Linear Equations Ko Solve Karna (Part 2)

(a) Sawaal (Problem Statement)

Gauss-Jordan method ka istemal karke yeh equations solve karo:

2x - 6y + 8z = 24 5x + 4y - 3z = 2 3x + y + 2z = 16

Gauss-Jordan Elimination Ke Steps

Sabse pehle, in equations ka augmented matrix banayenge:

[ 2  -6   8 |  24 ]
[ 5   4  -3 |   2 ]
[ 3   1   2 |  16 ]

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

Pehla element (R1,C1) abhi 2 hai, isko 1 banana hai.

R1 → R1 / 2 (Row 1 ko 2 se divide karo)

[ 1  -3   4 |  12 ]
[ 5   4  -3 |   2 ]
[ 3   1   2 |  16 ]

Step 2: Pehle pivot ke neeche zeros banana

Ab R1,C1 wale pivot (1) ke neeche ke elements (R2,C1 aur R3,C1) ko zero karenge.

R2 → R2 - 5*R1

R3 → R3 - 3*R1

[ 1  -3   4 |  12 ]
[ 0  19 -23 | -58 ]  <-- R2: [5-5*1, 4-5*(-3), -3-5*4 | 2-5*12] = [0, 19, -23 | -58]
[ 0  10 -10 | -20 ]  <-- R3: [3-3*1, 1-3*(-3), 2-3*4 | 16-3*12] = [0, 10, -10 | -20]

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

Dekho, Row 3 (R3) ko 10 se divide karke simplify kar sakte hain:

R3 → R3 / 10

[ 1  -3   4 |  12 ]
[ 0  19 -23 | -58 ]
[ 0   1  -1 |  -2 ]  <-- Simplified R3

Ab R2 aur R3 ko swap (badal) kar lete hain taaki R2,C2 mein 1 aa jaaye.

R2 ↔ R3

[ 1  -3   4 |  12 ]
[ 0   1  -1 |  -2 ]
[ 0  19 -23 | -58 ]

Ab R2,C2 wala pivot 1 ho gaya!

Step 4: Dusre pivot ke upar aur neeche zeros banana

Ab R2,C2 wale pivot (1) ke upar (R1,C2) aur neeche (R3,C2) zero banana hai.

R1 → R1 + 3*R2

R3 → R3 - 19*R2

[ 1   0   1 |   6 ]  <-- R1: [1, -3+3*1, 4+3*(-1) | 12+3*(-2)] = [1, 0, 1 | 6]
[ 0   1  -1 |  -2 ]
[ 0   0  -4 | -20 ]  <-- R3: [0, 19-19*1, -23-19*(-1) | -58-19*(-2)] = [0, 0, -4 | -20]

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

Ab R3,C3 wale element (-4) ko 1 banana hai.

R3 → R3 / (-4)

[ 1   0   1 |   6 ]
[ 0   1  -1 |  -2 ]
[ 0   0   1 |   5 ]

Step 6: Teesre pivot ke upar zeros banana

Ab R3,C3 wale pivot (1) ke upar (R1,C3 aur R2,C3) zero banana hai.

R1 → R1 - R3

R2 → R2 + R3

[ 1   0   0 |   1 ]  <-- R1: [1-0, 0-0, 1-1 | 6-5] = [1, 0, 0 | 1]
[ 0   1   0 |   3 ]  <-- R2: [0+0, 1+0, -1+1 | -2+5] = [0, 1, 0 | 3]
[ 0   0   1 |   5 ]

Yeh matrix ab Reduced Row Echelon Form (RREF) mein hai.

Hal (Solution)

RREF matrix se humein solution milta hai:

x = 1
y = 3
z = 5

Jaanch (Verification)

Ab x, y, aur z ki values ko original equations mein daal kar check karte hain:

Equation 1: 2x - 6y + 8z = 24

2(1) - 6(3) + 8(5) = 2 - 18 + 40 = -16 + 40 = 24 (Sahi hai!)

Equation 2: 5x + 4y - 3z = 2

5(1) + 4(3) - 3(5) = 5 + 12 - 15 = 17 - 15 = 2 (Sahi hai!)

Equation 3: 3x + y + 2z = 16

3(1) + (3) + 2(5) = 3 + 3 + 10 = 6 + 10 = 16 (Sahi hai!)

Solution bilkul sahi hai!