Lesson Exemplar in Senior High School Mathematics

Grade 12 • General Mathematics / STEM Linear Algebra • 90 Minutes

Content Topic Matrices and Determinants
Subtopic Determinants of 2×2 and 3×3 Matrices
Learning Competency The learners compute the determinant of a 2×2 and a 3×3 matrix.

I. OBJECTIVES

II. INSTRUCTIONAL RESOURCES & MATERIALS

References: Larson, R. (2017). Elementary Linear Algebra (8th ed.). Cengage Learning.

Materials: Whiteboard and markers, printable activity worksheets, evaluation sheets, visual slides.

III. LESSON PROCEDURES

Phase 1: Activating Prior Knowledge (10 Minutes)

Activity: "Match & Cross" (Group Activity)

Divide the class into groups of 4 to 5 students. Display Matrix A on the board:

A = [ 4 3 ] [ 1 5 ]

Instructions:

  1. Multiply the main diagonal entries: 4 × 5 = 20.
  2. Multiply the anti-diagonal entries: 1 × 3 = 3.
  3. Subtract the second product from the first product: 20 - 3 = 17.
Teacher Note: Connect this exercise directly to the definition of a 2×2 determinant: det(A) = ad - bc.

Phase 2: Establishing the Purpose of the Lesson (5 Minutes)

Explain that determinants provide key structural properties of matrices, including whether an inverse exists and whether a system of linear equations has a unique solution.

Phase 3: Presenting Examples (15 Minutes)

1. Determinant of a 2×2 Matrix

For a general 2×2 matrix: det([[a, b], [c, d]]) = ad - bc

Worked Example: Find det(B) for matrix B:

B = [ 6 -2 ] [ 3 4 ] det(B) = (6)(4) - (-2)(3) = 24 - (-6) = 30

2. Determinant of a 3×3 Matrix (Sarrus' Rule)

Copy the first two columns to the right. Sum down-right products (P1) and subtract up-right products (P2): det(C) = P1 - P2.

Worked Example: Find det(C) for matrix C:

C = [ 1 2 3 ] [ 0 1 4 ] [ 5 6 0 ] Extended Grid: 1 2 3 | 1 2 0 1 4 | 0 1 5 6 0 | 5 6 Down-right (P1): (1*1*0) + (2*4*5) + (3*0*6) = 0 + 40 + 0 = 40 Up-right (P2): (5*1*3) + (6*4*1) + (0*0*2) = 15 + 24 + 0 = 39 det(C) = 40 - 39 = 1

Phase 4: Discussing New Concepts — Cofactor Expansion (15 Minutes)

Demonstrate Cofactor Expansion along Row 1 using alternating signs (+ - +):

det(C) = 1 * det([[1, 4], [6, 0]]) - 2 * det([[0, 4], [5, 0]]) + 3 * det([[0, 1], [5, 6]]) det(C) = 1(0 - 24) - 2(0 - 20) + 3(0 - 5) det(C) = -24 + 40 - 15 = 1

Phase 5: Developing Mastery (15 Minutes)

Phase 6: Real-Life Application & Discussion (10 Minutes)

Prompt 1 (Computer Graphics): How do determinants apply to 2D/3D graphics transformations?

Expected Answer: Determinants serve as area or volume scaling factors. A determinant of zero means a shape collapses to a line or point.

Prompt 2 (Business Systems): What does a determinant of 0 mean in economic system modeling?

Expected Answer: It means no unique set of pricing or allocation solutions exists because the system equations are linearly dependent.

Phase 7: Generalization (5 Minutes)

  1. 2×2 determinant formula: ad - bc.
  2. 3×3 determinants can be calculated via Sarrus' Rule or Cofactor Expansion.
  3. A non-zero determinant guarantees that a matrix is invertible.

IV. EVALUATION — 15-ITEM ASSESSMENT & KEY

Part I: 2×2 Determinants

  1. Find det([[5, 2], [3, 4]])
    Answer: 14 | Solution: (5)(4) - (2)(3) = 20 - 6 = 14
  2. Find det([[7, -1], [4, 2]])
    Answer: 18 | Solution: (7)(2) - (-1)(4) = 14 + 4 = 18
  3. Find det([[-3, -5], [2, 6]])
    Answer: -8 | Solution: (-3)(6) - (-5)(2) = -18 + 10 = -8
  4. Find det([[0, 8], [3, -2]])
    Answer: -24 | Solution: (0)(-2) - (8)(3) = 0 - 24 = -24
  5. Find det([[9, 3], [3, 1]])
    Answer: 0 | Solution: (9)(1) - (3)(3) = 9 - 9 = 0
  6. Find det([[1/2, 4], [2, 10]])
    Answer: -3 | Solution: (1/2)(10) - (4)(2) = 5 - 8 = -3

Part II: 3×3 Determinants

  1. Find det([[1, 0, 0], [0, 4, 0], [0, 0, 3]])
    Answer: 12 | Solution: Diagonal matrix: 1 × 4 × 3 = 12
  2. Find det([[2, 1, 3], [0, 5, 1], [0, 0, -2]])
    Answer: -20 | Solution: Triangular matrix: (2)(5)(-2) = -20
  3. Find det([[3, 2, 1], [1, 4, 2], [2, 1, 3]])
    Answer: 25 | Solution: Down = 36 + 8 + 1 = 45; Up = 8 + 6 + 6 = 20 → 45 - 20 = 25
  4. Find det([[2, -1, 4], [1, 0, 3], [-3, 2, 1]])
    Answer: 6 | Solution: Down = 0 + 9 + 8 = 17; Up = 0 + 12 - 1 = 11 → 17 - 11 = 6
  5. Find det([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
    Answer: 0 | Solution: Down = 45 + 84 + 96 = 225; Up = 105 + 48 + 72 = 225 → 0
  6. Find det([[0, 3, 1], [2, -1, 5], [4, 0, 2]])
    Answer: 52 | Solution: Down = 0 + 60 + 0 = 60; Up = -4 + 0 + 12 = 8 → 60 - 8 = 52

Part III: Algebraic & Applied Determinants

  1. Find x if det([[x, 4], [3, x]]) = 0
    Answer: x = ±2√3 | Solution: x² - 12 = 0 → x² = 12 → x = ±2√3
  2. Find k if det([[k, 2], [5, 4]]) = 6
    Answer: k = 4 | Solution: 4k - 10 = 6 → 4k = 16 → k = 4
  3. Is N = [[1, 3, -2], [0, 2, 4], [0, 0, 5]] invertible?
    Answer: Invertible (det = 10 ≠ 0) | Solution: det(N) = 1 × 2 × 5 = 10. Since det ≠ 0, N is invertible.

Student Worksheet: Properties of Determinants

Name: ____________________________________ | Date: ____________ | Score: _____ / 20

Quick Reference: Key Properties of Determinants

  1. Transpose Property: det(AT) = det(A)
  2. Row Interchanging: Swapping two rows reverses the sign: det(A') = -det(A).
  3. Identical Rows/Cols: If two rows or columns are identical, det(A) = 0.
  4. Zero Row/Col: If an entire row or column is all zeros, det(A) = 0.
  5. Scalar Multiplication: Scaling one row by k scales the determinant by k. For an n×n matrix, det(kA) = kn · det(A).
  6. Row Addition: Adding a multiple of one row to another does not change the determinant.
  7. Multiplicative Property: det(AB) = det(A) · det(B)
  8. Inverse Property: det(A-1) = 1 / det(A)
  9. Triangular Matrix Property: Determinant is the product of main diagonal entries.

Part I: Conceptual & True/False Questions (1 point each)

State whether each statement is True or False. If false, state why.

  1. If det(A) = 5 for a 3×3 matrix A, then det(2A) = 10.
  2. Swapping the first and third rows of a matrix twice restores the determinant to its original value.
  3. If det(A) = 0, then matrix A has an inverse.
  4. For any square matrices A and B of the same size, det(A + B) = det(A) + det(B).

Part II: Property Identification (2 points each)

Without explicitly expanding the determinants, evaluate det(M) using properties.

  1. Evaluate det(A) for A = [[4, 12, -8], [0, 0, 0], [1, 5, 9]].
  2. Evaluate det(B) for B = [[3, -1, 5], [2, 7, 4], [3, -1, 5]].
  3. Evaluate det(C) for C = [[5, 0, 0], [-2, 3, 0], [4, 1, -2]].
  4. If matrix D is formed by replacing R₂ of M with (R₂ - 4R₁), and det(M) = -7, find det(D).

Part III: Applied Problem Solving (2 points each)

Given that A and B are 3×3 matrices with det(A) = 4 and det(B) = -3:

  1. Compute det(AT).
  2. Compute det(AB).
  3. Compute det(A-1).
  4. Compute det(3B).
  5. Compute det(A² B-1).

Worksheet Answer Key & Detailed Solutions

Part I Answers

1. False. For a 3×3 matrix, det(2A) = 2³ · det(A) = 8 × 5 = 40.
2. True. Swapping twice multiplies the sign by (-1)(-1) = +1.
3. False. A matrix is invertible if and only if det(A) ≠ 0.
4. False. Determinants are multiplicative, not additive in general.

Part II Answers

5. det(A) = 0. Zero Row Property (Row 2 consists of all zeros).
6. det(B) = 0. Identical Rows Property (Row 1 equals Row 3).
7. det(C) = -30. Lower Triangular Property: 5 × 3 × (-2) = -30.
8. det(D) = -7. Row Addition Property does not alter the determinant value.

Part III Answers

9. det(AT) = det(A) = 4
10. det(AB) = det(A) · det(B) = (4)(-3) = -12
11. det(A-1) = 1 / det(A) = 1/4
12. det(3B) = 3³ · det(B) = 27 × (-3) = -81
13. det(A² B-1) = [det(A)]² / det(B) = (4)² / (-3) = -16/3