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I would like to show that $\left(\begin{array}{ccc} 1 & s & s^2 \\ 1 & t & t^2 \\ 1 & u & u^2 \end{array}\right)$ has an inverse provided $s$, $t$ and $u$ are distinct.

I have tried to prove $A\cdot B\times C \neq 0$ without success.

I computed $A\cdot B\times C = tu^2-ut^2+st^2-su^2+s^2u-s^2t$. What to do next ?

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Read about Vandermonde determinant en.wikipedia.org/wiki/Vandermonde_matrix –  PAD Oct 24 '12 at 12:34
    
Thank you very much ! –  Bob Oct 24 '12 at 13:43

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$\det\begin{pmatrix} 1 & s & s^2 \\ 1 & t & t^2 \\ 1 & u & u^2\end{pmatrix}$

$=\det\begin{pmatrix} 1 & s & s^2 \\ 1-1 & t-s & t^2-s^2 \\ 1-1 & u-s & u^2-s^2\end{pmatrix}$ (applying $R_2'=R_2-R_1$ and $R_3'=R_3-R_1$)

$=\det\begin{pmatrix} 1 & s & s^2 \\ 0 & (t-s) & (t+s)(t-s) \\ 0 & (u-s) & (u+s)(u-s)\end{pmatrix}$

$=(t-s)(u-s)\det\begin{pmatrix} 1 & s & s^2 \\ 0 & 1 & (t+s) \\ 0 & 1 & (u+s)\end{pmatrix}$

$=(t-s)(u-s)(u+s-t-s)=(t-s)(u-s)(u-t)\ne 0$ if $t,s,u$ are distinct.

Now ,we can use this to find the inverse.

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Thank you very much ! –  Bob Oct 24 '12 at 13:25

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