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I am trying to show that this set P={ $p(\alpha,\beta,\gamma)=\pmatrix{1&\alpha&\beta\\0&1&\gamma\\0&0&1}$ $|$ $\alpha,\beta,\gamma$ $\in R$} is a group under matrix multiplication. I have already proved the closure, identity and associative properties. But with the inverse, Im stuck as to how I should go about proving it. I need to find $pp^{-1} = identity =p^{-1}p$.

So the inverse of $\pmatrix{1&x&y\\0&1&z\\0&0&1}$, I calculated it to be $\pmatrix{1&0&0\\-x&1&0\\xz-y&-z&1}$. But multiplying those 2 matrices it doesnt seem to be the identity matrix?

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    As an added point, since your inverse did not have the same form as the other elements of the group, you should have been immediately suspicious of it, especially since you have already proven closure.2011-11-15

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$\begin{align} \begin{pmatrix} 1 & x & y & \vdots & 1 & 0 & 0 \\ 0 & 1 & z & \vdots & 0 & 1 & 0 \\ 0 & 0 & 1 & \vdots & 0 & 0 & 1 \end{pmatrix} &\sim \begin{pmatrix} 1 & x & 0 & \vdots & 1 & 0 & -y \\ 0 & 1 & 0 & \vdots & 0 & 1 & -z \\ 0 & 0 & 1 & \vdots & 0 & 0 & 1 \end{pmatrix} \\ &\sim \begin{pmatrix} 1 & 0 & 0 & \vdots & 1 & -x & -y+xz \\ 0 & 1 & 0 & \vdots & 0 & 1 & -z \\ 0 & 0 & 1 & \vdots & 0 & 0 & 1 \end{pmatrix} \end{align}$

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    Alright, thanks!2011-11-12
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solve this $\pmatrix{1&x&y\\0&1&z\\0&0&1}*\pmatrix{1&\alpha&\beta\\0&1&\gamma\\0&0&1} = \pmatrix{1&0&0\\0&1&0\\0&0&1}$

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    Thanks! Initially I tried doing it another way though, but this way should work also.2011-11-12