Question

Obtain the value of Δ = `|(1+x,1,1),(1,1+y,1),(1,1,1+z)|` in terms of x, y and z. 

Further, if Δ = 0 and x, y, z are non-zero real numbers, prove that x^–1 + y^–1 + z^–1 = –1. 

Answer 1

Δ = `|(1+x,1,1),(1,1+y,1),(1,1,1+z)|`

By multiply and divide in R₁ by x, R₂ by y and R₃ by z, we get 

Δxyz = `|(1/x+1,1/x,1/x),(1/y,1/y+1,1/y),(1/2,1/2,1/2+1)|`

R₁ → R₁ + R₂ + R₃

Δ = `xyz|(1+1/x+1/y+1/z,1+1/x+1/y+1/z,1+1/x+1/y+1/z),(1/y,1/y+1,1/y),(1/2,1/z,1/z+1)|`

From R₁ take `(1+1/x+1/y+1/z)`,

Δ = `xyz(1+1/x+1/y+1/z)|(1,1,1),(1/y,1/y+1,1/y),(1/z,1/z,1/z+1)|`

C₂ → C₂ – C₁; C₃ → C₃ – C₁ 

Δ = `xyz(1+1/x+1/y+1/z)|(1,0,0),(1/y,1,0),(1/z,0,1)|`

Δ = Expanding along R₁ 

Δ = `xyz (1+1/x+1/y+1/z)`

Now, Δ = 0

⇒ `xyz(1+1/x+1/y+1/z)=0`

⇒ `1/x+1/y+1/z=-1`

⇒ x^–1 + y^–1 + z^–1 = –1