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