For which value(s) of λ , do the pair of linear equationsλx + y =

1.	For which value(s) of λ , do the pair of linear equationsλx + y = λ2 and x + λy = 1 have(i) no solution?(ii) infinitely many solutions?(iii) a unique solution?
Answer» The given pair of linear equations is λx + y = λ² and x + λy = 1 a₁ = λ, b₁= 1, c₁ = – λ² a₂ =1, b₂=λ, c₂=-1 The given equations are; λ x + y – λ² = 0 x + λ y – 1 = 0 Comparing the above equations with ax + by + c = 0; We get, a₁ = λ, b₁ = 1, c₁ = – λ²; a₂ = 1, b₂ = λ, c₂ = – 1; a₁ /a₂ = λ/1 b₁ /b₂ = 1/λ c₁ /c₂ = λ² (i) For no solution, a₁/a₂ = b₁/b₂≠ c₁/c₂ i.e. λ = 1/ λ ≠ λ² so, λ² = 1; and λ²≠ λ Here, we take only λ = – 1, Since the system of linear equations has infinitely many solutions at λ = 1, (ii) For infinitely many solutions, a₁/a₂ = b₁/b₂ = c₁/c₂ i.e. λ = 1/ λ = λ² so λ = 1/ λ gives λ = + 1; λ = λ² gives λ = 1,0; Hence satisfying both the equations λ = 1 is the answer. (iii) For a unique solution, a₁/a₂ ≠ b₁/b₂ so λ ≠1/ λ hence, λ² ≠ 1; λ ≠ + 1; So, all real values of λ except +1

For which value(s) of λ , do the pair of linear equationsλx + y = λ2 and x + λy = 1 have(i) no solution?(ii) infinitely many solutions?(iii) a unique solution?

Answer»

The given pair of linear equations is

λx + y = λ² and x + λy = 1

a₁ = λ, b₁= 1, c₁ = – λ²

a₂ =1, b₂=λ, c₂=-1

The given equations are;

λ x + y – λ² = 0

x + λ y – 1 = 0

Comparing the above equations with ax + by + c = 0;

We get,

a₁ = λ, b₁ = 1, c₁ = – λ²;

a₂ = 1, b₂ = λ, c₂ = – 1;

a₁ /a₂ = λ/1

b₁ /b₂ = 1/λ

c₁ /c₂ = λ²

(i) For no solution,

a₁/a₂ = b₁/b₂≠ c₁/c₂

i.e. λ = 1/ λ ≠ λ²

so, λ² = 1;

and λ²≠ λ

Here, we take only λ = – 1,

Since the system of linear equations has infinitely many solutions at λ = 1,

(ii) For infinitely many solutions,

a₁/a₂ = b₁/b₂ = c₁/c₂

i.e. λ = 1/ λ = λ²

so λ = 1/ λ gives λ = + 1;

λ = λ² gives λ = 1,0;

Hence satisfying both the equations λ = 1 is the answer.

(iii) For a unique solution,

a₁/a₂ ≠ b₁/b₂

so λ ≠1/ λ

hence, λ² ≠ 1;

λ ≠ + 1;

So, all real values of λ except +1