Gauss Elimination Method Se Hal

(a) Sawaal (Problem Statement)

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

10x + y + 2z = 13 3x + 10y + z = 14 2x + 3y + 10z = 15

Gauss Elimination Ke Steps

Pehle, augmented matrix banate hain:

[ 10   1   2 | 13 ]
[  3  10   1 | 14 ]
[  2   3  10 | 15 ]

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

Pehla element (R1,C1) 10 hai, isko 1 banana hai.

R1 → R1 / 10

[ 1   0.1   0.2 | 1.3 ]  <-- (10/10=1, 1/10=0.1, 2/10=0.2, 13/10=1.3)
[ 3    10     1   | 14  ]
[ 2     3    10   | 15  ]

Step 2: Pehle pivot ke neeche zeros banana

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

R2 → R2 - 3*R1

R3 → R3 - 2*R1

[ 1   0.1     0.2   |  1.3 ]
[ 0   9.7     0.4   | 10.1 ]  <-- R2: [3-3*1, 10-3*0.1, 1-3*0.2 | 14-3*1.3] = [0, 9.7, 0.4 | 10.1]
[ 0   2.8     9.6   | 12.4 ]  <-- R3: [2-2*1, 3-2*0.1, 10-2*0.2 | 15-2*1.3] = [0, 2.8, 9.6 | 12.4]

Step 3: Dusra pivot (R2,C2) ko 1 banana

Ab R2,C2 wala element 9.7 hai, isko 1 banana hai.

R2 → R2 / 9.7

[ 1   0.1       0.2       |  1.3     ]
[ 0   1         0.4/9.7   | 10.1/9.7 ]  <-- (0.4/9.7 ≈ 0.041237, 10.1/9.7 ≈ 1.041237)
[ 0   2.8       9.6       | 12.4     ]

Decimal values ko thoda round off karke likh sakte hain, ya fractions mein bhi rakh sakte hain accuracy ke liye. Hum yahan calculation mein exact values use karenge, display mein approximate.

[ 1   0.1       0.2        |  1.3      ]
[ 0   1         0.041237   |  1.041237 ]  (Approx values for display)
[ 0   2.8       9.6        | 12.4      ]

Step 4: Dusre pivot ke neeche zero banana

Ab R2,C2 wale pivot (1) ke neeche (R3,C2) zero banana hai.

R3 → R3 - 2.8*R2

[ 1   0.1        0.2        |  1.3      ]
[ 0   1          0.041237   |  1.041237 ]
[ 0   0          9.484536   |  9.484536 ]  <-- R3_C3: 9.6 - 2.8*(0.4/9.7) = (9.6*9.7 - 2.8*0.4)/9.7 = (93.12 - 1.12)/9.7 = 92/9.7 ≈ 9.484536
                                       R3_Const: 12.4 - 2.8*(10.1/9.7) = (12.4*9.7 - 2.8*10.1)/9.7 = (120.28 - 28.28)/9.7 = 92/9.7 ≈ 9.484536

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

Step 5: Teesra pivot (R3,C3) ko 1 banana

R3,C3 mein ab 92/9.7 hai.

R3 → R3 / (92/9.7) (Yaani R3 * (9.7/92))

[ 1   0.1       0.2       |  1.3 ]
[ 0   1         0.041237  |  1.041237 ] (0.4/9.7 | 10.1/9.7)
[ 0   0         1         |  1   ]

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 (accurate fractions use karte hue):

Equation 1: x + (1/10)y + (2/10)z = 13/10
Equation 2: y + (4/97)z = 101/97
Equation 3: z = 1

Equation 3 se:

z = 1

z ki value Equation 2 mein daalo:

y + (4/97)(1) = 101/97

y = 101/97 - 4/97

y = 97/97

y = 1

y aur z ki values Equation 1 mein daalo:

x + (1/10)(1) + (2/10)(1) = 13/10

x + 1/10 + 2/10 = 13/10

x + 3/10 = 13/10

x = 13/10 - 3/10

x = 10/10

x = 1

Hal (Solution)

x = 1
y = 1
z = 1

Jaanch (Verification)

Values ko original equations mein daal kar check karte hain:

Original Equation 1: 10x + y + 2z = 13

10(1) + (1) + 2(1) = 10 + 1 + 2 = 13 (Sahi hai!)

Original Equation 2: 3x + 10y + z = 14

3(1) + 10(1) + (1) = 3 + 10 + 1 = 14 (Sahi hai!)

Original Equation 3: 2x + 3y + 10z = 15

2(1) + 3(1) + 10(1) = 2 + 3 + 10 = 15 (Sahi hai!)

Solution bilkul sahi hai! Ekdum perfect!