Equations x + y + z = 6, x + 2y + 3z = 10 and x + 2y + λz = μ have infinite number of solutions if

  1. λ = μ ≠ 3, 10
  2. λ ≠ μ = 3, is any real number
  3. λ = 3, μ = 10
  4. none of the above

Answer (Detailed Solution Below)

Option 3 : λ = 3, μ = 10
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Detailed Solution

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Concept:

A = [aij]m × n = coefficient of matrix, X = Column matrix of the variables

B = column matrix of constants

The system AX = B has

1) A unique solution If Rank of A = Rank [A|B] and is equal to the number of variables.

2) Infinitely many solutions, If Rank of A = Rank of [A|B] < number of variables

3) no solution, If Rank of A ≠ Rank of [A|B], i.e. Rank of A < Rank of [A|B].

Calculation:

Let 

We can see that, at λ = 3, R2 = R3 and hence, |A| = 0.

For λ = 3 either infinite solutions exist or no solution exists.

The coefficient matrix of the given linear equation is

R3 → R3 - R2

⇒ 

Let the augmented matrix be

For no solution,  μ = 10 

Hence, λ = 3, μ = 10 is the correct answer.

Shortcut Trickx + y + z = 6

x + 2y + 3z = 10

x + 2y + λz = μ 

Hence, if we put λ = 3, μ = 10, two-equation will get coincide, which result in an infinite number of solution.

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