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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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You computed the inverse wrong: you know this, because when you multiplied the two matrices you did not get the identity! Obviously, you should therefore compute the inverse again, correctly. –  Mariano Suárez-Alvarez Nov 12 '11 at 2:23
    
thanks, i thought i did it correctly though, sorry for any confusion caused! –  Ray Nov 15 '11 at 3:15
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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. –  rcollyer Nov 15 '11 at 4:20

2 Answers 2

up vote 6 down vote accepted

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

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Ohh okay, thanks. Must have flipped it the other way around accidently! So is it sufficient to say that multiplying the 2 matrices produces the identity matrix, and hence proved the inverse property? –  Ray Nov 12 '11 at 2:29
    
Depends on whether you teacher will allow you to assume that one-sided matrix inverses for square matrices are automatically two-sided. Just to be safe I would show: $AA^{-1}=I_3$ and $A^{-1}A=I_3$. Also, don't forget to comment that you have not just demonstrated that there is an inverse, but that the inverse is upper-triangular (it belongs to $P$). –  Bill Cook Nov 12 '11 at 2:33
    
Alright, thanks! –  Ray Nov 12 '11 at 2:44

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. –  Ray Nov 12 '11 at 2:30

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