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