Determinants

Board: CBSE |Class:Class12

ComprehensivestudynotesforDeterminantsbyAjayYadav.

KeyConcepts

Definition

Determinantofa2×2:|A|=a₁₁a₂₂-a₁₂a₂₁.3×3:expandalongrow/columnusingminorsandcofactors.

Properties

|A’|=|A|.Interchangingrowsflipssign.Twoidenticalrows⇒|A|=0. Multiplying a row by k multiplies |A| by k. |AB| = |A||B|. Adding multiple of one row to another: |A| unchanged.

Area of Triangle

Area = ½|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)| = ½|det|. Collinear if det=0.

Adjoint and Inverse

adj(A) = C’ where C is cofactor matrix. A⁻¹ = adj(A)/|A|. Condition: |A|≠0. AA⁻¹ = A⁻¹A = I.

Cramer's Rule

For AX = B: x_j = |A_j|/|A| where A_j replaces column j with B.

Important Formulas

2×2 det	|A| = a11a22 - a12a21
|AB|	|AB| = |A||B|
Inverse	A⁻¹ = adj(A)/|A|
Area	Area = ½|det|

Solved Examples

Example 1: Find |A| for A=(2345).

Solution: 2(5)-3(4)=10-12=-2.

Example 2: Find inverse of A=(2314).

Solution: |A|=5. adj(A)=(4-3-12). A⁻¹ = (4/5-3/5-1/52/5).

Example 3: Find area of triangle with vertices (0,0),(4,0),(0,3).

Solution: det=|0(0-3)+4(3-0)+0(0-0)|=12. Area=12/2=6.

Practice Questions

1. Evaluate |1234|.
2. Find A⁻¹ for A=(3152).
3. Find area of triangle (1,2),(3,4),(5,6).
4. Solve using Cramer: 2x+3y=5, x-y=1.
5. Prove |1abc1bca1cab|=(a-b)(b-c)(c-a).

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