Question

Using properties of determinants, prove that:

`|(a,b,b+c),(c,a,c+a),(b,c,a+b)|` = (a+b+c)(a-c)² 

Answer 1

Let Δ = `|(a,b,b+c),(c,a,c+a),(b,c,a+b)|`

Operating C₃ → C₃ - C₂ , we obtain

Δ =`|(a,b,c),(c,a,c),(b,c,a+b-c)|`

Operating R₁ → R₁ + R₂ + R₃ , we obtain

Δ =`|(a+b+c,a+b+c,a+b+c),(c,a,c),(b,c,a+b-c)|`

Taking (a + b + c) common from R₁ , we obtain

= (a+b+c) `|(1,1,1),(c,a,c),(b,c,a+b-c)|`

Operating C₁ → C₁ - C₃ , C₂ → C₂ - C₃ , we obtain 

= (a+b+c) `|(0,0,0),(0,a-c,c),(c-a,2c-a-b,a+b-c)|`

Expanding along R₁ , we obtain

= (a+b+c){-(a-c)(a-c)}

= (a+b+c)(a-c)²