Class 9 Mathematics Chapter 4 Determinants Important Questions with Complete Solutions

Determinants have evolved as a high-weightage topic in CBSE Class 9 Mathematics, especially after the 2024–25 syllabus rationalization. The board expects students to not just calculate determinants but understand their geometric significance and practical applications. In the 2026–27 board exam pattern, determinants questions typically carry 8–12 marks across the paper—split between 1-mark MCQs, 2-mark short answers, and 3-5 mark problem-solving questions. The focus is on: (1) calculating 2×2 and 3×3 determinants using expansion methods, (2) applying properties of determinants to simplify expressions, (3) finding minors and cofactors correctly, (4) computing adjoints and inverses, and (5) solving systems of linear equations using Cramer's rule. Students who master the underlying logic of minors and cofactors—rather than memorizing formulas—score consistently high marks. This is because examiners often twist questions to test conceptual depth: asking you to prove a determinant equals zero, or to find the inverse of a matrix with parameters. Practising these important questions in the exact format and difficulty level of recent board papers ensures you're not caught off-guard on exam day.

MCQs in Class 9 determinants typically test quick recall of properties and basic calculations. Here are 5 representative questions with answers: **Q1:** If A = [3 2; 5 1], then |A| = ? (a) 13 (b) −7 (c) 7 (d) −13 **Answer:** (b) −7 **Explanation:** |A| = 3(1) − 2(5) = 3 − 10 = −7 **Q2:** If |A| = 5 and |B| = 2, then |AB| = ? (a) 10 (b) 7 (c) 3 (d) 2.5 **Answer:** (a) 10 **Explanation:** By the multiplicative property, |AB| = |A| × |B| = 5 × 2 = 10 **Q3:** The determinant of a 2×2 identity matrix is: (a) 0 (b) 1 (c) 2 (d) undefined **Answer:** (b) 1 **Explanation:** For I₂ = [1 0; 0 1], |I| = 1(1) − 0(0) = 1 **Q4:** If two rows of a determinant are identical, the determinant equals: (a) 1 (b) 0 (c) −1 (d) 2 **Answer:** (b) 0 **Explanation:** This is a fundamental property—identical rows make the determinant zero. **Q5:** The minor M₁₁ of element a₁₁ in matrix A = [2 3; 4 5] is: (a) 2 (b) 3 (c) 4 (d) 5 **Answer:** (d) 5 **Explanation:** M₁₁ is the determinant of the 1×1 matrix [5] obtained by deleting row 1 and column 1.

These questions require brief working and are designed to test conceptual understanding of properties and minors/cofactors: **Q1:** Find the determinant of A = [4 6; 2 3] using the formula |A| = ad − bc. **Answer:** |A| = 4(3) − 6(2) = 12 − 12 = 0 **Explanation:** When the determinant is zero, the matrix is singular (non-invertible). **Q2:** If |A| = 8, find |2A| where A is a 2×2 matrix. **Answer:** |2A| = 2² × |A| = 4 × 8 = 32 **Explanation:** For an n×n matrix, |kA| = kⁿ|A|. Here, n = 2, so k² = 4. **Q3:** Find all minors of matrix A = [1 2; 3 4]. **Answer:** M₁₁ = 4, M₁₂ = 3, M₂₁ = 2, M₂₂ = 1 **Explanation:** M₁₁ = minor of a₁₁ = det[4] = 4; M₁₂ = minor of a₁₂ = det[3] = 3; similarly for M₂₁ and M₂₂. **Q4:** State and verify the property: |A| = |Aᵀ| for A = [5 2; 7 3]. **Answer:** |A| = 5(3) − 2(7) = 15 − 14 = 1. Aᵀ = [5 7; 2 3], |Aᵀ| = 5(3) − 7(2) = 15 − 14 = 1. Hence |A| = |Aᵀ|. **Explanation:** The transpose property shows determinant is unchanged by transposition. **Q5:** If A = [2 3; 4 6], explain why this matrix has no inverse. **Answer:** |A| = 2(6) − 3(4) = 12 − 12 = 0. Since det(A) = 0, the inverse does not exist. **Explanation:** A matrix is invertible only if its determinant is non-zero.

These questions bridge calculation and conceptual understanding, requiring proper method and logical steps: **Q1:** Find the cofactor matrix and adjoint of A = [2 1; 3 4]. **Solution:** Step 1: Find minors: M₁₁ = 4, M₁₂ = 3, M₂₁ = 1, M₂₂ = 2 Step 2: Apply cofactor formula Cᵢⱼ = (−1)^(i+j) × Mᵢⱼ C₁₁ = (−1)² × 4 = 4, C₁₂ = (−1)³ × 3 = −3, C₂₁ = (−1)³ × 1 = −1, C₂₂ = (−1)⁴ × 2 = 2 Step 3: Cofactor matrix = [4 −3; −1 2] Step 4: Adjoint = transpose of cofactor matrix = [4 −1; −3 2] **Answer:** Cofactor matrix = [4 −3; −1 2], Adjoint(A) = [4 −1; −3 2] **Q2:** Find the inverse of A = [1 2; 3 4]. **Solution:** Step 1: Calculate |A| = 1(4) − 2(3) = 4 − 6 = −2 Step 2: Find cofactors: C₁₁ = 4, C₁₂ = −3, C₂₁ = −2, C₂₂ = 1 Step 3: Adjoint(A) = [4 −2; −3 1] Step 4: A⁻¹ = (1/|A|) × Adjoint(A) = (1/−2) × [4 −2; −3 1] = [−2 1; 1.5 −0.5] **Answer:** A⁻¹ = [−2 1; 3/2 −1/2] **Q3:** Verify that A × A⁻¹ = I for A = [2 0; 0 3]. **Solution:** Step 1: |A| = 2(3) − 0(0) = 6 Step 2: Cofactor matrix = [3 0; 0 2], Adjoint(A) = [3 0; 0 2] Step 3: A⁻¹ = (1/6)[3 0; 0 2] = [1/2 0; 0 1/3] Step 4: A × A⁻¹ = [2 0; 0 3] × [1/2 0; 0 1/3] = [1 0; 0 1] = I **Answer:** Verified. A × A⁻¹ = I. **Q4:** Prove using properties of determinants that |A| = |Aᵀ| without calculation for any 2×2 matrix. **Solution:** Let A = [a b; c d]. Then |A| = ad − bc. Aᵀ = [a c; b d]. Then |Aᵀ| = ad − bc. Since ad − bc = ad − bc, we have |A| = |Aᵀ|. **Explanation:** This property holds for all square matrices and is a consequence of how determinant is defined through row/column operations.

Long-answer questions test your ability to combine multiple concepts and present a complete argument: **Q1:** Solve the system of equations using the inverse matrix method: 2x + y = 5 x + 3y = 5 **Full Solution:** Step 1: Write in matrix form AX = B, where A = [2 1; 1 3], X = [x; y], B = [5; 5] Step 2: Calculate |A| = 2(3) − 1(1) = 6 − 1 = 5 Step 3: Find cofactors: C₁₁ = 3, C₁₂ = −1, C₂₁ = −1, C₂₂ = 2 Step 4: Adjoint(A) = [3 −1; −1 2] Step 5: A⁻¹ = (1/5)[3 −1; −1 2] Step 6: X = A⁻¹B = (1/5)[3 −1; −1 2][5; 5] = (1/5)[15 − 5; −5 + 10] = (1/5)[10; 5] = [2; 1] **Answer:** x = 2, y = 1 **Verification:** 2(2) + 1 = 5 ✓ and 2 + 3(1) = 5 ✓ **Q2:** If A = [1 2 3; 0 1 2; 0 0 1], prove that |A| = 1 and find its inverse. **Full Solution:** Step 1: Calculate |A| using expansion along the first column: |A| = 1 × |1 2; 0 1| − 0 + 0 = 1(1 − 0) = 1 Step 2: Since |A| = 1, the inverse exists. Step 3: For 3×3 matrices, find all 9 minors, then cofactors, then adjoint. M₁₁ = |1 2; 0 1| = 1, M₁₂ = |0 2; 0 1| = 0, M₁₃ = |0 1; 0 0| = 0 M₂₁ = |2 3; 0 1| = 2, M₂₂ = |1 3; 0 1| = 1, M₂₃ = |1 2; 0 0| = 0 M₃₁ = |2 3; 1 2| = 4 − 3 = 1, M₃₂ = |1 3; 0 2| = 2, M₃₃ = |1 2; 0 1| = 1 Step 4: Cofactor matrix = [1 0 0; −2 1 0; 1 −2 1] Step 5: Adjoint(A) = [1 −2 1; 0 1 −2; 0 0 1] Step 6: A⁻¹ = (1/1) × [1 −2 1; 0 1 −2; 0 0 1] = [1 −2 1; 0 1 −2; 0 0 1] **Answer:** |A| = 1, A⁻¹ = [1 −2 1; 0 1 −2; 0 0 1] **Q3:** State and prove the property: If k is a scalar, then |kA| = kⁿ|A| for an n×n matrix A. Illustrate with a 3×3 example. **Full Solution:** Step 1: **Property Statement:** For an n×n matrix A and scalar k, |kA| = kⁿ|A|. Step 2: **Proof:** When we multiply matrix A by scalar k, every element aᵢⱼ becomes k·aᵢⱼ. In the determinant expansion, each term is a product of n elements (one from each row/column). Thus, each term picks up a factor of k from n positions, giving kⁿ overall. Step 3: **Example with A = [1 2 3; 0 1 2; 0 0 1] (from Q2), n = 3, k = 2:** |A| = 1 (from Q2) 2A = [2 4 6; 0 2 4; 0 0 2] |2A| = 2 × |2 4; 0 2| × 1 = 2 × 2(2) = 8 Alternatively, |2A| = 2³ × |A| = 8 × 1 = 8 ✓ Step 4: **Conclusion:** The property holds; multiplication by scalar k scales the determinant by kⁿ. **Answer:** Property proven both theoretically and by example.

**Question (Case Study):** A construction company uses matrices to track costs of materials (wood, steel, cement) across three projects (A, B, C). The cost matrix is: C = [500 800 600; 700 600 900; 400 1000 500] (in thousands of rupees) where rows represent projects and columns represent materials. **Part (a):** Calculate the determinant of C and interpret whether the cost system is non-singular. **Part (b):** If the company applies a 10% cost reduction uniformly, form the new matrix C' and find |C'| in terms of |C|. **Part (c):** The company finds that one project's costs are exactly 1.5 times another project's costs. Without calculating the full determinant, predict what the determinant will be and justify your answer. **Full Solution:** **Part (a):** Calculate |C| using expansion along the first row: |C| = 500|600 900; 1000 500| − 800|700 900; 400 500| + 600|700 600; 400 1000| = 500[(600)(500) − (900)(1000)] − 800[(700)(500) − (900)(400)] + 600[(700)(1000) − (600)(400)] = 500[300000 − 900000] − 800[350000 − 360000] + 600[700000 − 240000] = 500(−600000) − 800(−10000) + 600(460000) = −300000000 + 8000000 + 276000000 = −16000000 **Interpretation:** Since |C| ≠ 0, the cost system is non-singular and invertible. This means the company can uniquely determine material costs from project costs if needed. **Part (b):** C' = 0.9C (10% reduction means multiply by 0.9) Using the property |kA| = kⁿ|A| where n = 3: |C'| = 0.9³ × |C| = 0.729 × (−16000000) = −11664000 **Answer:** |C'| = 0.729 × |C| **Part (c):** If one row is a scalar multiple of another row (say Row 2 = 1.5 × Row 1), then the rows are linearly dependent. **Property:** If rows are linearly dependent, |C| = 0. **Prediction:** The determinant will be zero. **Justification:** Linear dependence of rows means the rank is less than 3, and determinant becomes zero. The company's cost data would become singular (non-invertible), indicating redundant or inconsistent pricing.

Solving determinant questions once is not enough—mastery comes from drilling variations daily, understanding your specific weak points, and building speed under exam pressure. CBSETUTOR.ai's AI tutor is specifically trained on the 2024–25 CBSE Class 9 Mathematics curriculum and the patterns seen in recent board exams. Here's how the platform helps: **Personalized Question Generation:** The AI generates unlimited variations of these 17 questions, tweaking numbers, matrix sizes, and problem contexts so you never memorize answers. **Adaptive Difficulty:** Start with basic 2×2 determinant calculations, progress to 3×3 minors and cofactors, then tackle inverse matrices and system-solving. The platform adjusts speed and complexity based on your performance. **Real-Time Error Analysis:** When you make a sign error in cofactor calculations or forget the transpose in adjoint, the tutor pinpoints it instantly and explains the concept behind the mistake—not just the answer. **Step-by-Step Scaffolding:** For long-answer 5-mark questions, you can request hints at each step or see the full working. This builds problem-solving confidence without spoon-feeding. **Board-Exam Simulation:** Every two weeks, practice a full mock test with mixed determinant questions (MCQ, short-answer, long-answer) in exam time constraints. Review detailed analytics on time-per-question, accuracy, and conceptual gaps. **Live Doubt Clarification:** Beyond AI drilling, chat with expert tutors during live sessions to discuss tricky proofs (like why |A| = |Aᵀ|) or seek alternate solution methods. **Start a 3-day free trial at cbsetutor.ai** to experience personalized determinants practice right now—no credit card needed.