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Friday, October 31, 2008

Problem (2)

Solve the Flowing homogeneous system of linear equation by Gauss-Jordan elimination


Solution: -
The given system is

2I1 I2 + 3I3 + 4 I4 = 9

I1 - 2I3 + 7I4 = 11 System (1)

3I1 3I2 + I3 + 5I4 = 8

2I1 + I2 + 4I3 + 4I4 = 10

Let us represent the four linear equation of the system (1) by L1, L2, L3 and L4 respectively. Reduce the system to echelon form by elementary operation.

Apply L2 → L1 – 2L2 , L3 → 3L1 – 2L3 and L4 → L4 – L1

Thus we obtain the equivalent system is

2I1 I2 + 3I3 + 4I4 = 9

I2 - 7I3 - 10I4 = 13 System (2)

3I2 + 7 I3 + 2I4 = 11

2I2 + I3 + = 1

Apply L3 → 3L2 + L3 and L4 → 2L2 + L4

Thus we obtain the equivalent system is

2I1 I2 + 3I3 + 4I4 = 9

I2 - 7I3 - 10I4 = 13 System (3)

- 14 I3 - 28I4 = 50

- 13I3 - 20I4 = 27

Apply L4 → 13L3 – 14L4

Thus we obtain the equivalent system is

2I1 I2 + 3I3 + 4I4 = 9

I2 - 7I3 - 10I4 = 13 System (3)

- 14 I3 - 28I4 = 50

- 84I4 = 272

The system (3) is in echelon form. Now is has four equations with four variables. It has a unique solution.

Now in L4 of system (3)

-84I4 = 272

.: I4 = 68/21

And in L3 of system (3)

-14I3 – 272/3 =50

.: I3 = -211/21

In L2 of system (3)

I2 - 211/3 - 680/21 = 13

.: I2 = 810/7

And in L1 of system (3)

2I1 810/7 -211/7 + 272/21 = 9

.: I1 = 2980/21

So the solution of the given system is I1 = 2980/21, I2 = 810/7, I3 = -211/21, I4 = 68/21.

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