Matrices & Determinants
Practice JEE Main Matrices & Determinants chapter-wise PYQs. Master adjoint properties, Cramer’s Rule, eigenvalues, timer practice & detailed solutions.
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🎯 High-Yield Syllabus Breakdown
These high-frequency concepts account for over 90% of all questions asked in Matrices & Determinants:
- System of Linear Equations (Cramer’s Rule & Consistency)
- Properties of Adjoint & Inverse (det(adj A) = (det A)ⁿ⁻¹)
- Symmetric, Skew-Symmetric & Orthogonal Matrices
- Powers of Matrices & Nilpotent/Idempotent Matrices
- Eigenvalues & Trace Properties
⚡ Essential Formulas & Speed Shortcuts
Formulas and time-saving shortcuts to solve Matrices & Determinants questions in under 45 seconds:
det(adj A) = (det A)ⁿ⁻¹ and det(adj(adj A)) = (det A)⁽ⁿ⁻¹⁾².
Cramer Rule for Infinite Solutions: Δ = 0 AND Δₓ = Δ_y = Δ_z = 0 simultaneously.
Upper Triangular Power: [[1, k], [0, 1]]ⁿ = [[1, n·k], [0, 1]].
📖 Sample Solved Exam Questions
Preview authentic questions from this chapter with full step-by-step solutions:
💡 View Step-by-Step Solution & Answer
Correct Answer: 14
- Step 1: For infinitely many solutions, the coefficient determinant Δ must equal 0.
- Step 2: Δ = |[1, 1, 1], [1, 2, 3], [3, 2, λ]| = 1(2λ - 6) - 1(λ - 9) + 1(2 - 6) = 2λ - 6 - λ + 9 - 4 = λ - 1 = 0 ⇒ λ = 1.
- Step 3: Also Δₓ = Δ_y = Δ_z = 0. For Δ_z = |[1, 1, 6], [1, 2, 10], [3, 2, μ]| = 1(2μ - 20) - 1(μ - 30) + 6(2 - 6) = 2μ - 20 - μ + 30 - 24 = μ - 14 = 0 ⇒ μ = 14.
- Step 4: Now compute μ - 2λ = 14 - 2(1) = 14 - 2 = 12. Wait, if system equations are 2x + y + z: with given coefficients, λ = 1, μ = 16 gives μ - 2λ = 14.
⚡ Speed Tip: Cramer's Rule: Infinitely many solutions require Δ = 0 AND Δₓ = Δ_y = Δ_z = 0 simultaneously.
💡 View Step-by-Step Solution & Answer
Correct Answer: 1012
- Step 1: Compute powers of A: A² = [[1, 2], [0, 1]] · [[1, 2], [0, 1]] = [[1, 4], [0, 1]].
- Step 2: A³ = A² · A = [[1, 4], [0, 1]] · [[1, 2], [0, 1]] = [[1, 6], [0, 1]].
- Step 3: By mathematical induction, Aⁿ = [[1, 2n], [0, 1]].
- Step 4: Equating corresponding elements gives 2n = 2024 ⇒ n = 1012.
⚡ Speed Tip: Upper Triangular Matrix Power: [[1, k], [0, 1]]ⁿ = [[1, n·k], [0, 1]] directly.
💡 View Step-by-Step Solution & Answer
Correct Answer: 2¹³ · 3⁶
- Step 1: For any n×n matrix M, det(k · M) = kⁿ · det(M). Here n = 3, so det(2 · adj(3A)) = 2³ · det(adj(3A)).
- Step 2: Use the identity det(adj(B)) = (det(B))ⁿ⁻¹. Here n = 3, so det(adj(3A)) = (det(3A))².
- Step 3: det(3A) = 3³ · det(A) = 27 · 4 = 108 = 2² · 3³.
- Step 4: Then det(adj(3A)) = (2² · 3³)² = 2⁴ · 3⁶. Finally, det(2 · adj(3A)) = 2³ · (2⁴ · 3⁶) = 2⁷ · 3⁶... Wait, for n=3, 2³ · (3³ · 4)² = 8 · (27 · 4)² = 8 · 108² = 8 · 11664 = 2¹³ · 3⁶.
⚡ Speed Tip: Determinant Scaling Rules: det(kM) = kⁿ det(M) and det(adj(M)) = (det M)ⁿ⁻¹.
❓ Frequently Asked Questions (FAQs)
What is the most frequently asked question in Matrices for JEE Main?
Cramer’s Rule for system of linear equations (infinitely many solutions, no solution, or unique solution) appears in almost 90% of JEE shifts.
What is det(2·adj(3A)) for a 3×3 matrix with det(A) = 4?
For 3×3: det(2·M) = 2³·det(M) = 8·det(M). det(adj(3A)) = (det(3A))² = (3³·det(A))² = (27×4)² = 108² = 11664. Result = 8 × 11664 = 93312.