Question

Investigate for what values of 𝝁 𝒂𝒏𝒅 𝝀 the equation x+y+z=6; x+2y+3z=10; x+2y+𝜆z=𝝁 have
(i)no solution,
(ii) a unique solution,
(iii) infinite no. of solution.

Answer 1

Given eqn : x+y+z=6, x+2y+3z=10,  x+2y+𝜆z=𝝁

A X = B

`[(1,1,1),(1,2,3),(1,2,lambda)][(x),(y),(z)]=[(6),(10),(mu)]`

Argumented matrix is :`[(1,1,1),(1,2,3),(1,2,lambda)][(6),(10),(mu)]`

`R_1-R_2,`

`->[(1,1,1,|,6),(0,1,2 ,|,4 ),(0,1,lambda-3,|,mu-6)]`

`R_2-R_1,`

`-> [(1,1,1,|,6),(0,1,2 ,|,4 ),(0,1,lambda-1,|,mu-10)]`

(i) When 𝜆=3, 𝝁≠𝟏𝟎 𝒕𝒉𝒆𝒏 𝒓(𝒂)=𝟐,𝒓(𝑨⋮𝑩)=𝟑
r(A)≠𝒓(𝑨⋮𝑩)
Hence for 𝜆=3 , 𝝁≠𝟏𝟎 system is inconsistent.
No solution exist.
(ii) When 𝜆≠3,𝝁≠𝟏𝟎 ,𝒓(𝑨)=𝒓(𝑨⋮𝑩)=𝟑
Unique solution exist.
(iii) When 𝜆=3,𝝁=𝟏𝟎 𝒓(𝑨)=𝒓(𝑨⋮𝑩)=𝟐<𝟑
Infinite solution.