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<!-- ==================== LESSON EXEMPLAR ==================== -->
<div class="header-banner">
<h1>Lesson Exemplar in Senior High School Mathematics</h1>
<p>Grade 12 • General Mathematics / STEM Linear Algebra • 90 Minutes</p>
</div>
<table class="info-table">
<tr>
<td class="label">Content Topic</td>
<td>Matrices and Determinants</td>
</tr>
<tr>
<td class="label">Subtopic</td>
<td>Determinants of 2×2 and 3×3 Matrices</td>
</tr>
<tr>
<td class="label">Learning Competency</td>
<td>The learners compute the determinant of a 2×2 and a 3×3 matrix.</td>
</tr>
</table>
<h2>I. OBJECTIVES</h2>
<ul>
<li><strong>Knowledge:</strong> Identify and define the steps required to compute the determinants of 2×2 and 3×3 matrices using Sarrus' Rule and cofactor expansion.</li>
<li><strong>Skills:</strong> Compute the determinant of 2×2 and 3×3 matrices accurately and efficiently.</li>
<li><strong>Attitudes:</strong> Demonstrate perseverance, precision, and collaborative problem-solving during individual and group tasks.</li>
</ul>
<h2>II. INSTRUCTIONAL RESOURCES & MATERIALS</h2>
<p><strong>References:</strong> Larson, R. (2017). <em>Elementary Linear Algebra</em> (8th ed.). Cengage Learning.</p>
<p><strong>Materials:</strong> Whiteboard and markers, printable activity worksheets, evaluation sheets, visual slides.</p>
<h2>III. LESSON PROCEDURES</h2>
<h3>Phase 1: Activating Prior Knowledge (10 Minutes)</h3>
<p><strong>Activity: "Match & Cross" (Group Activity)</strong></p>
<p>Divide the class into groups of 4 to 5 students. Display Matrix <em>A</em> on the board:</p>
<div class="matrix-block">A = [ 4 3 ]
[ 1 5 ]</div>
<p><strong>Instructions:</strong></p>
<ol>
<li>Multiply the main diagonal entries: 4 × 5 = 20.</li>
<li>Multiply the anti-diagonal entries: 1 × 3 = 3.</li>
<li>Subtract the second product from the first product: 20 - 3 = 17.</li>
</ol>
<div class="callout">
<strong>Teacher Note:</strong> Connect this exercise directly to the definition of a 2×2 determinant: det(A) = ad - bc.
</div>
<h3>Phase 2: Establishing the Purpose of the Lesson (5 Minutes)</h3>
<p>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.</p>
<h3>Phase 3: Presenting Examples (15 Minutes)</h3>
<h4>1. Determinant of a 2×2 Matrix</h4>
<p>For a general 2×2 matrix: <code>det([[a, b], [c, d]]) = ad - bc</code></p>
<p><strong>Worked Example:</strong> Find det(B) for matrix B:</p>
<div class="matrix-block">B = [ 6 -2 ]
[ 3 4 ]
det(B) = (6)(4) - (-2)(3) = 24 - (-6) = 30
<h4>2. Determinant of a 3×3 Matrix (Sarrus' Rule)</h4>
<p>Copy the first two columns to the right. Sum down-right products (P<sub>1</sub>) and subtract up-right products (P<sub>2</sub>): <code>det(C) = P<sub>1</sub> - P<sub>2</sub></code>.</p>
<p><strong>Worked Example:</strong> Find det(C) for matrix C:</p>
<div class="matrix-block">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): (110) + (245) + (306) = 0 + 40 + 0 = 40 Up-right (P2): (513) + (641) + (002) = 15 + 24 + 0 = 39 det(C) = 40 - 39 = 1
<h3>Phase 4: Discussing New Concepts — Cofactor Expansion (15 Minutes)</h3>
<p>Demonstrate Cofactor Expansion along Row 1 using alternating signs (+ - +):</p>
<div class="matrix-block">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
<h3>Phase 5: Developing Mastery (15 Minutes)</h3>
<ul>
<li><strong>Problem 1:</strong> Find det(M) for M = [[5, -3], [-2, 4]].<br><em>Solution: (5)(4) - (-3)(-2) = 20 - 6 = 14</em></li>
<li><strong>Problem 2:</strong> Find det(N) for N = [[2, 0, 1], [3, -1, 2], [4, 1, 0]].<br><em>Solution: Down = 3, Up = 0 → det(N) = 3 - 0 = 3</em></li>
</ul>
<h3>Phase 6: Real-Life Application & Discussion (10 Minutes)</h3>
<p><strong>Prompt 1 (Computer Graphics):</strong> How do determinants apply to 2D/3D graphics transformations?</p>
<p><em>Expected Answer: Determinants serve as area or volume scaling factors. A determinant of zero means a shape collapses to a line or point.</em></p>
<p><strong>Prompt 2 (Business Systems):</strong> What does a determinant of 0 mean in economic system modeling?</p>
<p><em>Expected Answer: It means no unique set of pricing or allocation solutions exists because the system equations are linearly dependent.</em></p>
<h3>Phase 7: Generalization (5 Minutes)</h3>
<ol>
<li>2×2 determinant formula: <code>ad - bc</code>.</li>
<li>3×3 determinants can be calculated via Sarrus' Rule or Cofactor Expansion.</li>
<li>A non-zero determinant guarantees that a matrix is invertible.</li>
</ol>
<h2>IV. EVALUATION — 15-ITEM ASSESSMENT & KEY</h2>
<h4>Part I: 2×2 Determinants</h4>
<ol>
<li>Find det([[5, 2], [3, 4]])
<div class="item-solution"><strong>Answer: 14</strong> | Solution: (5)(4) - (2)(3) = 20 - 6 = 14</div>
</li>
<li>Find det([[7, -1], [4, 2]])
<div class="item-solution"><strong>Answer: 18</strong> | Solution: (7)(2) - (-1)(4) = 14 + 4 = 18</div>
</li>
<li>Find det([[-3, -5], [2, 6]])
<div class="item-solution"><strong>Answer: -8</strong> | Solution: (-3)(6) - (-5)(2) = -18 + 10 = -8</div>
</li>
<li>Find det([[0, 8], [3, -2]])
<div class="item-solution"><strong>Answer: -24</strong> | Solution: (0)(-2) - (8)(3) = 0 - 24 = -24</div>
</li>
<li>Find det([[9, 3], [3, 1]])
<div class="item-solution"><strong>Answer: 0</strong> | Solution: (9)(1) - (3)(3) = 9 - 9 = 0</div>
</li>
<li>Find det([[1/2, 4], [2, 10]])
<div class="item-solution"><strong>Answer: -3</strong> | Solution: (1/2)(10) - (4)(2) = 5 - 8 = -3</div>
</li>
</ol>
<h4>Part II: 3×3 Determinants</h4>
<ol start="7">
<li>Find det([[1, 0, 0], [0, 4, 0], [0, 0, 3]])
<div class="item-solution"><strong>Answer: 12</strong> | Solution: Diagonal matrix: 1 × 4 × 3 = 12</div>
</li>
<li>Find det([[2, 1, 3], [0, 5, 1], [0, 0, -2]])
<div class="item-solution"><strong>Answer: -20</strong> | Solution: Triangular matrix: (2)(5)(-2) = -20</div>
</li>
<li>Find det([[3, 2, 1], [1, 4, 2], [2, 1, 3]])
<div class="item-solution"><strong>Answer: 25</strong> | Solution: Down = 36 + 8 + 1 = 45; Up = 8 + 6 + 6 = 20 → 45 - 20 = 25</div>
</li>
<li>Find det([[2, -1, 4], [1, 0, 3], [-3, 2, 1]])
<div class="item-solution"><strong>Answer: 6</strong> | Solution: Down = 0 + 9 + 8 = 17; Up = 0 + 12 - 1 = 11 → 17 - 11 = 6</div>
</li>
<li>Find det([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
<div class="item-solution"><strong>Answer: 0</strong> | Solution: Down = 45 + 84 + 96 = 225; Up = 105 + 48 + 72 = 225 → 0</div>
</li>
<li>Find det([[0, 3, 1], [2, -1, 5], [4, 0, 2]])
<div class="item-solution"><strong>Answer: 52</strong> | Solution: Down = 0 + 60 + 0 = 60; Up = -4 + 0 + 12 = 8 → 60 - 8 = 52</div>
</li>
</ol>
<h4>Part III: Algebraic & Applied Determinants</h4>
<ol start="13">
<li>Find x if det([[x, 4], [3, x]]) = 0
<div class="item-solution"><strong>Answer: x = ±2√3</strong> | Solution: x² - 12 = 0 → x² = 12 → x = ±2√3</div>
</li>
<li>Find k if det([[k, 2], [5, 4]]) = 6
<div class="item-solution"><strong>Answer: k = 4</strong> | Solution: 4k - 10 = 6 → 4k = 16 → k = 4</div>
</li>
<li>Is N = [[1, 3, -2], [0, 2, 4], [0, 0, 5]] invertible?
<div class="item-solution"><strong>Answer: Invertible (det = 10 ≠ 0)</strong> | Solution: det(N) = 1 × 2 × 5 = 10. Since det ≠ 0, N is invertible.</div>
</li>
</ol>
<!-- SECTION DIVIDER -->
<div class="section-divider page-break"></div>
<!-- ==================== STUDENT WORKSHEET ==================== -->
<div class="header-banner" style="background-color: var(--secondary-color);">
<h1>Student Worksheet: Properties of Determinants</h1>
<p>Name: ____________________________________ | Date: ____________ | Score: _____ / 20</p>
</div>
<h2>Quick Reference: Key Properties of Determinants</h2>
<ol>
<li><strong>Transpose Property:</strong> det(A<sup>T</sup>) = det(A)</li>
<li><strong>Row Interchanging:</strong> Swapping two rows reverses the sign: det(A') = -det(A).</li>
<li><strong>Identical Rows/Cols:</strong> If two rows or columns are identical, det(A) = 0.</li>
<li><strong>Zero Row/Col:</strong> If an entire row or column is all zeros, det(A) = 0.</li>
<li><strong>Scalar Multiplication:</strong> Scaling one row by <em>k</em> scales the determinant by <em>k</em>. For an <em>n×n</em> matrix, det(kA) = k<sup>n</sup> · det(A).</li>
<li><strong>Row Addition:</strong> Adding a multiple of one row to another does not change the determinant.</li>
<li><strong>Multiplicative Property:</strong> det(AB) = det(A) · det(B)</li>
<li><strong>Inverse Property:</strong> det(A<sup>-1</sup>) = 1 / det(A)</li>
<li><strong>Triangular Matrix Property:</strong> Determinant is the product of main diagonal entries.</li>
</ol>
<h2>Part I: Conceptual & True/False Questions (1 point each)</h2>
<p>State whether each statement is <strong>True</strong> or <strong>False</strong>. If false, state why.</p>
<ol>
<li>If det(A) = 5 for a 3×3 matrix A, then det(2A) = 10.</li>
<li>Swapping the first and third rows of a matrix twice restores the determinant to its original value.</li>
<li>If det(A) = 0, then matrix A has an inverse.</li>
<li>For any square matrices A and B of the same size, det(A + B) = det(A) + det(B).</li>
</ol>
<h2>Part II: Property Identification (2 points each)</h2>
<p>Without explicitly expanding the determinants, evaluate det(M) using properties.</p>
<ol start="5">
<li>Evaluate det(A) for A = [[4, 12, -8], [0, 0, 0], [1, 5, 9]].</li>
<li>Evaluate det(B) for B = [[3, -1, 5], [2, 7, 4], [3, -1, 5]].</li>
<li>Evaluate det(C) for C = [[5, 0, 0], [-2, 3, 0], [4, 1, -2]].</li>
<li>If matrix D is formed by replacing R₂ of M with (R₂ - 4R₁), and det(M) = -7, find det(D).</li>
</ol>
<h2>Part III: Applied Problem Solving (2 points each)</h2>
<p><em>Given that A and B are 3×3 matrices with det(A) = 4 and det(B) = -3:</em></p>
<ol start="9">
<li>Compute det(A<sup>T</sup>).</li>
<li>Compute det(AB).</li>
<li>Compute det(A<sup>-1</sup>).</li>
<li>Compute det(3B).</li>
<li>Compute det(A² B<sup>-1</sup>).</li>
</ol>
<h2>Worksheet Answer Key & Detailed Solutions</h2>
<h4>Part I Answers</h4>
<div class="item-solution">
<strong>1. False.</strong> For a 3×3 matrix, det(2A) = 2³ · det(A) = 8 × 5 = 40.<br>
<strong>2. True.</strong> Swapping twice multiplies the sign by (-1)(-1) = +1.<br>
<strong>3. False.</strong> A matrix is invertible if and only if det(A) ≠ 0.<br>
<strong>4. False.</strong> Determinants are multiplicative, not additive in general.
</div>
<h4>Part II Answers</h4>
<div class="item-solution">
<strong>5. det(A) = 0.</strong> Zero Row Property (Row 2 consists of all zeros).<br>
<strong>6. det(B) = 0.</strong> Identical Rows Property (Row 1 equals Row 3).<br>
<strong>7. det(C) = -30.</strong> Lower Triangular Property: 5 × 3 × (-2) = -30.<br>
<strong>8. det(D) = -7.</strong> Row Addition Property does not alter the determinant value.
</div>
<h4>Part III Answers</h4>
<div class="item-solution">
<strong>9. det(A<sup>T</sup>) = det(A) = 4</strong><br>
<strong>10. det(AB) = det(A) · det(B) = (4)(-3) = -12</strong><br>
<strong>11. det(A<sup>-1</sup>) = 1 / det(A) = 1/4</strong><br>
<strong>12. det(3B) = 3³ · det(B) = 27 × (-3) = -81</strong><br>
<strong>13. det(A² B<sup>-1</sup>) = [det(A)]² / det(B) = (4)² / (-3) = -16/3</strong>
</div>