Gauss Elimination Method

(a) Sawaal (Problem Statement)

Gauss Elimination method ka istemal karke yeh system of equations solve karo:

6x₁ + 3x₂ + 2x₃ = 6 6x₁ + 4x₂ + 3x₃ = 0 20x₁ + 15x₂ + 12x₃ = 0

Gauss Elimination Ke Steps

Pehle, augmented matrix banate hain:

[  6   3   2 |  6 ]
[  6   4   3 |  0 ]
[ 20  15  12 |  0 ]

Step 1: Pehla pivot (R1,C1) ko 1 banana

Pehla element (R1,C1) 6 hai, isko 1 banana hai.

R1 → R1 / 6

[ 1   1/2   1/3 |  1 ]  <-- (6/6=1, 3/6=1/2, 2/6=1/3, 6/6=1)
[ 6   4     3   |  0 ]
[ 20 15    12   |  0 ]

Step 2: Pehle pivot ke neeche zeros banana

Ab R1,C1 wale pivot (1) ke neeche ke elements ko zero karenge.

R2 → R2 - 6*R1

R3 → R3 - 20*R1

[ 1   1/2       1/3       |  1 ]
[ 0   1         1         | -6 ]  <-- R2: [6-6*1, 4-6*(1/2), 3-6*(1/3) | 0-6*1] = [0, 1, 1 | -6]
[ 0   5         16/3      | -20]  <-- R3: [20-20*1, 15-20*(1/2), 12-20*(1/3) | 0-20*1] = [0, 5, (36-20)/3 | -20] = [0, 5, 16/3 | -20]

Yahaan R2,C2 mein already 1 aa gaya, toh accha hai!

Step 3: Dusre pivot ke neeche zero banana

Ab R2,C2 wala pivot 1 hai. Iske neeche (R3,C2) zero banana hai.

R3 → R3 - 5*R2

[ 1   1/2       1/3       |  1 ]
[ 0   1         1         | -6 ]
[ 0   0         1/3       |  10]  <-- R3: [0-5*0, 5-5*1, 16/3-5*1 | -20-5*(-6)] = [0, 0, (16-15)/3 | -20+30] = [0, 0, 1/3 | 10]

Matrix ab Row Echelon Form (REF) mein hai. Ab pivot elements ko 1 banana hai (teesre ko).

Step 4: Teesra pivot (R3,C3) ko 1 banana

R3 → R3 * 3

[ 1   1/2   1/3 |  1 ]
[ 0   1     1   | -6 ]
[ 0   0     1   | 30 ]

Matrix ab Row Echelon Form (REF) mein hai aur pivots 1 hain. Gauss Elimination yahan tak hota hai. Ab hum back-substitution karenge.

Back-Substitution (Peeche se values nikalna)

Matrix se equations wapas likhte hain:

Equation 1: x₁ + (1/2)x₂ + (1/3)x₃ = 1
Equation 2: x₂ + x₃ = -6
Equation 3: x₃ = 30

Equation 3 se:

x₃ = 30

x₃ ki value Equation 2 mein daalo:

x₂ + (30) = -6

x₂ = -6 - 30

x₂ = -36

x₂ aur x₃ ki values Equation 1 mein daalo:

x₁ + (1/2)(-36) + (1/3)(30) = 1

x₁ - 18 + 10 = 1

x₁ - 8 = 1

x₁ = 1 + 8

x₁ = 9

Hal (Solution)

x₁ = 9
x₂ = -36
x₃ = 30

Jaanch (Verification)

Values ko original equations mein daal kar check karte hain:

Original Equation 1: 6x₁ + 3x₂ + 2x₃ = 6

6(9) + 3(-36) + 2(30) = 54 - 108 + 60 = 114 - 108 = 6 (Sahi hai!)

Original Equation 2: 6x₁ + 4x₂ + 3x₃ = 0

6(9) + 4(-36) + 3(30) = 54 - 144 + 90 = 144 - 144 = 0 (Sahi hai!)

Original Equation 3: 20x₁ + 15x₂ + 12x₃ = 0

20(9) + 15(-36) + 12(30) = 180 - 540 + 360 = 540 - 540 = 0 (Sahi hai!)

Solution bilkul sahi hai!