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Friday, October 31, 2008

Problem (4)

For which value of a will be following system have no solution? Exactly one solution? Infinitely many solutions.

x + 2y- 3z = 4

3x - y + 5z = 2

4x + y +(a2 - 14)z = (a+2)

Solution: -

The given system is

x + 2y- 3z = 4

3x - y + 5z = 2 System (1)

4x + y +(a2 - 14)z = (a+2)

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

Apply L2 → L2 - 3L1 and L3→ L3 – 4L1

Thus we obtain the equivalent system is

x + 2y- 3z = 4

- 7y + 5z = -10 System (2)

- 7y + (a2 - 2) z = (a -14)

Now apply L3→ L3 - L2

Thus we obtain the equivalent system is

x + 2y- 3z = 4

- 7y + 5z = -10 System (3)

(a2 - 42) z = (a - 4)

The system is now

x + 2y- 3z = 4

- 7y + 5z = -10 System (4)

(a + 4) (a - 4) z = (a - 4)

The system (4) is in echelon form. Now it has three equations with three variables. If a ≠ 4 and a ≠ -4 then we get the unique solution.

If a = 4 we get two equations with three free variables. So then we get more then one solution.

And if a = -4 we get degenerate linear equation and in that case we get no solution.

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