Linear Equations Ko Solve Karna

(a) Sawaal (Problem)

Yeh equations ko Gauss-Jordan method ka istemal karke solve karo:

x + 4y - z = -5 x + y - 6z = -12 3x - y - z = 4

Gauss-Jordan Elimination Ke Steps

Sabse pehle, hum in equations ka augmented matrix (sanjojit avyuh) banayenge:

[ 1   4  -1 |  -5 ]
[ 1   1  -6 | -12 ]
[ 3  -1  -1 |   4 ]

Step 1: Pehle pivot ke neeche zeros banana

Pehla pivot R1,C1 mein 1 hai. Iske neeche ke elements (R2,C1 aur R3,C1) ko zero karenge.

R2 → R2 - R1 (Row 2 mein se Row 1 ko minus karo)

R3 → R3 - 3*R1 (Row 3 mein se Row 1 ka 3 guna minus karo)

[ 1   4  -1 |  -5 ]
[ 0  -3  -5 |  -7 ]  <-- (1-1=0, 1-4=-3, -6-(-1)=-5, -12-(-5)=-7)
[ 0 -13   2 |  19 ]  <-- (3-3*1=0, -1-3*4=-13, -1-3*(-1)=2, 4-3*(-5)=19)

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

Ab R2,C2 wale element (-3) ko 1 banana hai.

R2 → R2 / (-3) (Row 2 ko -3 se divide karo)

[ 1   4  -1         |  -5         ]
[ 0   1   5/3       |   7/3       ]
[ 0 -13   2         |  19         ]

Step 3: 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 - 4*R2 (Row 1 mein se Row 2 ka 4 guna minus karo)

R3 → R3 + 13*R2 (Row 3 mein Row 2 ka 13 guna add karo)

[ 1   0  -23/3      | -43/3      ] <-- R1: [1, 4-4*1, -1-4*(5/3) | -5-4*(7/3)] = [1,0,-23/3|-43/3]
[ 0   1   5/3       |   7/3      ]
[ 0   0   71/3      | 148/3      ] <-- R3: [0, -13+13*1, 2+13*(5/3) | 19+13*(7/3)] = [0,0,71/3|148/3]

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

Ab R3,C3 wale element (71/3) ko 1 banana hai.

R3 → R3 * (3/71) (Row 3 ko 3/71 se multiply karo)

[ 1   0  -23/3      | -43/3      ]
[ 0   1   5/3       |   7/3      ]
[ 0   0   1         | 148/71     ]

Step 5: Teesre pivot ke upar zeros banana

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

R1 → R1 + (23/3)*R3 (Row 1 mein Row 3 ka 23/3 guna add karo)

R2 → R2 - (5/3)*R3 (Row 2 mein se Row 3 ka 5/3 guna minus karo)

[ 1   0   0 |  117/71 ] <-- R1: [-23/3 + (23/3)*1 = 0], [-43/3 + (23/3)*(148/71) = 117/71]
[ 0   1   0 |  -81/71 ] <-- R2: [5/3 - (5/3)*1 = 0], [7/3 - (5/3)*(148/71) = -81/71]
[ 0   0   1 |  148/71 ]

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

Hal (Solution)

RREF matrix se humein solution milta hai:

x = 117/71
y = -81/71
z = 148/71

Jaanch (Verification)

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

Equation 1: x + 4y - z = -5

(117/71) + 4(-81/71) - (148/71) = (117 - 324 - 148) / 71 = (117 - 472) / 71 = -355 / 71 = -5 (Sahi hai!)

Equation 2: x + y - 6z = -12

(117/71) + (-81/71) - 6(148/71) = (117 - 81 - 888) / 71 = (36 - 888) / 71 = -852 / 71 = -12 (Sahi hai!)

Equation 3: 3x - y - z = 4

3(117/71) - (-81/71) - (148/71) = (351 + 81 - 148) / 71 = (432 - 148) / 71 = 284 / 71 = 4 (Sahi hai!)

Solution sahi hai.