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