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Friday, October 31, 2008

Problem (5)

For what value of λ, the following linear equations have a solution and solve them completely in each case:

x + y + z = 1

x + 2y + 4z = λ

x + 4y + 10z = λ2

Solution: -

The given system is

x + y + z = 1

x + 2y + 4z = λ System (1)

x + 4y + 10z = λ2

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 - L1 and L3→ L3 – L1

Thus we obtain the equivalent system is

x + y + z = 1

y + 3z = λ - 1 System (2)

3y + 9z = λ2 – 1

Now apply L3→ 3L2 - L3

x + y + z = 1

y + 3z = λ - 1 System (3)

0 = λ2 – 3λ + 2

If λ2 – 3λ + 2 ≠ 0, the given system will be in echelon form having two equations with three variables. So the system has 3-2 = 1 free variable which is z. So the system has non-zero solution for λ2 – 3λ + 2 = 0

that is (λ-1)(λ-2) = 0

So, λ = 1 or λ = 2

Thus the given system is consistent for λ = 1 and λ = 2.

Case İ when λ = 1

The system (3) will be

x + y + z = 1

y + 3z = 0

Which is in echelon form where z is a free variable.

Let z = a where a is arbitrary real number.

So y = -3a and x = 1 + 2a

Hence the given system of linear equations has infinite number of solutions for λ = 1. In particular, Let, a = 1, then x = 3, y = -3 and z = 1, is a particular solution of the given system.

Case İİ when λ = 2

The system (3) will be

x + y + z = 1

y + 3z = 1

Which is in echelon form where z is a free variable.

Let z = b where b is arbitrary real number. So y = 1- 3b, x = 2b. Hence the system has infinite number of solutions for solutions for λ = 2. In particular, Let, b = -1, then x = 4, y = -2 and z = -1, is a particular solution of the given system.

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