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If $G=\{\begin{pmatrix}x & x\\0 & 0\end{pmatrix}:x\in \mathbb R\setminus \{0\}\}$ show that G is abelian (commuative) group. I have to show "." is a Binary opreation, Associative and commuative. but as for a Identity element and Inverse, Is this $E=\begin{pmatrix}1 & 1\\0 & 0\end{pmatrix}$ and $A^{-1}=\begin{pmatrix}1/x & 1/x\\0 & 0\end{pmatrix}$ correct answer ?

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  • $\begingroup$ Yes, it is the correct answer. It only remains to show the operation is associative and commutative. Hint: If you forget about the matrices and think about the entries of the matrix, it looks a lot like the commutative group $(\mathbb{R}\setminus\{0\},\cdot)$ $\endgroup$ – Darío G Mar 23 '16 at 10:29
  • $\begingroup$ Please do not use pictures for critical portions of your post. Pictures cannot be searched and are inaccessible to those using screen readers. I have edited your question to reflect this principle $\endgroup$ – Shahab Mar 23 '16 at 10:34
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Since matrix multiplication is associative, we only have to show that the multiplication is commutative. This follows from

$$\pmatrix{x&x\\0&0}\pmatrix{y&y\\0&0}=\pmatrix{xy&xy\\0&0}$$

This multiplication also shows that the product of two elements of the set is in the set again.

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