Question

for what value of k , the following system of linear equation has no solution - 3x+y=1 2(k-1)x+(k-1)y=(2k+1)

Answer 1

Hiii friend,

3X + Y = 1

3X +Y -1 = 0........(1)


2(K-1)X+(K-1)Y= (2K+1)

2(K-1)X+(K-1)Y -K+1 = 0.......(2)

These equations are in the form of

A1X+B1Y+C1= 0 and A2X+B2Y+C2=0

Where,

A1 = 3 , B1 = 1 and C1 = -1

And,

A2 = (2K-1) , B2 = (K-1) and C2 = -2K+1


Therefore,

A1/A2= 3/(2K-1), B1/B2 = 1/(K-1) and C1/C2 = -1/(2K+1)

The given equations have no solution.

Then for no solution we must have,


A1/A2 = B1/B2 not = C1/C2

=> 3/(2K-1) = 1/(K-1) # 1/(2K+1)

=> 3/(2K-1) = 1/(K-1) and 1/(K-1) # 1/(2K+1)

=> 3K-3 = 2K-1 and 1/(K-1) # 1/(2K+1)


=> 3K-2K = -1+3 and 1/(K-1) # 1/(2K+1)

=> K = 2 and 1/(K-1) # 1/(2K+1)

Clearly,

When K = 2

Then,

1/(K-1) # 1/(2K+1)

=> 1/(2-1) # 1/(4+1)


Hence,

K = 2.


HOPE IT WILL HELP YOU...... :-)