Gauss-Jordan Method (x,y,z Variables)

(a) Sawaal (Problem Statement)

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

x + 2y + z = 3 2x + 3y + 3z = 10 3x - y + 2z = 13

Gauss-Jordan Elimination Ke Steps

Sabse pehle, augmented matrix banayenge:

[ 1   2   1 |  3 ]
[ 2   3   3 | 10 ]
[ 3  -1   2 | 13 ]

Pehla pivot (R1,C1) already 1 hai, bahut accha!

Step 1: Pehle pivot ke neeche zeros banana

R2 → R2 - 2*R1

R3 → R3 - 3*R1

[ 1   2   1 |  3 ]
[ 0  -1   1 |  4 ]  <-- R2: [2-2*1, 3-2*2, 3-2*1 | 10-2*3] = [0, -1, 1 | 4]
[ 0  -7  -1 |  4 ]  <-- R3: [3-3*1, -1-3*2, 2-3*1 | 13-3*3] = [0, -7, -1 | 4]

Step 2: Dusra pivot (R2,C2) ko 1 banana

Ab R2,C2 wale element (-1) ko 1 banana hai.

R2 → R2 * (-1)

[ 1   2   1 |  3 ]
[ 0   1  -1 | -4 ]
[ 0  -7  -1 |  4 ]

Step 3: Dusre pivot ke upar aur neeche zeros banana

R1 → R1 - 2*R2

R3 → R3 + 7*R2

[ 1   0   3 |  11 ]  <-- R1: [1-2*0, 2-2*1, 1-2*(-1) | 3-2*(-4)] = [1, 0, 3 | 11]
[ 0   1  -1 |  -4 ]
[ 0   0  -8 | -24 ]  <-- R3: [0+7*0, -7+7*1, -1+7*(-1) | 4+7*(-4)] = [0, 0, -8 | -24]

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

Ab R3,C3 wale element (-8) ko 1 banana hai.

R3 → R3 / (-8)

[ 1   0   3 |  11 ]
[ 0   1  -1 |  -4 ]
[ 0   0   1 |   3 ]

Step 5: Teesre pivot ke upar zeros banana

R1 → R1 - 3*R3

R2 → R2 + R3

[ 1   0   0 |   2 ]  <-- R1: [1-3*0, 0-3*0, 3-3*1 | 11-3*3] = [1, 0, 0 | 2]
[ 0   1   0 |  -1 ]  <-- R2: [0+0, 1+0, -1+1 | -4+3] = [0, 1, 0 | -1]
[ 0   0   1 |   3 ]

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

Hal (Solution)

RREF matrix se humein solution milta hai:

x = 2
y = -1
z = 3

Jaanch (Verification)

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

Equation 1: x + 2y + z = 3

(2) + 2(-1) + (3) = 2 - 2 + 3 = 0 + 3 = 3 (Sahi hai!)

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

2(2) + 3(-1) + 3(3) = 4 - 3 + 9 = 1 + 9 = 10 (Sahi hai!)

Equation 3: 3x - y + 2z = 13

3(2) - (-1) + 2(3) = 6 + 1 + 6 = 7 + 6 = 13 (Sahi hai!)

Solution bilkul sahi hai! Ekdum mast!