Question

Using properties of determinants prove the following: `|[1,x,x^2],[x^2,1,x],[x,x^2,1]|=(1-x^3)^2`

Answer 1

The given determinant is `|[1,x,x^2],[x^2,1,x],[x,x^2,1]|`

Applying the transformation c₁ → c₁ + c₂ + c₃, we get

`|[1,x,x^2],[x^2,1,x],[x,x^2,1]|=|[1+x+x^2,x,x^2],[x^2+1+x,1,x],[x+x^2+1,x^2,1]|=(1+x+x^2)|[1,x,x^2],[1,1,x],[1,x^2,1]|`

Again applying the transformation R₁ → R₁ − R₂ and R₂ → R₂ − R₃, we get

`(1+x+x^2)|[1,x,x^2],[1,1,x],[1,x^2,1]|=(1+x+x^2)|[0,x-1,x^2-x],[0,1-x^2,x-1],[1,x^2,1]|=(1+x+x^2)(x-1)^2|[0,1,x],[0,-x-1,1],[1,x^2,1]|`

`=(x^3-1)(x-1){0-0+(1+x+x^2)}=(x^3-1)(x-1)(x^2+x+1)`

`=(x^3-1)(x^3-1)=(x^3-1)^2=(1-x^3)^2`

hence ` |[1,x,x^2],[x^2,1,x],[x,x^2,1]|=(1-x^3)^2`