Question

Show that `|(0,ab^2,ac^2),(a^2b,0,bc^2),(a^2c,b^2c,0)| = 2a^3b^3c^3`

Answer 1

LHS = `|(0,ab^2,ac^2),(a^2b,0,bc^2),(a^2c,b^2c,0)|`

Taking a₂, b₂ and c₂ common from C₁, C₂ and C₃ respectively we get,

LHS = `a^2b^2c^2 |(0,a,a),(b,0,b),(c,c,0)|`

Again taking a,b,c common from R₁, R₂ and R₃ respectively we get,

LHS = `a^3b^3c^3 |(0,1,1),(1,0,1),(1,1,0)|`

Expanding along R₁ we get

LHS = `a^3b^3c^3 [0|(0,1),(1,0)| - |(1,1),(1,0)| + 1|(1,0),(1,1)|]`

= a³b³c³ [- 1(0 - 1) + 1 (1 - 0)]

= a³b³c³ (1 + 1)

= 2a³b³c³ = RHS.

Hence proved.