Question

By using properties of determinants, prove that `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|` = 0.

Answer 1

L.H.S. = `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|` 

Applying R₁ → R₁ + R₂, we get

L.H.S. = `|(x + y + z, x + y + z, x + y + z),(z, x , y),(1, 1, 1)|`

Taking (x + y + z) common from R₁, we get

L.H.S. = `(x + y + z)|(1, 1, 1),(z, x, y),(1, 1, 1)|`

= (x + y + z) (0)          …[∵ R₁ and R₃ are identical]
= 0
= R.H.S.