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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