix Types:

Matrix Order

The order of a matrix is its dimension, expressed as (number of rows × number of columns).

Visual Order Representation:

Examples:

  • Row matrix: (123)(1 \quad 2 \quad 3) - Order: 1 × 3
  • Column matrix: (123)\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} - Order: 3 × 1
  • Square matrix: (1234)\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - Order: 2 × 2

Equal Matrices

Two matrices are equal if they have the same order and corresponding elements are equal.

Equality Conditions:

Example: If A = B, then:

(abcd)=(efgh)\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} e & f \\ g & h \end{pmatrix}

This means:

  • a = e
  • b = f
  • c = g
  • d = h

Matrix Operations Visualization

Matrix Operations Flow:

Matrix Operations

1. Addition and Subtraction: Can only be performed on matrices of the same order.

Operation Visualization:

Addition:

A+B=(aij+bij)A + B = \begin{pmatrix} a_{ij} + b_{ij} \end{pmatrix}

Subtraction:

AB=(aijbij)A - B = \begin{pmatrix} a_{ij} - b_{ij} \end{pmatrix}

Example:

(2314)+(1234)=(3548)\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} + \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 3 & 5 \\ 4 & 8 \end{pmatrix}

2. Scalar Multiplication: Multiply each element by the scalar.

Operation Visualization:

kA=(kaij)kA = \begin{pmatrix} ka_{ij} \end{pmatrix}

Example:

3×(1234)=(36912)3 \times \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 3 & 6 \\ 9 & 12 \end{pmatrix}

3. Matrix Multiplication: Matrix A (m × n) can be multiplied by matrix B (n × p) if the number of columns in A equals the number of rows in B. The result is matrix C (m × p).

Multiplication Process:

C=AB=(k=1naikbkj)C = AB = \begin{pmatrix} \sum_{k=1}^{n} a_{ik}b_{kj} \end{pmatrix}

Step-by-Step Example:

(1234)×(5678)=(1×5+2×71×6+2×83×5+4×73×6+4×8)=(19224350)\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \times \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix} = \begin{pmatrix} 1×5+2×7 & 1×6+2×8 \\ 3×5+4×7 & 3×6+4×8 \end{pmatrix} = \begin{pmatrix} 19 & 22 \\ 43 & 50 \end{pmatrix}

Multiplication Rules:

  • Order matters: AB ≠ BA in general
  • Associative: (AB)C = A(BC)
  • Distributive: A(B + C) = AB + AC

Matrix Multiplication Examples

3×3 Matrix Multiplication:

(123456789)×(987654321)=(30241884695413811490)\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} \times \begin{pmatrix} 9 & 8 & 7 \\ 6 & 5 & 4 \\ 3 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 30 & 24 & 18 \\ 84 & 69 & 54 \\ 138 & 114 & 90 \end{pmatrix}

Identity Matrix

The identity matrix (I) is a square matrix with 1s on the main diagonal and 0s elsewhere. AI = IA = A.

Identity Matrix Properties:

Example (2 × 2):

I=(1001)I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

Example (3 × 3):

I=(100010001)I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}

Identity Matrix Multiplication:

(1234)×(1001)=(1234)\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \times \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}

Matrix Inverse

For a square matrix A, the inverse A⁻¹ is the matrix that satisfies AA⁻¹ = A⁻¹A = I.

Inverse Properties:

For 2 × 2 matrix: If A = (\begin{pmatrix} a & b \ c & d \end{pmatrix}), then:

A1=1adbc(dbca)A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

The inverse exists only if ad - bc ≠ 0 (determinant ≠ 0).

Inverse Calculation Example:

A=(2134)A = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} det=(2)(4)(1)(3)=83=5det = (2)(4) - (1)(3) = 8 - 3 = 5 A1=15(4132)=(4/51/53/52/5)A^{-1} = \frac{1}{5} \begin{pmatrix} 4 & -1 \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} 4/5 & -1/5 \\ -3/5 & 2/5 \end{pmatrix}

Determinant Calculation

Determinant Properties:

  • Only defined for square matrices
  • Determinant of 2×2: det(A)=adbcdet(A) = ad - bc
  • Determinant of 3×3: det(A)=a(eifh)b(difg)+c(dheg)det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
  • If det = 0, matrix has no inverse (singular matrix)

Determinant Visualization:

Solving Systems of Equations

Matrix Method for 2×2 Systems:

System: ax + by = p, cx + dy = q Matrix form: (\begin{pmatrix} a & b \ c & d \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} p \ q \end{pmatrix})

Solution: (\begin{pmatrix} x \ y \end{pmatrix} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \begin{pmatrix} p \ q \end{pmatrix})

Solution Method:

  1. Write system in matrix form: AX = B
  2. Find inverse of coefficient matrix A⁻¹
  3. Calculate solution: X = A⁻¹B
  4. Verify by substitution

Generalization for 3×3 Systems:

(abcdefghi)(xyz)=(pqr)\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} p \\ q \\ r \end{pmatrix} (xyz)=A1(pqr)\begin{pmatrix} x \\ y \\ z \end{pmatrix} = A^{-1} \begin{pmatrix} p \\ q \\ r \end{pmatrix}

Problem-Solving Matrix Method

Matrix Operations Workflow:

Step-by-Step Solved Examples

Example 1: Basic Matrix Operations

Problem: Given matrices A and B:

A=(231402),B=(123014)A = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 0 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \end{pmatrix}

Find: a) A + B b) A - B c) 3A

Operation Check:

Solution: a) A + B:

(2+13+21+34+00+12+4)=(354416)\begin{pmatrix} 2+1 & 3+2 & 1+3 \\ 4+0 & 0+1 & 2+4 \end{pmatrix} = \begin{pmatrix} 3 & 5 & 4 \\ 4 & 1 & 6 \end{pmatrix}

b) A - B:

(213213400124)=(112412)\begin{pmatrix} 2-1 & 3-2 & 1-3 \\ 4-0 & 0-1 & 2-4 \end{pmatrix} = \begin{pmatrix} 1 & 1 & -2 \\ 4 & -1 & -2 \end{pmatrix}

c) 3A:

3×(231402)=(6931206)3 \times \begin{pmatrix} 2 & 3 & 1 \\ 4 & 0 & 2 \end{pmatrix} = \begin{pmatrix} 6 & 9 & 3 \\ 12 & 0 & 6 \end{pmatrix}

Verification:

  • Addition: Result is same order (2×3)
  • Scalar multiplication: Each element scaled correctly
  • Result matrices maintain original structure

Answer: a) (\begin{pmatrix} 3 & 5 & 4 \ 4 & 1 & 6 \end{pmatrix}), b) (\begin{pmatrix} 1 & 1 & -2 \ 4 & -1 & -2 \end{pmatrix}), c) (\begin{pmatrix} 6 & 9 & 3 \ 12 & 0 & 6 \end{pmatrix})

Example 2: Matrix Multiplication

Problem: Given matrices A and B:

A=(1234),B=(2013)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}

Find AB and BA.

Multiplication Check:

Solution: AB:

(1×2+2×11×0+2×33×2+4×13×0+4×3)=(461012)\begin{pmatrix} 1×2+2×1 & 1×0+2×3 \\ 3×2+4×1 & 3×0+4×3 \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 10 & 12 \end{pmatrix}

BA:

(2×1+0×32×2+0×41×1+3×31×2+3×4)=(241014)\begin{pmatrix} 2×1+0×3 & 2×2+0×4 \\ 1×1+3×3 & 1×2+3×4 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 10 & 14 \end{pmatrix}

Key Observations:

  • AB ≠ BA (Matrix multiplication is not commutative)
  • Both results are 2×2 matrices
  • Order matters in matrix multiplication

Answer: AB = (\begin{pmatrix} 4 & 6 \ 10 & 12 \end{pmatrix}), BA = (\begin{pmatrix} 2 & 4 \ 10 & 14 \end{pmatrix})

Example 3: Finding Matrix Inverse

Problem: Find the inverse of matrix A:

A=(2314)A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}

Inverse Calculation Steps:

Solution: Step 1: Find determinant det(A) = ad - bc = 2×4 - 3×1 = 8 - 3 = 5

Step 2: Apply inverse formula

A1=15(4312)=(4/53/51/52/5)A^{-1} = \frac{1}{5} \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 4/5 & -3/5 \\ -1/5 & 2/5 \end{pmatrix}

Verification:

AA1=(2314)(0.80.60.20.4)=(1001)=IAA^{-1} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} 0.8 & -0.6 \\ -0.2 & 0.4 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I

Answer: (A^{-1} = \begin{pmatrix} 0.8 & -0.6 \ -0.2 & 0.4 \end{pmatrix})

Example 4: Solving System of Equations

Problem: Use matrix method to solve: 2x + 3y = 7 x + 4y = 6

Solution: Step 1: Write in matrix form

(2314)(xy)=(76)\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 6 \end{pmatrix}

Step 2: Find inverse of coefficient matrix From Example 3: (\begin{pmatrix} 2 & 3 \ 1 & 4 \end{pmatrix}^{-1} = \begin{pmatrix} 0.8 & -0.6 \ -0.2 & 0.4 \end{pmatrix})

Step 3: Multiply both sides by inverse

(xy)=(0.80.60.20.4)(76)=(0.8×7+(0.6)×6(0.2)×7+0.4×6)=(5.63.61.4+2.4)=(21)\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0.8 & -0.6 \\ -0.2 & 0.4 \end{pmatrix} \begin{pmatrix} 7 \\ 6 \end{pmatrix} = \begin{pmatrix} 0.8×7 + (-0.6)×6 \\ (-0.2)×7 + 0.4×6 \end{pmatrix} = \begin{pmatrix} 5.6 - 3.6 \\ -1.4 + 2.4 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}

Answer: x = 2, y = 1

Example 5: Real-world Application - Business

Problem: A company produces two products A and B. The production costs are:

  • Product A: RM2 material + RM3 labor per unit
  • Product B: RM1 material + RM4 labor per unit

If total material cost is RM7 and total labor cost is RM12, find the number of units produced.

Solution: Let x = units of A, y = units of B

Equations: 2x + y = 7 (material cost) 3x + 4y = 12 (labor cost)

Matrix form:

(2134)(xy)=(712)\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 12 \end{pmatrix}

Find inverse: det = 2×4 - 1×3 = 8 - 3 = 5

(2134)1=15(4132)=(0.80.20.60.4)\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}^{-1} = \frac{1}{5} \begin{pmatrix} 4 & -1 \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} 0.8 & -0.2 \\ -0.6 & 0.4 \end{pmatrix}

Solve:

(xy)=(0.80.20.60.4)(712)=(0.8×7+(0.2)×12(0.6)×7+0.4×12)=(5.62.44.2+4.8)=(3.20.6)\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0.8 & -0.2 \\ -0.6 & 0.4 \end{pmatrix} \begin{pmatrix} 7 \\ 12 \end{pmatrix} = \begin{pmatrix} 0.8×7 + (-0.2)×12 \\ (-0.6)×7 + 0.4×12 \end{pmatrix} = \begin{pmatrix} 5.6 - 2.4 \\ -4.2 + 4.8 \end{pmatrix} = \begin{pmatrix} 3.2 \\ 0.6 \end{pmatrix}

Answer: 3.2 units of A and 0.6 units of B (fractional units suggest planning values)

Real-world Applications

1. Computer Science

  • Graphics: Transformations, scaling, rotation
  • Data Processing: Image processing, compression
  • Machine Learning: Data representation, neural networks

2. Engineering

  • Structural Analysis: Force and stress calculations
  • Electrical Engineering: Circuit analysis
  • Control Systems: State-space representations

3. Economics and Business

  • Input-Output Models: Economic relationships
  • Inventory Management: Supply chain optimization
  • Financial Analysis: Portfolio management

4. Statistics and Data Science

  • Data Analysis: Principal component analysis
  • Regression Analysis: Multiple linear regression
  • Machine Learning: Pattern recognition

Important Terms

TermDefinitionExample
MatrixRectangular array of numbers(\begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix})
OrderDimensions (rows × columns)2 × 2 matrix
ElementIndividual number in matrixa1a_1₁ = 1 (row 1, column 1)
Equal MatricesSame order with equal elementsA = B
Identity MatrixSquare matrix with 1s on diagonal(\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix})
Matrix InverseMatrix that gives identity when multipliedAA⁻¹ = I
DeterminantScalar value for square matricesad - bc for 2×2
Coefficient MatrixMatrix of system equation coefficients(\begin{pmatrix} a & b \ c & d \end{pmatrix})

Summary Points

  • Matrix Order: (rows × columns)
  • Addition/Subtraction: Same order required
  • Matrix Multiplication: Inner dimensions must match
  • Identity Matrix: AI = IA = A
  • Inverse: AA⁻¹ = A⁻¹A = I, exists if det ≠ 0
  • 2×2 Inverse Formula: (\frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix})
  • System Solution: X = A⁻¹B for AX = B

Practice Tips for SPM Students

1. Master Matrix Operations

  • Practice matrix addition, subtraction, and multiplication
  • Understand the conditions for each operation
  • Learn to calculate determinants correctly

2. Inverse Calculation

  • Memorize the 2×2 inverse formula
  • Practice finding determinants
  • Remember that inverse exists only if det ≠ 0

3. System of Equations

  • Learn to convert systems to matrix form
  • Practice matrix method for solving
  • Verify solutions by substitution

4. Common Mistakes to Avoid

  • Forgetting order requirements for operations
  • Incorrect matrix multiplication order
  • Wrong determinant calculation
  • Confusing matrix and scalar operations

SPM Exam Tips

Paper 1 (Multiple Choice)

  • Look for matrix operation types and conditions
  • Remember inverse existence condition (det ≠ 0)
  • Practice quick determinant calculations
  • Use elimination method for difficult questions

Paper 2 (Structured)

  • Show all matrix operation steps
  • Demonstrate inverse calculations clearly
  • Write complete matrix equations
  • Verify matrix solutions by substitution

Did You Know? Matrices were first systematically studied in the 19th century, but their origins date back to ancient Chinese mathematics. Today, matrices are essential in computer graphics, quantum mechanics, artificial intelligence, and countless other applications!

Next Chapter: In Chapter 3, you'll explore consumer mathematics focusing on insurance, including risk assessment, types of insurance, and premium calculations.