Engineering Mathematics - Linear Algebra

Question

If the equations (λ+1)x + 8y = 4λ and λx + (λ+3)y = 3λ-1 have no solution, then the number of values of λ is :

(A) one

(B) two

(C) three

(D) more than three

Answer 1

Ans will be A) 1

First put $\lambda =1$

U will get equation

$2x+8y=4$

$x+4y=2$

These two will give same lines

So, here it will have many solution

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Now, put $\lambda =3$

$4x+8y=12$

$3x+6y=8$

In this case we will get parallel lines

So, no solution will get here

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No more solution possible

Comments

mam we will have only one solution for this. if u can cross check once.

Answer 2

let us consider two lines having equation :

ax+by=c...........@1

dx+ey=f............@2

now we can say that we have no solution only when both the lines are parallel to each other --> condition for parallel lines==> (a/d)=(b/e)≠(c/f)

putting the values from original equation we have:

(ƛ+1)/ƛ=8/(ƛ+3)

ƛ²+4ƛ+3=8ƛ

ƛ²-4ƛ+3=0 on solving the equation we get:

ƛ= 1 ,3 solution

but for ƛ=1 we get parralel lines. so we will have many solutions.

so accepted answer is only 3. hence the answer is only option A.

 

Comments

Hi  @arvin, the following condition is to be met for parallel lines :

a1 / a2= b1 / b2 ≠ c1 / c2

Here if ƛ =1, then we will have 

a1 / a2= b1 / b2 = c1 / c2

 which is'nt right.

Therefore only 1 value is possible. 

ƛ = 3

 

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hey thanks a ton :) see now updated :)