Question

Answer the following question:

By using properties of determinant 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)|`

By R₁ + R₂ , we get,

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

By 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 ≡ R₃]

= 0

= R.H.S.