Chapter 4 — Determinants — covers evaluation, properties, area of triangles, adjoint, inverse of matrices, and solving linear equations using Cramer’s rule. Carries 8-10 marks — one of the highest-scoring chapters.
Key Concepts
Determinant Evaluation
2×2: |a b; c d| = ad − bc
3×3 (expansion along R₁):
|A| = a₁₁(A₁₁) − a₁₂(A₁₂) + a₁₃(A₁₃)
Where Aᵢⱼ = (−1)^(i+j) × Mᵢⱼ (cofactor), Mᵢⱼ = minor
3×3 (expansion along R₁):
|A| = a₁₁(A₁₁) − a₁₂(A₁₂) + a₁₃(A₁₃)
Where Aᵢⱼ = (−1)^(i+j) × Mᵢⱼ (cofactor), Mᵢⱼ = minor
Properties of Determinants
- |Aᵀ| = |A|
- Interchanging rows/columns changes sign
- Two identical rows/columns → |A| = 0
- Scalar multiplication of a row: multiplies determinant by that scalar
- |kA| = kⁿ|A| for n×n matrix
- Row/column operation Rᵢ → Rᵢ + kRⱼ doesn’t change value
- |AB| = |A| × |B|
Area of Triangle
Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|
Or using determinant: Area = ½|[x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1]|
Points are collinear if Area = 0
Or using determinant: Area = ½|[x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1]|
Points are collinear if Area = 0
Adjoint and Inverse
adj(A) = transpose of cofactor matrix
A⁻¹ = (1/|A|) × adj(A) (exists only when |A| ≠ 0)
Properties: A × adj(A) = |A| × I
|adj(A)| = |A|ⁿ⁻¹ (for n×n matrix)
(AB)⁻¹ = B⁻¹A⁻¹
A⁻¹ = (1/|A|) × adj(A) (exists only when |A| ≠ 0)
Properties: A × adj(A) = |A| × I
|adj(A)| = |A|ⁿ⁻¹ (for n×n matrix)
(AB)⁻¹ = B⁻¹A⁻¹
System of Linear Equations (AX = B)
X = A⁻¹B (unique solution when |A| ≠ 0)
Cramer’s Rule:
x = D₁/D, y = D₂/D, z = D₃/D
Where D = |A|, D₁ = |A with column 1 replaced by B|, etc.
If |A| = 0: system is either inconsistent (no solution) or has infinite solutions
Cramer’s Rule:
x = D₁/D, y = D₂/D, z = D₃/D
Where D = |A|, D₁ = |A with column 1 replaced by B|, etc.
If |A| = 0: system is either inconsistent (no solution) or has infinite solutions
Solved Examples
Example 1
Q: Find the inverse of A = [2 3; 1 4]
Solution:
|A| = 8−3 = 5 ≠ 0 → inverse exists
adj(A) = [4 −3; −1 2]
A⁻¹ = (1/5)[4 −3; −1 2] = [4/5 −3/5; −1/5 2/5]
Example 2
Q: Using determinants, find the area of triangle with vertices (1,0), (6,0), (4,3).
Solution: Area = ½|1(0−3) + 6(3−0) + 4(0−0)| = ½|−3 + 18 + 0| = ½(15) = 7.5 sq units
Quick Revision Points
- 2×2: ad−bc; 3×3: expand along any row/column
- |kA| = kⁿ|A|; |AB| = |A||B|; |Aᵀ| = |A|
- A⁻¹ = adj(A)/|A|; exists only if |A| ≠ 0 (non-singular)
- Cramer’s rule: x = D₁/D, y = D₂/D, z = D₃/D
- Area of triangle using determinant = ½|det|; collinear if = 0
Chapter Navigation
Previous: Matrices Class 12 Notes
Next: Continuity and Differentiability Class 12 Notes
Related Chapters in Class 12 Maths
- Matrices Class 12 Notes
- Relations and Functions Class 12 Notes
- Continuity and Differentiability Class 12 Notes
Practice What You Learned
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