Question

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

Answer 1

Answer:

The given pair of linear equations is

λx + y = λ2

and x + λy = 1

a1 = λ, b1= 1, c1 = – λ

2

a2 =1, b2=λ, c2=-1

The given equations are;

λ x + y - λ

2 = 0

x + λ y - 1 = 0

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

We get,

a1 = λ, b1 = 1, c1 = - λ 2 ;

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 ≠ λ 2

so, λ

2 = 1;

and λ

2 ≠ 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 = λ 2

so λ = 1/ λ gives λ = + 1;

λ = λ 2

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, λ

2 ≠ 1;

λ ≠ + 1;

So, all real values of λ except +1