Question

Using properties of determinants, prove that `|(1,1,1+3x),(1+3y, 1,1),(1,1+3z,1)| = 9(3xyz + xy +  yz+ zx)`

Answer 1

Let `triangle = |(1,1,(1+3z)),(1+3y, 1,1),(1,1+3z,1)| = 9(3xyz + xy + yz + zx)`

Applying R₂→R₂ − R₁, R₃→R₃ − R₁

`=>triangle = |(1,1,1+3x),(3y,0,-3x),(0,3z,-3x)|`

Expanding along R₁ ,we get

`triangle = 1(0 + 9xz) - 1(-9xy - 0) + (1+3x)(9yz  -0)`

`= 9xz + 9xy + 9yz + 27xyz`

`= 9(3xyz +xy +yz+zx)`