If S is the set of distinct values of 'b' for which the following system of linear equations
`x+y+z=1`
`x+ay +z=1`
`ax+by+z=0`
has no solution then S is

Options

A. an empty set
B. an infinite set
C. a finite set containing two or more elements
D. a singleton set

Answer

D

Explanation

The given system of equation can be written as
`{:[(1,1,1),(1,a,1),(a,b,1)][(x),(y),(z)]=[(1),(1),(0)]:}`
or, AX=B, where `{:A=[(1,1,1),(1,a,1),(a,b,1)],X=[(x),(y),(z)]andB=[(1),(1),(0)]:}`
`{:absA=abs((1,1,1),(1,a,1),(a,b,1)):}=a-b-1+a+b-a^2=-(a^2-2a+1)=-(a-1)^2`
We oberve that
`absA=0rArr -(a-1)^2=0rArr a=1`.
For this value of a, we find that first two equations represent the same plane. The system of equations will have no solution if the planes represented by first and third equation are parellel. For which, we must have
`1/a=1/b=1/1rArra=b=1`.
Hence, S=[1] is a singleton set.