Question

Prove that: `|(y^2z^2, yz, y + z),(z^2x^2, zx, z + x),(x^2y^2, xy, x + y)|` = 0

Answer 1

`|(y^2z^2, yz, y + z),(z^2x^2, zx, z + x),(x^2y^2, xy, x + y)|` 

[Multiplying R₁, R₂, R₃ by x, y, z respecctively]

= `1/(xyz) |(xy^2z^2, xyz, xy + xz),(x^2yz^2, xyz, yz + xy),(x^2y^2z, xyz, xz + yz)|`

[Taking (xyz) common from C₁ and C₂]

= `1/(xyz) (xyz)^2 |(yz, 1, xy + xz),(xz, 1, yz + xy),(xy, 1, xz + yz)|`

[Applying C₃ → C₃ + C₁]

= `xyz|(yz, 1, xy + yz + zx),(xz, 1, xy + yz + zx),(xy, 1, xy + yz + zx)|`

[Taking (xy + yz + zx) common from C₃]

= ` xyz(xy + yz + zx) |(yz, 1, 1),(xz, 1, 1),(xy, 1, 1)|`

= 0 ....[∵ C₂ and C₃ are identical]