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Friday, October 31, 2008

Problem (3)

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

2x1 + 3x2 + 5x3 – x4 = 8

3x1 + 4x2 + 2x3 – 3x4 = 8

x1 + 2x2 + 8x3 – x4 = 8

7x1 + 9x2 + x3 – 8x4 = 8

Solution: -

The given system is

2x1 + 3x2 + 5x3 – x4 = 8

3x1 + 4x2 + 2x3 – 3x4 = 8 System (1)

x1 + 2x2 + 8x3 – x4 = 8

7x1 + 9x2 + x3 – 8x4 = 8

Let us represent the three linear equation of the system (1) by L1, L2, L3 and L4respectively. Reduce the system to echelon form by elementary operation.

Apply L1↔L3

Then we have the equivalent system

x1 + 2x2 + 8x3 – x4 = 8

3x1 + 4x2 + 2x3 – 3x4 = 8 System (2)

2x1 + 3x2 + 5x3 – x4 = 8

7x1 + 9x2 + x3 – 8x4 = 8

Now Apply L2 → L2 - 3L1, L3→ L3 – 2L1 and L4→ L4 – 7L1.

Thus we obtain the equivalent system

x1 + 2x2 + 8x3 - x4 = 8

- 2x2 -22x3 + 6x4 = -26 System (3)

- x2 - 11x3 + 3x4 = -13

- 5x2 - 55x3 +15x4 = -56

Apply L2 → L2/2

x1 + 2x2 + 8x3 - x4 = 8

- x2 -11x3 + 3x4 = -13 System (4)

- x2 - 11x3 + 3x4 = -13

- 5x2 - 55x3 +15x4 = -56

Apply the operations L3 → L2 + L3 and L4 → 7L2 + L4

Thus we obtain the equivalent system

x1 + 2x2 + 8x3 - x4 = 8

x2 +11x3 -3x4 = 13 System (5)

0 = 0

0 = 9

The given system has been reduced to echelon form and contains an equation of the form 0 = 9 which

is not true: hence the given system is inconsistent. i.e. the system has no solution.

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