Determinants 🔢

Har square matrix ek number se associated hoti hai — determinant. Yeh number bahut kuch batata hai!

📐 Area aur Determinant ka Connection

Kya tumhe pata hai ki ek triangle ki area nikalne ka formula ek determinant se aata hai? Agar teen points (x₁,y₁), (x₂,y₂), (x₃,y₃) hain, toh unka triangle ka area = ½|det| ek 3×3 matrix ka! Determinant ek ऐसा magical number hai jo matrix ki "strength" batata hai — kya system solvable hai, kya matrix invertible hai, kya points collinear hain — sab kuch!

1. Determinant of 2×2 Matrix

A = [[a,b],[c,d]] ke liye: |A| = det(A) = ad - bc

Yeh ek simple formula hai — diagonal multiply karo aur subtract karo.

✅ Example: det([[3,1],[2,4]])

= 3×4 - 1×2 = 12 - 2 = 10

2. Determinant of 3×3 Matrix — Cofactor Expansion

3×3 determinant ko hum pehli row ke saath expand karte hain (cofactor expansion):

|A| = a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃ Jahan Cᵢⱼ = (-1)^(i+j) × Mᵢⱼ Mᵢⱼ = minor (i-th row, j-th column remove karne ke baad 2×2 det) Sign pattern: | + - + | | - + - | | + - + |

✅ Worked Example: 3×3 Determinant

A = [[1,2,3],[4,5,6],[7,8,9]]

M₁₁ = |[[5,6],[8,9]]| = 45-48 = -3

M₁₂ = |[[4,6],[7,9]]| = 36-42 = -6

M₁₃ = |[[4,5],[7,8]]| = 32-35 = -3

|A| = 1×(-3) - 2×(-6) + 3×(-3) = -3 + 12 - 9 = 0

|A| = 0 → Matrix is singular (not invertible)!

3. Adjoint aur Inverse Matrix

Adjoint: adj(A) = transpose of cofactor matrix Inverse: A⁻¹ = adj(A) / |A| (only when |A| ≠ 0) Steps to find A⁻¹: 1. Calculate |A| 2. Find all cofactors Cᵢⱼ 3. Form cofactor matrix [Cᵢⱼ] 4. Take transpose → adj(A) 5. A⁻¹ = adj(A)/|A|

4. System of Linear Equations — Matrix Method

Equations AX = B mein: X = A⁻¹B (agar |A| ≠ 0)

Yeh method 2 ya 3 variables wale linear equations efficiently solve karta hai.

✅ Solve: 2x+y=5, x+3y=10

Matrix form: A = [[2,1],[1,3]], X = [x,y]ᵀ, B = [5,10]ᵀ

|A| = 6-1 = 5 ≠ 0 ✓

adj(A) = [[3,-1],[-1,2]]

A⁻¹ = (1/5)[[3,-1],[-1,2]]

X = A⁻¹B = (1/5)[[3,-1],[-1,2]][5,10]ᵀ = (1/5)[5,5]ᵀ = [1,1]ᵀ

∴ x = 1, y = 1

5. Area of Triangle using Determinant

Area = ½ |det| where det = |x₁ y₁ 1| |x₂ y₂ 1| |x₃ y₃ 1| Points collinear ⟺ det = 0 (area = 0)

6. Practice Questions

Q1. Find det([[2,-1,0],[1,3,-2],[0,1,4]])

Solution

Expand along row 1: 2×(12+2) - (-1)×(4-0) + 0 = 2×14 + 4 = 28+4 = 32

Q2. Find A⁻¹ for A = [[1,2],[3,7]].

Solution

|A|=7-6=1. adj(A)=[[7,-2],[-3,1]]. A⁻¹=[[7,-2],[-3,1]]

Q3. Area of triangle with vertices (1,2), (3,4), (5,2).

Solution

det = |1 2 1; 3 4 1; 5 2 1| = 1(4-2)-2(3-5)+1(6-20) = 2+4-14 = -8. Area = ½|−8| = 4 sq units.